On the Algebraic Theory of Fields

Home , Algebraic element, Algebraic function, Complete field, Ground field, Radical extension

Lectures on the Algebraic Theory of Fields

By K.G. Ramanathan

Tata Institute of Fundamental Research, Bombay 1956 Lectures on the Algebraic Theory of Fields

By K.G. Ramanathan

Tata Institute of Fundamental Research, Bombay 1954 Introduction

There are notes of course of lectures on Field theory aimed at pro- viding the beginner with an introduction to algebraic extensions, algebraic function fields, formally real fields and valuated fields. These lectures were preceded by an elementary course on group theory, vector spaces and ideal theory of rings—especially of Noetherian rings. A knowledge of these is presupposed in these notes. In addition, we assume a familiarity with the elementary topology of topological groups and of the real and complex number fields. Most of the material of these notes is to be found in the notes of Artin and the books of Artin, Bourbaki, Pickert and Van-der-Waerden. My thanks are due to Mr. S. Raghavan for his help in the writing of these notes.

K.G. Ramanathan

Contents

1 General extension ﬁelds 1 1 Extensions...... 1 2 Adjunctions...... 3 3 Algebraic extensions ...... 5 4 Algebraic Closure ...... 9 5 Transcendental extensions ...... 12

2 Algebraic extension fields 17 1 Conjugate elements ...... 17 2 Normal extensions ...... 18 3 Isomorphisms of fields ...... 21 4 Separability ...... 24 5 Perfectfields ...... 31 6 Simple extensions ...... 35 7 Galois extensions ...... 38 8 Finitefields ...... 46

3 Algebraic function ﬁelds 49 1 F.K. Schmidt’s theorem ...... 49 2 Derivations ...... 54 3 Rational function ﬁelds ...... 67

4 Norm and Trace 75 1 Normandtrace ...... 75 2 Discriminant ...... 82

v vi Contents

5 Composite extensions 87 1 Kronecker product of Vector spaces ...... 87 2 Compositeﬁelds ...... 93 3 Applications...... 97

6 Special algebraic extensions 103 1 Rootsofunity...... 103 2 Cyclotomic extensions ...... 105 3 Cohomology ...... 113 4 Cyclic extensions ...... 119 5 Artin-Schreier theorem ...... 126 6 Kummer extensions ...... 128 7 Abelian extensions of exponent p ...... 133 8 Solvable extensions ...... 134

7 Formally real fields 149 1 Orderedrings ...... 149 2 Extensions of orders ...... 152 3 Realclosedfields ...... 156 4 Completion under an order ...... 166 5 Archimedian ordered fields ...... 170

8 Valuated fields 175 1 Valuations...... 175 2 Classification of valuations ...... 177 3 Examples ...... 180 4 Completefields ...... 184 5 Extension of the valuation of a complete...... 194 6 Fields complete under archimedian valuations ...... 201 7 Extension of valuation of an incomplete field ...... 205

Appendix 209 1 Decomposition theorem ...... 209 2 Characters and duality ...... 213 3 Pairing of two groups ...... 217 Chapter 1

General extension ﬁelds

1 Extensions

A field has characteristic either zero or a prime number p. 1 Let K and k be two fields such that K k. We shall say that K ⊃ is an extension field of k and k a subfield of K. Any field T such that K T k is called an intermediary field, intermediate between K and ⊃ ⊃ k. If K and K′ are two fields, then any homomorphism of K into K′ is either trivial or it is an isomorphism. This stems from the fact that only ideals in K are (o) and K. Let K have characteristic p , o. Then the mapping a ap of K into itself is an isomorphism. For, → (a b)p = ap bp ± ± (ab)p = ap bp · and ap = bp = (a b)p = o = a = b. In fact for any integer e 1, e ⇒ − ⇒ ≥ a ap is also an isomorphism of K into itself. → Let now Z be the ring of rational integers and K a field whose unit element we denote by e. The mapping m me of Z into K obviously a → homomorphism of the ring Z into K. The kernel of the homomorphism is the set of m is Z such that me = 0 in K. This is an ideal in Z and as Z is a principal ideal domain, this ideal is generated by integer say p. Now p is either zero or else is a prime. In the first case it means that K contains

1 2 1. General extension ﬁelds

a subring isomorphic to Z and K has characteristic zero. Therefore K contains a subfield isomorphic to the field of rational numbers. In the 2 second case K has characteristic p and since Z/(p) is a finite field of p elements, K contains a subfield isomorphic to Z/(p). Hence the Theorem 1. A field of characteristic zero has a subfield isomorphic to the field of rational numbers and a field of characteristic p > o has a subfield isomorphic to the finite of p residue classes of Z modulo p. The rational number filed and the finite field of p elements are called prime fields. We shall denote them by Γ. When necessary we shall denote the finite field of p elements by Γp. Let K/k be an extension field of k. We shall identity the elements of K and k and denote the common unit element by 1. Similarly for the zero element. K has over k the structure of a vector space. For, α, β K, ∈ λ k = α + β K, λα K. Therefore K δ has over k a base αλ in ∈ ⇒ ∈ ∈ { } the sense that every α K can be uniquely written in the form ∈ α = aλαλ aλ k ∈ Xλ and aλ = 0 for almost all λ. If the base αλ consists only of a finite { } number of elements we say that K has a finite base over k. The extension K/k is called a finite or infinite extension of k according as K has over k a finite or an infinite base. The number of basis elements we call the degree of K over k and denote it by (K : k). If (K : k) = n then there exist n elements ω,...ωn in K which are linearly independent over k and every n + 1 elements of K linearly dependent over k. Let K be a finite field of q elements. Obviously K has characteristic p , o. Therefore K contains a subfield isomorphic to Γp. Call it also Γp. K is a finite dimensional vector space over Γp. Let (K : Γp) = n 3 Then obviously K has pn elements. Thus Theorem 2. The number of elements q in a finite field is a power of the characteristic.

Let K T k be a tower of ﬁelds. K/T has a base αλ and T/K ⊃ ⊃ { } has a base βν . This means that for α K { } ∈ α = tλαλ Xλ 2. Adjunctions 3 tλ T and tλ = 0 for almost all λ. Also tλ being in T we have ∈

tλ = aλµβµ µ X aλµ = 0 for almost all µ. Thus

α = aλµ(αλβµ) Xλµ Thus every element α of K can be expressed linearly in terms of αλβµ . On the other hand let { }

(αλβµ) = 0 Xλµ aλµ k and aλµ = 0 for almost all λ, µ. Then ∈

0 = ( aλµβµ)αλ µ Xλ X But aλµβµ T and since the αλ from a base of K/T we have µ ∈ { } P aλµβµ = 0 for all λ. Xλ But βµ form a base of T/k so that aλµ = 0 for all λ, µ. We have { } thus proved that αλβµ is a base of K/k. In particular if (K : k) is ﬁnite { } then (K : T) and (T : k) are ﬁnite and

(K : k) = (K : T)(T : k)

As special cases, (K : k) = (T : k) = K = T (T is an intermediary ⇒ ﬁeld of K and k). (K : k) = (K : T) = T = k. ⇒ 2 Adjunctions

Let K/k be an extension filed and Kα a family of intermediary extension fields. Then Kα is again an intermediary field but, in general, Kα 4 α α T S 4 1. General extension fields

is not a field. We shall define for any subset S of K/k the field k(S ) is called the field generated byS over k. It is trivial to see that

k(S ) = T T S \⊃

i.e., it is the intersection of all intermediary fields T containing S . k(S ) is said to be got from k by adjunction of S to k. If S contains a finite number of elements, the adjunction is said to be finite otherwise infinite. In the former case k(S ) is said to be finitely generated over k. If (K : k) < then obviously K is finitely generated over k but the converse is ∞ not true.

Obviously k(S US ′) = k(S )(S ′) because a rational function of S US ′ is a rational function of S ′ over k(S ). Let K/k be an extension ﬁeld and α K. Consider the ring k[x] of ∈ polynomials in x over k. For any f (x) k[x], f (x) is an element of K. ∈ Consider the set G of polynomials f (x) k[x] for which f (α) = o. G ∈ is obviously a prime ideal. There are now two possibilities, G = (0), G , (o). In the former case the inﬁnite set of elements 1,α,α2,... are all linearly independent over k. We call such an element α of K, transcendental over k. In the second case G , (o) and so G is a principal ideal generated by an irreducible polynomial ϕ(x). Thus 1,α,α2,... are linearly dependent. We call an element α of this type algebraic over k. We make therefore the

Definition. Let K/k be an extension field. α K is said to be algebraic ∈ over k if α is root of a non zero polynomial in k[x]. Otherwise it is said to be transcendental. 5 If α is algebraic, the ideal G defined above is called the ideal of α over k and the irreducible polynomial ϕ(x) which is a generator of G is called the irreducible polynomial of α over k. ϕ(x) may be made by multiplying by a suitable element of k. This monic polynomial we shall call the minimum polynomial of α. 3. Algebraic extensions 5

3 Algebraic extensions

Suppose α K is algebraic over k and ϕ(x) its minimum polynomial ∈ over k. Let f (x) k[x] and f (α) , o. f (x) and ϕ(x) are then coprime ∈ and so there exist polynomials g(x), h(x) in k[x] such that

f (x)g(x) = 1 + ϕ(x)h(x) which means that ( f (α)) 1 = g(α) k[α]. Thus k[α] = k(α). On the − ∈ other hand suppose α K such that k[α] = k(α) then there is a g(α) in ∈ k[α] such that αg(α) = 1 or that α satisfies xg(x) 1 in k[x] so that α is − algebraic. Hence 1) α K algebraic over k k[α] is a field. ∈ ⇐⇒ We now define an extension K/k to be algebraic over k if every of K is algebraic over k. In the contrary case K is said to be transcendental extension of k We deduce immediately

2) K/k algebraic every ring R with k R K is a ﬁeld ⇐⇒ ⊂ ⊂ If R is a ring and α in R then k[α] R then k[α] R. But α is ⊂ ⊂ algebraic so that α 1 k[α] R so that R is a ﬁeld. The converse − ∈ ⊂ follows from (1).

3) (K : k) < = K/k algebraic. ∞ ⇒ For let (K : k) = n then for any for α K, the n + 1 elements ∈ 1,α,α2,...αn are linearly dependant over k so that α is algebraic. 6 The converse is not true Let K/k be an extension ﬁeld and α K algebraic over k. Let ϕ(x) ∈ be the minimum polynomial of α over k and let degree of ϕ(x) be n. 2 n 1 Then 1,α,α ,...,α − are linearly independent over k so that

(k(α) : k) n. ≥ On the other hand any β in k(α) is a polynomial in α over k. Let β = b + b α + + b αm. Put ψ(x) = b + b x + + b xm. o 1 ··· m o 1 ··· m 6 1. General extension ﬁelds

Then ψ(x) = ϕ(x)h(x) + R(x) where R(x) = 0 or deg R(x) < n. Hence ψ(α) = β = R(α) and so n 1 every β cab be expressed linearly in terms of 1,α,...,α − . Thus k(α) : k n. ≤ We have hence 4) If α K is algebraic over K, k(α)/k is an algebraic extension of ∈ degree equal to the degree of the minimum polynomial of α over k. We shall call k(α) : k the degree of α over k 5) If α is algebraic over k then for any L, k L K,α is algebraic over ⊂ ⊂ L. For, the ideal of α over k (which is , (0) since α is algebraic) is contained in the ideal of α in L[X] k[x]. ⊃ Therefore k(α) : k (L(α) : L) ≥ 7 Note that the converse is not true. For let z be transcendental over k and consider the field k(z) of rational functions of z. k(z) : k is not finite. But k(z) : k(z2) is finite as z is a root of x2 z2 over k(z2). − 6) If α1,...,αn in K are algebraic over k then k(α1,...,αn) is algebraic over k.

For, put Ko = k, Ki = k(α1,...,αi),

Kn = k(α1,...,αn)

Then Ki/Ki 1 is algebraic and is a ﬁnite extension. Now − (Kn : k) = πi(Ki : Ki 1) − which is also ﬁnite. Hence Kn is algebraic over k. We deduce immediately 3. Algebraic extensions 7

7) K/T algebraic, T/k algebraic = K/k algebraic. ⇒ For if α K, α is a root of ϕ(x) = Xn + a xn+1 + + a in T[x]. ∈ 1 ··· n Thus α is algebraic over k(a1 ..., an). Hence

k(a , a ,... a ,α) : k(a , a ,..., a ) < . 1 2 n 1 2 n ∞ k(a , a ,..., a ) : k < 1 2 n ∞ k(a , a ,..., a ,α) : k < 1 2 n ∞ which proves the contention. If follows that if K/k is any extension, then the set L of elements α of K algebraic over k is a field L which is algebraic over k. L is called the algebraic closure of k in K We shall now show how it is possible to construct algebraic extensions of a field. If k is a field and ϕ(x) a polynomial in k[x], an element α of an extension field K is said to be root of ϕ(x) if ϕ(α) = o. It then follows that ϕ(x), has in K at most n roots, n being degree of ϕ(x). 8 Let f (x) be an irreducible polynomial in k[x] The ideal generated by f (x) in k[x] is a maximal ideal since f (x) is irreducible. Therefore the residue class ring K of k[x]/( f (x)) is a field. Let σ denote the natural homomorphism of k[x] onto K. σ then maps k onto a subfield of K. We shall identify this subfield with k itself (note that k[x] and ( f (x)) are vector spaces over k). Let ξ in K be the element into which x goes by σ ξ = σx Then K = k(ξ). In the first place k(ξ) K. Any element in K is the ⊂ image, by σ, of an element say ϕ(x) in k[x]. But

ϕ(x) = h(x) f (x) + ψ(x)

So ϕ(ξ) K and ϕ(ξ) = ψ(ξ). But ψ(x) above has degree degree ∈ ≤ of f . Thus k(ξ) K k[ξ] k(ξ) ⊂ ⊂ ⊂ This shows that K = k(ξ) and that (K : k) is equal to the degree of f (x). Also ξ in K satisﬁes f (ξ) = o. We have thus proved that for 8 1. General extension ﬁelds

every irreducible polynomial f (x) in k[x] there exists an extension field in which f (x) has a root. Let now g(x) be any polynomial in k[x] and f (x) an irreducible factor of g(x) in k[x]. Let K be an extension of k in which f (x) has a root ξ. Let in K g(x) = (x ξ)λψ(x). − Then ψ(x) k[x]. We again take an irreducible factor of ψ(x) and ∈ construct K′ in which ψ(x) has a root. After finite number of steps we arrive at a field L which is an extension of k and in which g(x) splits 9 completely into linear factors. Let α1,...,αn be the distinct roots of g(x) in L. We call k(α1,...,αn) the splitting field of g(x) in L. Obviously (k(α ,...,α ) : k) n! 1 n ≤ We have therefore the important

Theorem 3. If k is a field and f (x) k[x] then f (x) has a splitting field ∈ K and (K : k) n!, n being degree of f (x). ≤ It must be noted however that a polynomial might have several splitting fields. For instance if D is the quaternion algebra over the rational number field Γ, generated by 1, i, j, k then Γ(i), Γ( j), Γ( f ), Γ(k) are all splitting fields of x2 + 1 in Γ[x]. These splitting fields are all distinct. Suppose k and k′ are two fields which are isomorphic by means of an isomorphism σ. Then σ can be extended into an isomorphismσ ¯ of k[x] on k′[x] by the following prescription

i i σ¯ ( a x ) = (σa )x a k σa k′ i i i ∈ i ∈ Let now f (x)X be a polynomialX in k[x] which is irreducible. Denote σ¯ σ¯ by f (x), its image in k′[x] by means of the isomorphismσ ¯ . Then f (x) 1 is again irreducible in k′[x]; for if not one can by means ofσ ¯ − obtain a nontrivial factorization of f (x) in k[x]. σ¯ Let now α be a root of f (x) over k and β a root of f (x) over k′. Then σ¯ k(α) k[x]/( f (x)), k′(β) k′[x]/( f (x)) ≃ ≃ σ Let τ be the natural homomorphism of k′[x] on k′[x]/( f (x)). Con- sider the mapping τ σ on k[x]. Since σ is an isomorphism, it follows · 4. Algebraic Closure 9 that τ σ is a homomorphism of k[x] on k [x]/(Fσ(x)). The kernel of the · ′ homomorphism is the set of ϕ(x) in k[x] such that

ϕσ(x) f σ(x) . ∈ 10 This set is precisely ( f (x)). Thus

σ¯ k[x]/( f (x)) k′[x]/( f (x)) ≃

By our identification, the above fields contain k and k′ respectively as subfields so that there is an isomorphism µ of k(α) on k′(β) and the restriction of µ to k is σ. In particular if k = k , then k(α) and k(β) are k isomorphic i.e., they ′ − are isomorphic by means of an isomorphism which is identity on k. We have therefore

Theorem 4. If f (x) k[x] is irreducible and α and β are two roots of ∈ it (either in the same extension field of k or in different extension fields), k(α) and k(β) are k isomorphic. − Note that the above theorem is false if f (x) is not irreducible in k[x].

4 Algebraic Closure

We have proved that every polynomial over k has a splitting ﬁeld. For a given polynomial this ﬁeld might very well coincide with k itself. Sup- pose k has the property that every polynomial in k has a root in k. Then it follows that the only irreducible polynomials over k are linear polynomials. We make now the

Deﬁnition. A ﬁeld Ω is algebraically closed if the only irreducible polynomials in Ω[x] are linear polynomials.

We had already defined the algebraic closure of a field k contained in a field K. Let us now make the

Definition. A field Ω/k is said to be an algebraic closure of k if 10 1. General extension fields

1) Ω is algebraically closed

2) Ω/k is algebraic. 11 We now prove the important Theorem 5. Every ﬁeld k admits, upto k-isomorphism, one and only one algebraic closure.

Proof. 1) Existence. Let M be the family of algebraic extensions Kα of k. Partially order M by inclusion. Let Kα be a totally ordered { } subfamily of M. Put K = Kα for Kα in this totally ordered family. α Now K is a ﬁeld; for β K, β K means β Kα for some α and 1 ∈S 2 ∈ 1 ∈ β Kβ for some β. Therefore β , β in Kα or Kβ whichever is larger 2 ∈ 1 2 so that β + β K. Similarly β β 1 K. Now K/k is algebraic 1 2 ∈ 1 2− ∈ since every λ K is in some Kα and so algebraic over k. Thus ∈ K M and so we can apply Zorn’s lemma. This proves that M has a ∈ maximal element Ω.Ω is algebraically closed; for if not let f (x) be an irreducible polynomial in Ω[x] and ρ a root of f (x) in an extension Ω(ρ) of Ω. Then since Ω/k is algebraic. Ω(ρ) is an element of M. This contradicts maximality of Ω. Thus Ω is an algebraic closure of k.

2) Uniqueness. Let k and k′ be two ﬁelds which are isomorphic by means of an isomorphism σ. Consider the family M of triplets (K, K ,σ)α with the property 1) Kα is an algebraic extension of { ′ } k, Kα′ of k′, 2) σα is an isomorphism of Kα on Kα′ extending σ. By theorem 4, M is not empty. We partially order M in the following manner

(K, K′,σ)α < (K, K′,σ)β

If 1) Kα Kβ, Kα′ Kβ′ , 2) σβ is an extension of σα. Let K, K′,σ)α ⊂ ⊂ = = { } be a simply ordered subfamily. Put K Kα, K′ Kα′ , α α These are then algebraic over k and kS′ respectively.S 12 Deﬁneσ ¯ on K by 4. Algebraic Closure 11

σ¯ x = σα x where x Kα. (Note that every x K is in some Kα in the simply ∈ ∈ ordered subfamily). It is easy to see thatσ ¯ is well - defined. Suppose x Kβ and Kβ Kα then σα is an extension of σρ and so σα x = σβx. ∈ ⊂ This proves thatσ ¯ is an isomorphism of K on K′ and extends σ. Thus the triplet (K, K′, σ¯ ) is in M and is an upper bound of the subfamily. By Zorn’s lemma there exists a maximal triplet (Ω, Ω′, τ). We assert that Ω is algebraically closed; for if not let ρ be a root of an irreducible polynomial f (x) Ω[x]. Then f ζ (x) Ω [x] is also irreducible. Let ρ ∈ ∈ ′ ′ be a root of f τ(x). Then τ can be extended to an isomorphismτ ¯ of Ω(ρ) on Ω′(ρ). Now (Ω(ρ), τ¯ is in M and hence leads to a contradiction. Thus Ω is an algebraic closure of k, Ω′ of k′ and τ an isomorphism of Ω on Ω′ extending σ. In particular if k = k′ and σ the identity isomorphism, then Ω and Ω′ are two algebraic closures of k and τ is then a k-isomorphism. Out theorem is completely demonstrated. Let f (X) be a polynomial in k[x] and K = k(α1,...,αn, a splitting field of f (x), so that α1,...,αn are the distinct roots of f (x) in K. Let K′ be any other splitting field and β1,...βm the distinct roots of f (x) in K′. Let Ω be an algebraic closure of K and Ω′ of K′. Then Ω and Ω′ are two algebraic closures of k. There exists therefore an isomorphism σ of Ω Ω = = on ′ which is identity on k. Let σK K1. Then K1 k(σα1 ,...,σαn ). Since α1,...αn are distinct σα1 ...,σαn are distinct and are roots of f (x). Thus K1 is a splitting field of f (x) in Ω′. This proves that 13

K′ = K1.

β1,...,βm are distinct and are roots of f (x) in Ω′. We have m = n and βi = σα2 in some order. Therefore the restriction of σ to K is an isomorphism of K on K′. We have

Theorem 6. Any two splitting fields K, K′ of a polynomial f (x) in k[x] are k isomorphic. − Let K be a finite field of q elements. Then q = Pn where n is an integer 1 and p is the characteristic of K. Also n = (K : Γ), Γ being ≥ 12 1. General extension fields

the prime ﬁeld. Let K∗ denote the abelian group of non-zero elements of K. Then K being a ﬁnite group of order q 1, ∗ − q 1 α − = 1

for all α K . Thus K is the splitting field of the polynomial ∈ ∗ xq x − in Γ[x]. It therefore follows Theorem 7. Any two finite fields with the same number of elements are isomorphic. A finite field cannot be algebraically closed; for if K is a finite field of q elements and a K the polynomial ∈ ∗ f (x) = x (X b) + a − b K Y∈ ∗ is in K[x] and has no root in K.

5 Transcendental extensions

We had already deﬁned a transcendental extension as one which contains at least on transcendental element. Let K/k be a transcendental extension and Z1,..., Zn any n elements of K. Consider the ring R = k[x1,... xn] of polynomials over k in n 14 variables. Let Y be the subset of R consisting of those polynomials f (x1,... xn) for which f (Z1,... Zn) = 0.

Y is obviously an ideal of R. If Y = (o) we say that Z1,... Zn are algebraically independent over k. If Y , (o), they are said to be algebraically dependent. Any element of K which is algebraic over k(Z1,..., Zn) is therefore algebraically dependent on Z1,... Zn. We now deﬁne a subset S of K to be algebraically independent over k if every ﬁnite subset of S is algebraically independent over k. If K/k is transcendental there is at least one such non empty set S . 5. Transcendental extensions 13

Let K/k be a transcendental extension and let S , S ′ be two subset of K with the properties

i) S algebraically independent over k

ii) S ′ algebraically independent over k(s)

Then S and S ′ are disjoint subsets of K and S US ′ are algebraically independent over k. That S and S ′ are disjoint is trivially seen. Let now Z ,... Z S and Z ,... Z S be algebraically dependent. This will 1 m ∈ 1′ n′ ∈ ′ mean that there is a polynomial f ,

f = f (x1,..., xm+n) in m + n variables with coeﬃcients in k, such that

= f (Z1,..., Zm, Z1′ ,... Zn′ ) 0.

Now f can be regarded as a polynomial in xm+1,..., xm+n with coefficients in k(x1,..., xm). If all these coefficients are zero then Z1,... Zm, ffi Z1′ ,... Zn′ are algebraically independent over k. If some coe cient is , 0, then f (Z1,... Zm, xm+1,..., xm+n) is a non zero polynomial over = = k(S ) which vanishes for xm+1 Z1′ ,... xm+n Zn′ which contradicts the fact that S ′ is algebraically independent over k(S ). Thus f = o identically and our contention is proved. 15 The converse of the above statement is easily proved. An extension field K/k is said to be generated by a subset M of K if K/k(M) is algebraic. Obviously K itself is a set of generators. A subset B of K is said to be a transcendence base of K if

1) B is a set of generators of K/k

2) B algebraically independent over k.

If K/k is transcendental, then, it contains algebraically independent elements. We shall prove that K has a transcendence base. Actually much more can be proved as in 14 1. General extension ﬁelds

Theorem 8. Let K/k be a transcendental extension generated by S and A a set of algebraically independent elements contained in S . Then there is a transcendence base B of K with

A B S ⊂ ⊂ Proof. Since S is a set of generators of K, K/k(S ) is algebraic. Let M be the family of subsets Aα of K with

1) A Aα S ⊂ ⊂ 2) Aα algebraically independent over k.

The set M is not empty since A is in M. Partially order M by inclusion. Let Aα be a totally ordered subfamily. Put B0 = Aα. Then { } α B S . Any finite subset of B will be in some Aα for large α and so o ⊂ o S Bo satisfies 2) also. Thus using Zorn’s lemma there exists a maximal element B in M. Every element x of S depends algebraically on B for otherwise BUx will be in M and will be larger than B. Thus k(S )/k(B) is algebraic. Since K/k(S ) is algebraic, it follows that B satisfies the conditions of the theorem. 16 The importance of the theorem is two fold; firstly that every set of elements algebraically independent can be completed into a transcendence base of K and further more every set of generators contains a base. We make the following simple observation. Let K/k be an extension, Z1,... Zm, m elements of K which have the property that K/k(Z1,... Zm) is algebraic, i.e., that Z ,..., Z is a set of generators. If Z K then Z 1 m ∈ depends algebraically on Z1,..., Zm i.e., k(Z, Z1,..., Zm)/k(Z1,..., Zm) is algebraic. We may also remark that if in the algebraic relation con- necting Z, Z1,... Zm, Z1 occurs then we can say that

k(Z, Z1,... Zm)/k(Z, Z2,..., Zm)

is algebraic which means that Z, Z2,... Zm is again a set of generators. We now prove the 5. Transcendental extensions 15

Theorem 9. If K/k has a transcendence base consisting of a ﬁnite number n of elements, every transcendence base has n elements.

Proof. Let Z1,..., Zn and Z1′ ,..., Zm′ be two transcendence bases consisting of n and m elements respectively. If n , m let n < m. Now

K/k(Z1,..., Zn) is algebraic. Z1′ is transcendental over k and depends algebraically on Z1,..., Zn so that if Z1 appears in the algebraic relation, by the remark above, Z1′ , Z2,..., Zn is again a set of generators, Z2′ depends algebraically on Z1′ ,..., Zn. In this algebraic relation at least one of Z2,..., Zn has to appear since Z1′ , Z2′ , are algebraically independent. If Z2 appears then Z2′ , Z1′ , Z3,..., Zn′ is a set of generators. We repeat this process n times, and ﬁnd, that Z1′ , Z2′ , Z3′ ,..., Zn′ is a set of generators which means that Zn′+1,..., Zm′ depend algebraically on Z1′ ,..., Zn′ . This 17 is a contradiction. So n m. We interchange n and m and repeat the ≥ argument and get n m. This proves that n = m. ≤ The unique integer n will be called the dimension of K/k.

n = dimk K

It is also called the transcendence degree. A similar theorem is true even if K has infinite transcendence base but we don’t prove it. Let k L K be a tower of extensions and let B be a transcen- ⊂ ⊂ 1 dence base of L/k and B2 that over K/L. We assert that B1UB2 is a transcendence base of K/k. In the first place B1UB2 is algebraically independent over k. Now k(B1UB2) is a subfield of L(B2). Every element in L(B2) is a ratio of two polynomials in B2 with coefficients in L. The elements of L are algebraic over K(B1). Thus L(B2) is algebraic over k(B1UB2). But K/L(B2) is algebraic. Thus K/k(B1UB2) is algebraic. This proves our assertion. In particular it proves

Theorem 10. If k L K then ⊂ ⊂

dimk K = dimk L + dimL K. 16 1. General extension ﬁelds

A transcendental extension K/k is said to be purely transcendental if there exists a base B with K = k(B). Note that this does not mean that every base has this property. For instance if k(x) is the field of rational functions of x then x2 is also transcendental over x but k(x2) is a proper subfield of k(x). = = Let K k(x1,... xn) and K′ k(x1′ ,... xn′ ) be two purely transcen- 18 dental extensions of dimension n. Consider the homomorphism σ defined by = σ f (x1,... xn) f (x1′ ,..., xn′ ) Where f (x ,..., x ) k[x ,..., x ]. It is then easy to see that this is 1 n ∈ 1 n an isomorphism of K on K′. This proves Theorem 11. Two purely transcendental extensions of the same dimension n over k k-isomorphic.

This theorem is true even if the dimension is inﬁnite. Chapter 2

Algebraic extension ﬁelds

1 Conjugate elements

Let Ω be an algebraic closure of k and K an intermediary field. Let Ω′ 19 be an algebraic closure of K and so of k. Then there is an isomorphism τ of Ω′ on Ω which is trivial on k. The restriction of this isomorphism to K gives a field τK in Ω which is k-isomorphic to K. Conversely suppose K and K′ are two subfields of Ω which are k-isomorphic. Since Ω is a common algebraic closure of K and K′, there exists an automorphism of Ω which extends the k-isomorphism of K and K′. Thus

1) Two subﬁelds K, K′ of Ω/k are k-isomorphic if and only if there exists a k-automorphism σ of Ω such that σK = K′.

We call two such ﬁelds K and K′ conjugate ﬁelds over k.

We deﬁne two elements ω,ω′ of Ω/k, to be conjugate over k if there exists a k-automorphism σ of Ω such that

σω = ω′

The automorphisms of Ω which are trivial on k form a group and so the above relation of conjugacy is an equivalence relation. We can therefore put elements of Ω into classes of conjugate elements over k. We then have

17 18 2. Algebraic extension ﬁelds

2) Each class of conjugate elements over k contains only a ﬁnite number of elements.

20 Proof. Let C be a class of conjugate elements and ω C. Let f (x) ∈ be the minimum polynomial of ω in k[x]. Let σ be an automorphism of Ω/k. Then σω C. But σω is a root of f σ(x) = f (x). Also if ∈ ω C then ω = σω for some automorphism σ of Ω/k. In that case ′ ∈ ′ σω = ω′ is again a root of f (x). Thus the elements in C are all roots of the irreducible polynomial f (x). Our contention follows.

Notice that if α, β are any two roots, lying in Ω, of the irreducible polynomial f (x), then k(α) and k(β) are k-isomorphic. This isomorphism can be extended into an automorphism of Ω. Thus Theorem 1. To each class of conjugate elements of Ω there is associated an irreducible polynomial in k[x] whose distinct roots are all the elements of this class.

If α Ω we shall denote by Cα the class of α. Cα is a ﬁnite set. ∈ 2 Normal extensions

Suppose K is a subfield of Ω/k and σ an automorphism of Ω/k. Let σK K. We assert that σK = K. For let α K and denote by C¯α the ⊂ ∈ set Cα K ∩ Since σK K we have σα K so σα C¯α. Thus ⊂ ∈ ∈ σC¯α α ⊂ C¯α is a finite set and σ is an isomorphism of K into itself. Thus σC¯α = C¯α 21 which means α σK. Thus K = σK. ∈ We shall now study a class of fields K Ω/k which have the prop- ⊂ erty σK K, ⊂ 2. Normal extensions 19 for all automorphisms σ of Ω/k. We shall call such fields, normal extensions of k in Ω. Let K/k be a normal extension of k and Ω algebraic closure of k containing K. Let α K and Cα the class of α. We assert that Cα K. ∈ ⊂ For if β is an element of Cα, there is an automorphism σ of Ω/k for which β = σα. Since σK K, it follows that β K. Now any element α ⊂ ∈ in K is a root of an irreducible polynomial in k[x]. Since all the elements of Cα are roots of this polynomial, it follows that if f (x) is an irreducible polynomial with one root in K, then all roots of f (x) lie in K. Conversely let K be a subfield of Ω/k with this property. Let σ be an automorphism of Ω/k and α/K. Let σ be an automorphism of Ω/k and α K. Let Cα be the class of α. Since Cα K,σα K. But α is ∈ ⊂ ∈ arbitrary in K. Therefore σK K ⊂ and K is normal. Thus the Theorem 2. Let k K Ω. Then σK = K for all automorphisms σ of ⊂ ⊂ Ω/k every irreducible polynomial f (x) k[x] which has one root ⇐⇒ ∈ in K has all roots in K. Let f (x) be a polynomial in k[x] and K its splitting field. Let Ω be an algebraic closure of K. Let α1,...,αn be the distinct roots of f (x) in 22 Ω. Then K = k(α1,...,αn)

Let σ be an automorphism of Ω/k. σα j = α j for some j. Thus σ takes the set α1,...,αn onto itself. Since every element of K is a rational function of α ,...,α , it follows that σK K. Thus 1 n ⊂ i) The splitting ﬁeld of a polynomial in k[x] is a normal extension of k.

Let Kα be a family of normal subﬁelds of Ω/k. Then Kα is { } α trivially normal. Consider k( Kα). This again is normal since for α T Ω any automorphism σ of /k.S

σk Kα k σKα k Kα ⊂ ⊂  α   α   α  [ [ [             20 2. Algebraic extension ﬁelds

Let now fα(x) be a set of polynomials in k[x] and Kα their splitting { } ﬁelds, then K( Kα) is normal. Also it is easy to see that α S

L = k Kα  α  [    is the intersection of all subﬁelds of Ω/k in which every one of the polynomials fα(x) splits completely. Thus

ii) if fα(x) is a set of polynomials in k[x], the subﬁeld of Ω generated { } by all the roots of fα(x) is normal. { } We also have

iii) If K/k is normal and k L K then K/L is also normal. ⊂ ⊂ 23 For if σ is an L- automorphism of Ω, then σ is also a k-automorphism of Ω and so σK K. ⊂ The k automorphisms of Ω form a group G(Ω/k). From what we − have seen above, it follows that a subﬁeld K of Ω/k is normal if and only if σK = K for every σ G(Ω/k). Now a k automorphism ∈ − of K can be be extended into an automorphism of Ω/k, because every such automorphism is an isomorphism of K in Ω. It therefore follows

iv) K/k is normal if and only if every isomorphism of K in Ω/k is an automorphism of K over k.

As an example, let Γ be the field of rational numbers and f (x) = x3 2. Then f (x) is irreducible in Γ[x]. Let α = √3 2 be one of its roots. − Γ(α) is of degree 3 over Γ and is not normal since it does not contain = 1+ √ 3 Γ Γ ρ where ρ − 2 − . However the field (α, ρ) of degree 6 over is normal and is the splitting field of x3 2. − If K is the field of complex numbers, consider K(z) the field of rational function of 2 over K. Consider the polynomial x3 z in K(z)[x]. This 1 − is irreducible. Let ω = z 3 be a root of this polynomial. Then K(z)(ω) is of degree 3 K(z) and is the splitting field of the polynomial x3 z. − 3. Isomorphisms of fields 21

3 Isomorphisms of ﬁelds

Let K/k be an algebraic extension of k and W any extension of K and so of k. A mapping σ of K into W is said to be k linear if for α, β K 24 − ∈ σ(α + β) = σα + σβ

σα W and if λ k,σ(λα) = λσα. If σ is a k linear map of K into ∈ ∈ − W we deﬁne ασ for α in W by

(ασ)β = ασβ for β K. This again is a k linear map and so the k linear maps of K ∈ − − into W form a vector space V over W. A k isomorphism σ of K into W is obviously a k linear map and − − so σ V. We shall say, two isomorphisms σ, τ of K into W (trivial on ∈ k) are distinct if there exists at least one ω K,ω , 0 such that ∈ σω , τω

Let S be the set of mutually distinct isomorphisms of K into W. We then have

Theorem 3. S is a set of linearly independent elements of V over W.

Proof. We have naturally to show that every ﬁnite subset of S is linearly independent over W. Let on the contrary σ1,...,σn be a ﬁnite subset of S satisfying a non trivial linear relation

αiσi = 0 Xi α W. We may clearly assume that no proper subset of σ ,...,σ i ∈ 1 n is linearly dependent. Then in the above expression all αi are diﬀerent from zero. Let ω be any element of K. Then

αiσiω = 0 Xi 22 2. Algebraic extension ﬁelds

If we replace ω by ωω′ we get, since σi′ s are isomorphisms, 25

αiσiω′.σiω = o Xi for every ω K. This means that σ ,...,σ satisfy another linear rela- ∈ i n tion αiσiω′.σiω = o Xi Since the isomorphisms are mutually distinct, we can choose ω′ in K in such a way that σ1ω′ , σnω′ We then get from the two linear relations, the expression

n 1 − αi σiω′αi σi = o. = αn − σnω′αi Xi 1 This relation is non trivial since the coefficient of σ1 is different from zero. This leads to a contradiction and our theorem is proved. Suppose dim V < then it would mean that S is a finite set. But ∞ the converse is false. We have however the Theorem 4. If (K : k) < , then dim V = (K : k) ∞ Proof. Let (K : k) = n and ω1,...,ωn a basis of K/k.

Consider the k-linear mappings σ1,...,σn deﬁned by

σi(ω j) = o i , j =  1 i , j  Then σ1,...,σn are linearly independent elements of V over W. For, let αiσi = o, αi W. Then i ∈ P αiσi ωi = o Xi 26 for j = 1,..., n. This proves that α j = o. Now let σ be any k-linear 3. Isomorphisms of ﬁelds 23

mapping. It is uniquely determined by its eﬀects on ω1,...,ωn. Put αi = σωi and let τ be given by

τ = σ α σ − i i Xi = = = Then τ(ω j) σω j i αiσi(ω j) o so that τ o. Our contention is established. From this we obtainP the very important

Corollary. If (K : k) < then K has in Ω/k at most (K : k) distinct ∞ k-isomorphisms.

Let α Ω. Consider the field k(α)/k. Let α(1)(= α),...,α(n) be the ∈ distinct conjugates of α over k. An isomorphism σ of k(α)/k is determined completely by its effect on α. Since every isomorphism comes from an automorphism of Ω/k, it follows that k(α(i)) are all the distinct isomorphic images of k(α). Thus 1) Number of distinct k-isomorphisms of k(α) in Ω is equal to the number of distinct roots in Ω of the minimum polynomial of α. Let K/k be an algebraic extension and Ω an algebraic closure of k containing K. Let K have the property that K/k has only finitely many distinct k-isomorphisms in Ω. Let K(1)(= K), K(2),..., K(n) be the distinct isomorphic fields. Let α Ω and let α have over K exactly m ∈ distinct conjugates α(1)(= α),...,α(m). This means that if f (x) is the minimum polynomial of α over K, then f (x) has in Ω, m distinct roots. We claim that K(α) has over k exactly mn distinct isomorphisms in Ω. 27 For, let σi(i = 1,..., n) be the k-isomorphisms defined by

(1) (i) σiK = K

Let f σi (x) be the image of the polynomial f (x) in K[x] by means of the above isomorphism. Let the roots of f σi (x) by α(i1),...,α(in) these being the distinct ones. There exists then an isomorphism σi j( j = (1) (1) (i) (i j) 1,..., m) extending σi of K (α ) on K (α ). Since i has n values, it follows that there are at least mn distinct isomorphisms of K(α) over k. 24 2. Algebraic extension ﬁelds

Let now σ be any automorphism of Ω/k. Let σK = K(i). Then it (1) (i j) σ σ takes α into a root α of f (x) = f i (x) where σi is the isomorphism which coincides with σ on K. Thus since every isomorphism of K(α) over k comes from an automorphism of Ω/k, our contention is established. Let now K = k(ω1,...,ωn) and Ki = k(ω1,...,ωi) so that Ko = k and Kn = K. Let Ki have over Ki 1 exactly Pi distinct Ki 1-isomorphisms. Then K/k has exactly p − p distinct k-isomorphisms− in Ω. 1 ··· n Hence 2) If K L k be a tower of ﬁnite extensions and K has ever L, ⊃ ⊃ n distinct L-isomorphisms in Ω and L has over k, m distinct k-isomorphisms then K has over k precisely mn distinct k-isomorphisms. 28 In particular let (K : k) < and let K have in Ω exactly (K : k) ∞ distinct isomorphisms. Let L be any intermediary ﬁeld. Let a be the number of distinct L-isomorphisms of K and b the number of distinct k-isomorphisms of L. Then (K : k) = ab (K : L)(L : k) = (K : k) ≤ But a (K : L), b (L : k). Thus a = (K : L) and b = (L : k). ≤ ≤ 4 Separability

Let Ω be an algebraic closure of k and ω Ω. Let φ(x) be the minimum ∈ polynomial of ω in k. Suppose k(ω)/k has exactly (k(ω) : k) distinct k-isomorphisms in Ω. Then from the last article it follows that all the roots of φ(x) are distinct. Conversely let the irreducible polynomial φ(x) be of degree n and all its n roots distinct. Then k(ω)/k has n distinct k-isomorphisms ω being a root of φ(x). But it can have no more. Let us therefore make the Definition. An element ω Ω is said to be separably algebraic or sepa- ∈ rable over k if its minimum polynomial has all roots distinct. Otherwise it is said to be inseparable. 1) Let W/k be any extension field and ω W separable over k. Let L ∈ be an intermediary field. Then ω is separable over L. 4. Separability 25

For, the minimum polynomial of ω over L divides that over k.

2) ω Ω separable over k k(ω)/k has in Ω(k(ω) : k) distinct k- 29 ∈ ⇔ isomorphisms.

Let now K = k(ω1,...,ωn) and let ω1,...ωn be all separable over k. Put Ki = k(ω1,...,ω) so that Ko = k and Kn = K. Now Ki 1(ωi) and − ωi is separable over Ki 1 so that Ki over Ki 1 has exactly (Ki : Ki 1) − − − distinct Ki 1- isomorphisms. This proves that K has over k −

(Kn : Kn 1) ... (K1 : Ko) = (Kn : Ko) = (K : k) − distinct k-isomorphisms. If therefore ω K, Then by previous ar- ∈ ticle k(ω) has exactly (k(ω) : k) distinct isomorphisms and hence ω is separable over k. Conversely if K/k is ﬁnite and every element of K is separable over k, then K/k has exactly (K : k) distinct k- isomorphisms. Hence

3) (K : k) < , K/k has (K : k) distinct k-isomorphisms every ∞ ⇔ element of K is separable over k. Let us now make the

Deﬁnition. A subﬁeld K of Ω/k is said to be separable over k if every element of K is separable over k.

From 3) and the deﬁnition, it follows that

4) K/k is separable for every subﬁeld L of K with (L : K) < , L ⇔ ∞ has exactly (L : K) distinct isomorphisms over k.

5) K/L, L/k separably algebraic K/k separable. ⇔ For, let ω K. Then ω is separable over L. Let ω ,...,ω be ∈ 1 n the coefficients in the irreducible polynomial satisfied by ω over L. 30 Then ω has over K1 = k(ω1,...,ωn) exactly (K1(ω) : K1) distinct K1-isomorphisms. Also K1/k is finite separable. Thus K1(ω) has over k exactly (K1(ω) : k) distinct isomorphisms which proves that ω is separable over k. The converse follows from 2). 26 2. Algebraic extension fields

6) If Kα is a family of separable subﬁelds of Ω then { } 2) Kα and b)k( Kα) are separable. α α

a)T follows easily becauseS every element of Kα is separable over k. b) follows since every element of k( Kα) is a rational function of a α ﬁnite number of elements and as eachS of these is separable the result follows from 3).

7) Let K/k be any extension-not necessarily algebraic. The set L of elements of k separably algebraic over K is a ﬁeld. This is evident. We call L the separable closure of k in K. We had already deﬁned an algebraic element ω to be inseparable if its minimum polynomial has repeated roots. Let us study the nature of irreducible polynomials. Let f (x) = a + a x + + a xn be an irreducible polynomial in k[x]. o 1 ··· n If it has a root ω Ω which is repeated, then ω is a root of ∈ 1 n 1 f (x) = a + 2a x + + na x − . 1 2 ··· n Thus f (x) f 1(x) which can happen only if |

iai = o, i = 1,..., n.

31 Let k have characteristic zero. Then ia = o a = 0 that is f (x) is i ⇒ i a constant polynomial. Thus

8) Over a ﬁeld of characteristic zero, every non constant irreducible polynomial has all roots distinct.

Let now k have characteristic p , o. if pχi then iai = o ai = o. 1 ⇒ Thus for f (x) to be identically zero we must have ai = o for pχi. In this case f (x) = a + a xp + o p ··· or that f (x) k[xp]. Let e be the largest integer such that f (x) ∈ e+1 e ∈ k[xp] but not in k[xp ]. Consider the polynomial φ(y) with φ(xp ) = 4. Separability 27

f (x). Then φ(y) is irreducible in k[y] and φ(y) has no repeated roots. Let β1,...,βt be the roots of φ(y) in Ω. Then

e e f (x) = (xp β ) (xp β ). − 1 ··· − t

Thus n = t pe. The polynomial xPe β has in Ω all roots identical · − i to one of them say αi. Then

e e e e xP β = xP ∆αp = (x α )p − i − i − i Thus e f (x) = (x α ) (x α ) p { − 1 ··· − t }

Moreover since β1 ...βt are distinct, α1,...,αt are also distinct. Hence

9) Over a ﬁeld of characteristic p , o, the roots of an irreducible polynomial are repeated equally often, the multiplicity of a root being pe, e o. ≥ It is important to note that (x α ) ... (x α ) is not a polynomial in 32 − 1 − t k[x] and t is not necessarily prime to p. We call t the reduced degree of f (x) (or of any of its roots ) and pe, its degree of inseparability. Thus Degree of: Reduced degree X-degree of inseparability If ω Ω then we had seen earlier that k(ω)/k has as many distinct ∈ isomorphisms in Ω as there are distinct roots in Ω of the minimum k(ω) polynomial of ω over k. If we call the reduced degree of as the k reduced degree of ω we have

10) Reduced degree of ω = Number of distinct roots of the minimum polynomial of ω over k. We may now call a polynomial separable if and only if every root of it in Ω is separable. In particular if f (x) k[x] is irreducible then ∈ f (x) is separable if one root of it is separable. 28 2. Algebraic extension ﬁelds

Let ω Ω and f (x) the minimum polynomial of ω in k[x]. If t = ∈ reduced degree of ω, then

e f (x) = (x ω ) ... (x ω ) p { − 1 − t } e n = t pe. Let ω = ω. Consider ωp = β . Then − 1 1 1 e e f (x) = (xp β ) (xp β ) − 1 ··· − t

and β1,...βt are separable over k. Consider the ﬁeld k(β1) which is a subﬁeld of k(ω).β being of degree t over k, (k(β1) : k) = t. This means that (k(ω) : k(β)) = pe.

But the interesting fact to note is that k(ω) has over k(β) only the identity isomorphism or that k(ω) is fixed by every k(β)-automor- 33 phism of Ω/k(β). Also since every element of k(ω) is a rational function of ω over k(β), it follows that e λp k(β) ∈ for every λ k(ω). Thus the integer e has the property that for every ∈e λ k(ω), λp k(β) and there is at least one λ (for instance ω) for ∈ e ∈ which λp < k(β). e is called the exponent of ω, equivalently of k(ω). We define the exponent of an algebraic element α over k to be the e e 1 integer e o such that αp is separable but not αp − . Hence ≥ 11) Exponent of α is zero α is separable over k. ⇔ We shall now extend this notion of exponent and reduced degree to any finite extension. Let K/k be finite so that K = k(ω1,...,ωn). Put as before Ko = k, Ki = k(ω1,...,ωi) so that Kn = K. Let ωi have reduced degree di and exponent ei over Ki 1. Then −

ei (Ki : Ki 1) = di p − 4. Separability 29

From the deﬁnition of di, it follows that the number of distinct k- isomorphisms of K/k is d1 ... dn. We put

d = d d 1 ··· n and call it the reduced degree of K/k. Then

(K : k) = d.... p f where f = e + + e . We call p f the degree of inseparability of K/k. 1 ··· n In order to be able to give another interpretation to the integer d we 34 make the following considerations. Let Ω K k and let K/k have the property that every k automor- ⊃ ⊃ phism of Ω/k acts like identity on K. Thus if σ G(Ω/k) and ω K, ∈ ∈ then σω = ω All elements of k have this property. Let ω K, ω < k. Then by ∈ deﬁnition, ω has in Ω only one conjugate. The irreducible polynomial of ω has all roots equal. Thus the minimum polynomial of ω is

m xp a − where a k. i.e., ωpm k. On the other hand let K be an extension of k ∈ ∈ in Ω with the property that for every ω K ∈ m ωp k ∈ for some integer m o. Let σ be an automorphism of Ω/k. Then m m ≥ σωp = ωp . But

m m m o = σωp ωp = (σω ω)p − − which shows that σω = ω. ω being arbitrary in K, it follows that every element of G(Ω/k) is identity on K. Hence for ω Ω the following three statements are equivalent ∈ 1) σω = ω for all σ G(Ω/k) ∈ 30 2. Algebraic extension ﬁelds

m 2) ωp k for some m o depending on ω ∈ ≥ 3) The irreducible polynomial of ω over k is of the form xpm a, a k. − ∈ We call an element ω Ω which satisﬁes any one of the above 35 ∈ conditions, a purely inseparable algebraic element over k. Let us make the

Deﬁnition . A subﬁeld K/k of Ω/k is said to be purely inseparable if element ω of K is purely inseparable.

From what we have seen above, it follows that K/k is purely inseparable is equivalent to the fact that every k-automorphism of Ω is identity on K. Let K/k be an algebraic extension and L the maximal separable subfield of K/k. For every ω K,ωpe is separable for some e o which e ∈ ≥ means that ωp L. Thus K/L is a purely inseparable extension field. ∈ This means that every k-isomorphism of L/k can be extended uniquely to a k-isomorphism of K/k. Let, in particular, K/k be a finite extension and L the maximal separable subfield of K. Then K/L is purely inseparable and K/L has no L-isomorphism other than the identity. Thus the number of distinct isomorphism other than the identity. Thus the number of distinct isomorphisms of K/k equals (L : k). But from what has gone before

(K : L) = p f , (L : k) = d.

For this reason we shall call d also the degree of separability of K/k and denote it by [K : k]. We shall denote the degree of inseparability, p f , by K : k . Then

(K : k) = [K : k] K : k .

Note. In case k has characteristic zero, every algebraic element over k is separable.

36 If Ω is the algebraic closure of k and L the maximal separable subfield of Ω then Ω/L is purely inseparable. Ω coincides with L in case k 5. Perfect fields 31 has characteristic zero. But it can happen that L is a proper subfield of Ω. Let K/k be an algebraic extension and L the maximal separable subfield. Consider the exponents of all elements in K. Let e be the maximum of these if it exists. we call e the exponent of the extension K/k. It can happen that e is finite but K/L is infinite. If K/k is a finite extension then K/L has degree p f so that the maximum e of the exponents of elements of K exists. If e is the exponent of K/k then e f ≤ It can happen that e < f . For instance let k have characteristic p , 0 and let α k be not a pth power in k. Then k(α1/p) is of degree p over ∈ k. Let β in k be not a pth power in k. Then k(α1/p, β1/p) is of degree p2 over k, β < k(α1/p) and for every λ k(α1/p, β1/p), λp k. ∈ ∈ We may for instance take k(x, y) to be the field of rational functions of two variables and K = k(x1/p, y1/p). Then (K : k(x, y)) = p2 and λp k(x, y) for every λ K. ∈ ∈

5 Perfect ﬁelds

Let k be a field of characteristic p > 0. Let Ω be its algebraic closure. p Let ω k. Then there is only one element ω′ Ω such that ω′ = ω. ∈ ∈ 1 We can therefore write ω1/p without any ambiguity. Let kp− be the field generated in Ω/k by the pth roots of all elements of k. Similarly from 37 2 kp− ,... Let n K = kp− n 0 [≥ n Obviously K is a field; for if α, β K, α, β kp− for some large n. ∈ ∈ We denote K by kp−∞ We shall study kp−∞ in relation to k and Ω . kp−∞ is called the root field of k. Let ω kp−∞ . Then ω kp−∞ for some n so that ωp∞ k or ω is ∈ ∈ ∈ purely inseparable. On the other hand let ω Ω be purely inseparable. n ∈ Then ωp k for some n i.e., ω kp−∞ kp−∞ . Thus ∈ ∈ ⊂ 32 2. Algebraic extension fields

1) kp−∞ is the largest purely inseparable subfield of Ω/k. Therefore every automorphism of Ω/k is identity on kp−∞ . The set of elements of Ω which are fixed under all the k automorphisms of − Ω/k form a field called the fixed of G(Ω/k). Since every such element ω, fixed under G(Ω/k) is purely inseparable, Ω kp−∞ . Hence ∈ 2) kp−∞ is the fixed field of the group of k automorphism of Ω/k. − Let f (x) be an irreducible polynomial in kp−∞ [x]. We assert that this p p p is separable. For if not f (x) k −∞ [x ]. Thus f (x) = ao + a1 x + ∈ t p + a xnp. Since a kp−∞ [X p], it is in some kp− and so a = b for ··· n i ∈ i i b kp−∞ . Hence i ∈ f (x) = (b + b x + + b xn)p o 1 ··· n which is contradicts the fact that f (x) is irreducible. Hence

38 3) Ω/kp−∞ is a separable extension. We now make the

Deﬁnition. A ﬁeld k is said to be perfect if every algebraic extension of k is separable.

It follows from the deﬁnition that

1) An algebraically closed field is perfect 2) A field of characteristic zero is perfect. We shall now prove 3) A field k of characteristic p > 0 is perfect if and only if k = kp−∞ . Let k have no inseparable extension. Then for a k, a1/p k ∈ ∈1 also; for, otherwise k(a1/p) is inseparable over k. Thus k = kp− = = kp−∞ . The converse has already been proved. ··· We deduce immediately 4) A finite field is perfect. For if k is a finite field of characteristic p > r then a ap is an → automorphism of k. 5. Perfect fields 33

5) Any algebraic extension of a perfect field is perfect. For let K/k be algebraic and k be perfect. If α is inseparable over K, then it is already so over k. An example of an imperfect field is the field of rational functions of one variable x over a finite field k. For if k has characteristic p, then x1/p < k(x) and k(x1/p) is a purely inseparable extension over k(x).

Note 1. If α Ω is inseparable over k, it is not true that it is inseparable ∈ over every intermediary ﬁeld, whereas this is true if α is separable. 39

Note 2. If K/k is algebraic and K kp−∞ contains k properly then K is an ∩ inseparable extension. But the converse of this is not true, that is, if K/k is an inseparable extension, it can happen that there are no elements in K which are purely inseparable over k. We give to this end the following example due to Bourbaki. Let k be a ﬁeld of characteristic p > 2 and let f (x) by in irreducible polynomial n n 1 f (x) = x + a x − + + a 1 ··· n in k[x]. If α1,...,αt are the distinct roots of f (x) in Ω then

pe f (x) = (x α ) ... (x α ) e 1. − 1 − t ≥ n o where n = t. pe. Put φ(x) = f (xp). Then · e φ(x) = (xp α ) ... (xp α ) p − 1 − t } = 1/p If βi αi then

e+1 φ(x) = (x β ) ... (x β ) p − 1 − t and β,...,βt are distinct since α1 ...αt are distinct. Suppose φ(x) is reducible in k[x] and let ψ(x) be an irreducible factor of φ(x) in k[x]. Then pµ ψ(x) = (x β ) (x β ) − 1 ··· − t n o 34 2. Algebraic extension ﬁelds

for µ 0 and ℓ t. (This is because roots of ψ(x) occur with the same ≥ ≤ multiplicity). We can write

µ 1 p p p − ψ(x) = (x α ) ... (x αℓ) − 1 − (µ has to be 1) since otherwise, it will mean that (x α1) (x αt) ≥ − ··· − ∈ k[x]). Now this will mean that φ(x) = ψ(x). W(x) in k[xp] so that ℓ = t. Hence µ ψ(x) = (x β ) (x β ) p ,µ 1. { − 1 ··· − t } ≥ 40 Since ψ(x) is irreducible and

µ 1 ψ(x) = (xp α ) ... (xp α ) p − { − 1 − t } we see that µ 1 = e or µ = e + 1. Thus − φ(x) = f (xp) = ψ(x) p { } Thus if f (xp) is reducible, it is the pth power of an irreducible poly- = p = nomial. In this case ai bi , bi k, i 1,... n. p ∈ Conversely if a = b , b k, i = 1,... n. then f (xp) is reducible. i i i ∈ Hence f (xp) is reducible f (x) kp[x]. ∈ Let k now be a field of characteristic p > 2 given by k =Γ(x, y), the field of rational functions in two variables x, y over the prime field Γ if p elements. Consider the polynomial f (z) = z2p + xzp + y, in k[z]. Since x1/p, y1/p do not lie in k, f (z) is irreducible in k. Let ϑ be a root of f (z). Then (k(ϑ) : k) = 2p. Let β be in k(ϑ) and not in k such that βp k. Then k(ϑ) k(β) k. Also (k(β) : k) = p. In k(β)[x] ∈ ⊃ ⊃ the polynomial f (z) cannot be irreducible, since k(ϑ) : k(β) = 2. It is reducible and so k(β) will contain x1/p and y1/p. But then k(x1/p, y1/p) ⊃ k(β). Thus p2 = k(x1/p, y1/p) : k k(β) : k ϑ) : k = 2p ≤ ≤ but this is impossible. Thus there is no element β in k(ϑ) with βp k. ∈ All the same k(ϑ) is inseparable. 6. Simple extensions 35

6) If k is not perfect then kp−∞ is an inﬁnite extension of k.

n n For, if kp−∞ /k is ﬁnite then, since kp−∞ = kp− we have kp− = n p (n+1) k − for some n, S pn 1 Applying the mapping a aa we ﬁnd k = kp− . But this is false. 41 → Thus (n+1) n kp− : kp− > 1 which proves our contention.

6 Simple extensions

An algebraic extension K/k is said to be simple if there is an ω K such ∈ that K = k(ω). Obviously (K : k) is finite. We call ω a primitive element of K. The primitive element is not unique for, ω + λ, λ k also is ∈ primitive. We now wish to find conditions when an algebraic extension would be simple. We first prove Lemma. Let k be an infinite field and α, β elements in an algebraic closure Ω of k such that α is separable over k. Then k(α, β) is a simple extension of k. Proof. Let f (x) and φ(x) be the irreducible polynomials of α and β respectively in k[x], so that f (x) = (x α ) (x α ) − 1 ··· − n φ(x) = (x β ) (x β ). − 1 ··· − m

Since α is separable, α1,...αn are all distinct. We shall put α = α1, and β = β1. Construct the linear polynomials

βi + Xα j(i = 1,..., m; j = 1,..., n) These mn polynomials are in Ω, the algebraic closure of k and since k is inﬁnite, there exists an element λ k such that ∈ βi + λα j , βi, +λα j,α , J′ 36 2. Algebraic extension ﬁelds

Put γ = β1 + λα1 = β + λα 42 Obviously λ can be chosen so that γ , 0 Now k(γ) k(α, β). We shall will now show that α k(γ). From ⊂ ∈ deﬁnition of γ, β also will be in k(γ) and that will mean

k(γ) k(α, β) k(γ). ⊂ ⊂ In order to do this consider the polynomial φ(γ λ.x)in k(γ)[x]. Also − it vanishes for x = α. Furthermore by our choice of λ, φ(γ λα ) , 0 − i for i > 1. In the algebraic closure Ω, therefore, f (x) and φ(γ λx) have − just x α as a factor. But f (x) and φ(γ λx) are both polynomials in − − k(γ)[x]. So x α is the greatest common divisor of φ(γ λx) and f (x) − − in k(γ)[x]. Thus α k(γ). Our lemma is demonstrated. ∈ We have therefore

Corollary . If α ,...,α are separably algebraic and β Ω then 1 n ∈ k(β,α,...αn) is a simple extension. We deduce immediately

Corollary. A ﬁnite separable extension is simple.

Let Γ be the ﬁeld of rational numbers and Γ(ω, ρ) the splitting ﬁeld of the polynomial x3 2 in Γ[x]. Then Γ(ω, ρ) is simple. A primitive − element γ is given by ω + ρ = γ. Then Γ(γ) is of degree 6 over Γ. It is easy to see that γ has over Γ the minimum polynomial

(x3 3x 3)2 + 3x(x + 1)(x3 3x 3) + 9x2(x + 1)2 − − − − Let now K/k be a ﬁnite extension and L the maximal separable subﬁeld of K/k. Then (K : L) = p f the degree of inseparability and (L : k) = d the reduced degree. If we consider the exponents of elements of K, these have a maximum e and

e f. ≤ 43 We had given an example of e < f . We shall now prove the 6. Simple extensions 37

Theorem 5. If e is the exponent and p f the degree of inseparability of finite extension K of k, then e = f K/k is simple. ⇐⇒ Proof. Let K = k(ω). Let Ko be the maximal separable subfield of K/k. f f Now (K : Ko) = p . But p is degree of ω. Thus e = f Let now K/k be a finite extension and e = f . There exists then a ω pe pt in K such that ω is separable and in Ko but for no, t < e ω is in Ko. Thus K = Ko(ω). Ko being a finite separable extension by our lemma, Ko = k(β) for β separable. Thus

K = k(ω, β).

Using the lemma again, our contention follows. We now investigate the number of intermediary fields between K and k where K is an algebraic extension of k. Let us first consider a simple extension K = k(ω). Let φ(x) be the minimum polynomial of ω in k[x]. Let L be any intermediary field. Let f (x) be the minimum polynomial of ω over L. Then f (x) divides φ(x). Let f (x) have coefficients a0,... an in L. Put L f the field k(a0,..., an). Then f (x) is minimum polynomial of ω over L f . Thus

(K : L) = (K : L f )

But L L. This proves that L = L , and so for every intermediary f ⊂ f field there is a unique divisor of φ(x). Since φ(x) has in Ω only finitely many factors, K/k has only a finite number of intermediary fields. We will now prove that the converse is also true. We shall assume k 44 is infinite. Let now K/k be an algebraic extension having only a finite number of intermediary fields. Let α, β K. Consider the elements α + λβ for ∈ λ k. Since k is infinite, the fields k(α + λβ) are infinite in number and ∈ cannot be all distinct. Let for λ = λ1, λ2 λ1 , λ2

k(α + λ1β) = k(α + λ2β) = k(γ).

Then α + λ β, α + λ β are in k(γ). Thus (λ λ ) k(γ). Hence 1 2 1 − 2 ∈ β k(γ) because λ λ k. This means that α k(γ). ∈ 1 − 2 ∈ ∈ 38 2. Algebraic extension ﬁelds

Therefore k(α, β) k(γ) k(α, β). ⊂ ⊂ Hence every subfield of K, generated by 2 and hence by a finite number of elements is simple. Let K be a maximal subfield of K/k which is simple. (This exists since K/k has only finitely many intermediary fields). Let K = k(ω). Let β K and β < K . Then k(γ) = k(ω, β) K 0 ∈ 0 ⊂ and K k(γ) contradicting maximality of K . Thus β(γ) = K. We have 0 ⊂ 0 proved

Theorem 6. K/k is simple K/k has only ﬁnitely many intermediary ⇐⇒ ﬁelds.

We deduce

Corollary. If K/k is simple, then every intermediary ﬁeld is simple.

Note 1. We have the fact that if K/k is infinite there exist infinitely many intermediary fields.

Note 2. Theorem 6 has been proved on the assumption that k is an in- 45 finite field. If k is finite the theorem is still true and we give a proof later.

7 Galois extensions

Let K/k be an algebraic extension and G the group of automorphism of K which are trivial on k. Let L be the subset of all elements of K which are fixed by G. L is then a subfield of K and is called the fixed field of G. We shall now consider the class of algebraic extensions K/k which are such that the group G(K/k) of automorphisms of K which are trivial on k, has k as the fixed field. We call such extensions galois extensions, the group G(K/k) itself being called the galois group of K/k. We now prove the

Theorem 7. k is the ﬁxed ﬁeld of the group of k automorphisms of K K/k is a normal and separable extension. ⇐⇒ 7. Galois extensions 39

Proof. Let k be the ﬁxed ﬁeld of the group G(K/k) of k-automorphisms of K. Let ω K. Let ω (= ω),...ω be all the distinct conjugates of ω ∈ 1 n that lie in K. Consider the polynomial

f (x) = (x ω ) (x ω ) − 1 ··· − n If σ is an element of G(K/k),σ permutes ω1,...,ωn so that σ leaves the polynomial f (x) unaltered. The coefficients of this polynomial are fixed under all elements of G and hence since k is the fixed field G, f (x) ∈ k[x]. Hence the minimum polynomial roots of the minimum polynomial of ω, since ω1,...ωn are conjugates. Thus f (x)/φ(x). Therefore K splitting field of φ(x), φ(x) has all roots distinct. Thus K/k is normal and separable.

Suppose now K/k is normal and separable. Consider the group 46 G(K/k) of k-automorphisms of K. Let α K. Since K/k is separable, ∈ all conjugates of α are distinct. Also since K/k is normal K contains all the conjugates. If α is fixed under all σ G(K/k), then α is a purely ∈ inseparable element of K and hence is in k. Our theorem is thus proved. We thus see that galois extensions are identical with extension fields which are both normal and separable. Examples of Galois extensions are the splitting fields of polynomials over perfect fields. Let k be a field of characteristic , 2 and let K = k( √α) for α k ∈ and √α < k.( √α)2 = α k. Every element of K is uniquely of the form ∈ a + √α b, a, b k. If σ is an automorphism of K which is trivial on k, · ∈ then its effect on K is determined by its effect on √α. Now

α = σ ( √α)2 = σ( √α).σ( √α) or that σ( √α)/ √α = λ isn such thato λ2 = 1. Since λ K, λ = 1. Thus ∈ ± σ is either th identity or the automorphism

σ( √α) = √α − Thus G(K/k) is a group of order 2. K/k is normal and separable. We shall obtain some important properties of galois extensions. 40 2. Algebraic extension ﬁelds

1) If K/k is a galois extension and k L K, then K/L is a galois ⊂ ⊂ extension also. 47 For, K/L is clearly separable. We had already seen that it is normal. If we denote by G(K/L) the galois group of K over L, than G(K/L) is a subgroup of G(K/k).

2) If k L L K then G(K/L ) is a subgroup of G(K/L ). This ⊂ 1 ⊂ 2 ⊂ 2 1 is trivial.

3) If Kα is a family of galois extensions of k contained in Ω then Kα { } α and k( Kα) are galois. α T This followsS from the fact that this is already true for normal and also for separable extensions.

4) If K/k is galois and L and L′ are two intermediary ﬁelds of K/k which are conjugate over k, then G(G/L) and G(K/L′) are conjugate subgroups of G(K/k) and conversely.

Proof. Since L and L′ are conjugate over k let σ be an automorphism of K/k so that σL = L . Let τ G(K/σL). Then for every ω σL ′ ∈ ∈ τω = ω

But ω = σω for ω L. Thus ′ ′ ∈ 1 σ− τσω′ = ω′

Since this is true for every ω L, it follows that ′ ∈ 1 σ− G(K/σL)σ G(K/L) ⊂ In a similar manner one proves that σG(K/L)σ 1 G(K/σL) which − ⊂ proves our contention.

Conversely suppose that L and L′ are two subﬁelds such that G(K/L) 1 and G(K/L′) are conjugate subgroups of G(K/k). Let G(K/L′) = σ− 48 G(K/L)σ. Let ω L and τ G(K/L). Then σ 1τσ G(K/L ) and so ∈ ′ ∈ − ∈ ′ 7. Galois extensions 41

1 σ− τσω = ω or τ(σω) = σω. This being true for all τ, it follows that σω L for all ∈ ω in L . Thus σL L . We can similarly prove that σ 1L L which ′ ′ ⊂ ′ − ⊂ ′ proves our statement. In particular let L/k be a normal extension of k which is contained in K. Then σL = L for all σ G(K/k). This means that G(K/L) is a ∈ normal subgroup of G(K/k). On the other hand if L is any subﬁeld such that G(K/L) is a normal subgroup of G(K/k) then by above σL = L for all σ G(K/k) which proves that L/k is normal. Thus ∈ 5) Let k L K. Then L/k is normal G(K/L) is a normal ⊂ ⊂ ⇐⇒ subgroup of G(K/k)

6) If L/k is normal, then G(K/k) G(L/k)/G(K/L). ≃ Let σ G(K/k) andσ ¯ the restriction of σ to L. Thenσ ¯ is an auto- ∈ morphism of L/k so thatσ ¯ G(L/k). Now σ σ¯ is a homomorphism ∈ → fo G(K/k) into G(L/k). For στω = στω = σ(τω) = σ¯ τω¯ for all ω L. Thusσ ¯ τ¯ = στ ∈ The homomorphism is onto since every automorphism of L/k can be 49 extended into an automorphism of K/k. Nowσ ¯ is identity if and only if

σω¯ = ω for all ω L. Thus σ G(K/L). Also every σ G(K/L) has this ∈ ∈ ∈ property so that the kernel of the homomorphism is G(K/L). Let K/k be a ﬁnite galois extension. Every isomorphism of K/k in Ω is an an automorphism. Also K/k being separable, K/k has exactly (K : k) district isomorphisms. This shows that

(K : k) = order of G.

We shall now prove the converse

7) If G is a finite group of automorphisms of K/k having k as the fixed field then (K : k) = order of G. 42 2. Algebraic extension fields

Proof. Let σ1,...,σn be the n elements of G and ω1,...ωn+1 any n + 1 elements of K. Denote by VK the vector space over K of n dimensions formed by n- tuples (α1,...,αn). Deﬁne n + 1 vectors Ω1,..., Ωn+1 by

Ωi = (σi(ωi),...,σn(ωi))i = 1,..., n + 1

Among these vectors there exists m n vectors linearly independent ≤ over K. Let Ω1,..., Ωm be independent. Then

m Ωm+1 = aiΩi ai K = ∈ Xn 1 This equation gives, for the components of the Ω′ s,

m σℓ(ωm+1) = aiσℓ(ωi)α = 1,..., n. = X1 i

Since σ1,...,σn form a group then σnσ1,...,σnσn are again the 50 elements σ1,...,σn in some order. Thus

m σhσℓ(ωm+1) = σh(ai σhσℓ(ωi) = | Xi 1 which means m σℓ(ωm+1) = σh(ai)σℓ(ωi) i = 1,..., n = Xi 1 subtracting we have

m (σh(ai) aiσℓ(ωi) = 0 = − Xi 1 which means that m σ (a ) a Ω = 0 h i − i i i=1 X 7. Galois extensions 43

From linear independence, it follows that σh(ai) = ai for all i. But h is arbitrary. Thus a k. We therefore have by taking σ to be the i ∈ 1 identity element of G m ωm+1 = aiωi = Xi 1 ai are in k, not all zero. Hence

(K : k) n. ≤ But every element of G is an isomorphism of K/k. Thus

(K : k) order of G = n. ≥ Our assertion is established. Suppose K/k is a galois extension. For every subfield L of K/k, the extension K/L is galois. We denote its galois group by G(L) and this is a subgroup of G(K/k). Suppose g is any subgroup of G(K/k) and let F(g) be its fixed field. Then F(g) is a subfield of K. The galois group G(F(g)) of K/F(g) contains g. In general one has only 51

g G(F(g)) ⊂ Let now K/k be a finite galois extension. Let g be a subgroup of G(K/k) = G and F(g), the fixed field of g. Then by above

K : F(g) = order of g. and so g = G K/F(g)

If g1 and g2 are two subgroups of G with g1 g2 then F(g1) and ⊂ F(g2) are distinct. For if F(g1) and F(g2) are identical, then by above g1 = G(K/F(g2)) = g2. We thus have the Main Theorem(of finite galois theory). Let K/k be a finite galois extension with galois group G. Let M denote the class of all subgroups of G and N the class all subfield of K/k. Let φ be the mapping which 44 2. Algebraic extension fields

assigns to every subgroup g M, the fixed field F(g) of g in N. Then φ ∈ is a mapping of M onto N which is biunivocal. In order to restore this property even for infinite extensions, we de- velop a method due originally to Krull. Let K/k be a galois extension with galois group G(K/k). For every ω K, we denote by Gω the galois group G(K/k(ω)). This then is ∈ a subgroup of G(K/k). We make G(K/k) into a topological group by prescribing the Gω as a fundamental system of neighbourhoods of the { } identity element e G(K/k). Obviously Gα = (e). For σ α Gα ∈ α ∈ ⇒ σα = α for all α K so that σ = e. It is easy to verify that Gα satisfy ∈ T { T} 52 the axioms for a fundamental system of open sets containing the identity elements e. Any open set in G is therefore a union of sets of type σGα or a { } finite intersection of such. Also since Gα are open subgroups they are closed; for, σGα is open for all σ and hence

σGα σ,e [

is also open. Therefore Gα is closed. This proves that the topology on G makes it totally disconnected. We call the topology on G the Krull topology. If g is a subgroup of G,g ¯ the closure of g is also a subgroup. We now prove the

Lemma. Let g be a subgroup of G a and L its ﬁxed ﬁeld. Then

G(K/L) = g¯ (the closure of g).

Proof. Let ω be an element in K and f (x) its minimum polynomial in k. Consider f (x) as a polynomial over L and let L′ be its splitting field over L. Then L′/L is a galois extension. The restriction of elements of g to L′ are automorphisms of L′ with L as fixed field (by definition of L). By finite galois theory these are all the elements of the galois group of L′/L. This means that every automorphism of L′/L comes from an elements of g. 7. Galois extensions 45

Let σ G(K/L). Let ω be any element in K and G the group ∈ G(K/k(ω)). The restriction of σ to L′ is an automorphism of L′ with L as fixed field. There is thus an element τ g, which has on L the ∈ ′ same effect as σ. Hence τ 1σ is identity on L and since ω L we get − ′ ∈ ′ 1 τ− σω = ω 1 1 This means that τ σ Gω by definition of Gω. Hence σ τ Gω 53 − ∈ − ∈ or τ σGω. But σGω is an open set containing σ. Therefore since ∈ σGω for all Gω form a fundamental system of neighbourhoods of σ { } we conclude that σ g¯. ∈ Thus G(K/L) g¯ ⊂ Let now σ g¯. Then σGω is a neighbourhood of σ and so intersects ∈ g in a non empty set. Let ω L. Let τ σGω g. Let σ in Gω such ∈ ∈ ∩ ′ that σσ′ = τ

By definition of τ, τω = ω. But τω = σσ ω = ω. Therefore σσ Gω. ′ ′ ∈ But σ′ is already is in Gω. Therefore σω = ω. Also ω being arbitrary σL = L which means that σ G(K/L). Thus ∈ g¯ G(K/L) ⊂ and our contention is established. From the lemma, it follows that if L is an intermediary field, the galois group G(K/L) is a closed subgroup of G(K/k). On the other hand if g is a closed subgroup of G and F(g) its fixed field then G K/F(g) = g¯ = g. we have hence the fundamental Theorem 8. Let K/k be a galois extension and G(K/k) the galois group with the Krull - topology. Let M denote the set of closed subgroups of G and N the set of intermediary fields of K/k. Let φ be the mapping which assigns to every g M, the fixed field F(g) of g in N. Then φ is a 54 ∈ biunivocal mapping of M on N. 46 2. Algebraic extension fields

Suppose now that K/k is a galois extension and L is an intermediary field. Let G(K/k) and G(K/L) be the galois groups. On G(K/L) there are two topologies, one that is induced by the topology on G(K/k) and 55 the other the topology that G(K/L) possesses as a galois extension. If L′ is a subfield of K/L so that L′/L is finite, then G(K/L′) is an open set in the inherent topology on G(K/L). On the other hand L′ = L(ω) since L′/L is a separable extension. Thus G(K/L′) Gω G(K/L) ⊂ ∩ which proves that the two topologies are equivalent. Here we have used the fact that a finite separable extension of L is simple. We gave already seen the truth of this statement if L is infinite. In case L is finite it is proved in the next section. ‘ In a similar manner if K/k is galois and L is a normal extension of k in K, then on G(L/k) there are two topologies, one the inherent one and the other the topology of the quotient group G(K/k)/G(K/L). One can prove that the two topologies are equivalent. We call an extension K/k abelian or solvable according as G(K/k) is abelian or a solvable group. If K/k is a galois extension and G(K/k) its galois group with the Krull topology let H denote the closure of the algebraic commutator subgroup of G(K/k). H is called the topological commutator subgroup. If L is its fixed field, then since H is normal, L/k is a galois extension. Its galois group is isomorphic to G/H which is abelian. From the property of the commutator subgroup, it follows that L is the maximal abelian subfield of K/k.

8 Finite ﬁelds

Let K be a finite field of q elements, q = pn where p is the characteristic of K. Let Γ be the prime field of p elements. Then

(K : Γ) = n

K∗ the group of non-zero elements of K is an abelian group of order q 1. For α K we have − ∈ ∗ q 1 α − = 1 8. Finite ﬁelds 47

1 being the unit element of K. The q 1 elements of K∗ are roots of q 1 − x − 1. Also − q 1 x − 1 = (x α). − − α K Y∈ ∗ Let α K . Let d be its order as an element of the ﬁnite group K . ∈ ∗ ∗ Then αd = 1. Consider the polynomial xd 1. It has in K at most d − ∗ roots. Also d/q 1. But − q 1 d q 1 d x − 1 = (x 1)(x − − + ) − − ··· Since xd 1 and xq 1 d + both have respectively at most d and − − − ··· q 1 d roots in K and they together q 1 roots in K it follows that − − ∗ − ∗ for every divisor d of q 1, xd 1 has exactly d roots in K . These roots − − ∗ from a group of order d. If it is cyclic then there is an element of order d and there are exactly φ(d) elements of order d. Also

φ(d) = q 1 − d/q 1 X− which proves that for every divisor d of q 1 there are φ(d) 1 elements 56 − ≥ of order d. Thus

1) The multiplicative group of a finite field is cyclic. Let k be a finite of q elements and K a finite extension of k of degree n n. Then K has q elements. Since K∗ is cyclic, let ρ be a generator of K∗. Then K = k(ρ) which proves

2) Every finite extension of a finite field is simple. For a finite field of characteristic p, a ape is an automorphism of → K. Since k has q elements we have

aq = a

for every a k. ∈ 48 2. Algebraic extension ﬁelds

Now a aq is an automorphism of K/k which ﬁxes elements of k. → Call this automorphism σ. σ is determined uniquely by its eﬀect on a generator ρ of K∗. Consider the automorphisms

2 n 1 1,σ,σ ,...,σ −

These are distinct. For,

i i 1 i 1 q qi σ ρ = σ − (σρ) = σ − (ρ ) = ρ

Hence ρq = 1 qi o( mod qn) or i = o (Since i < n). But K/k ⇐⇒ ≡ being of degree n cannot have more than n automorphisms. We have

3) The galois group of a finite extension of a finite field is cyclic. The generator σ of this cyclic group, defined by

σa = aq

is called the Frobenius automorphism. It is deﬁned without any ref- 57 erence to a generator of K∗. Let L be an intermediary ﬁeld of K/k. Then (L : k) is a divisor of (K : k) = n. If d = (L : k) then L has qd elements. Also since K/k has a cyclic galois group, there is one and only one subgroup of a given order d. Hence

4) The number of intermediary ﬁelds of K/k is equal to the number of divisors of n. Chapter 3

Algebraic function ﬁelds

1 F.K. Schmidt’s theorem

Let K/k be an extension ﬁeld and x1,..., xn+1 any n + 1 elements of 58 K. Let R = k[z1,..., zn+1] be the ring of polynomials in n + 1 variables over k. Let Y be the ideal in R of polynomials f (z1,..., zn+1) with the property

f (x1,..., xn+1) = 0.

Then clearly Y is a prime ideal of R. Also since R is a Noetheeian ring, Y is ﬁnitely generated. Note that Y = (o) if and only if x1,..., xn+1 are algebraically independent over k. Y is called the ideal of the set x1,..., xn+1. We shall consider the case where the set x1,..., xn+1 has dimension n over k, that is that k(x1,..., xn+1) is of transcendence degree n over k. We prove

Theorem 1. Y is a principal ideal generated by an irreducible polynomial.

Proof. Without loss in generality we may assume that x1,..., xn are algebraically independent over k and that xn+1 is algebraic over k(x1,..., xn), so that Y , (o).

49 50 3. Algebraic function ﬁelds

Consider the degrees of the polynomials f (z1,..., zn+1) (in Y ) in the variable zn+1. These degrees have a minimum greater than zero since Y , (o), and x1,..., xn are algebraically independent over k. Let ϕ be a polynomial in Y of smallest degree in zn+1. Put = λ + λ 1 + + ϕ Aozn+1 A1zn− ... Aλ

59 where A0, A1,..., Aλ are polynomials in z1,... zn with coeﬃcients in k. We may assume that A0, A1,..., Aλ have no common factor in k[z1,..., zn]. For if A(z1,..., zn) is a common factor of A0,..., Aλ then

ϕ(z1,..., zn+1) = A(z1,..., zn) ϕ1(z1,..., zn+1)

and so, since x1,..., xn are algebraically independent,

ϕ1(x1,..., xn+1) = o

and ϕ1 will serve our purpose. So we can take ϕ to be a primitive polynomial in zn+1 over R′ = k[z1,..., zn]. Clearly ϕ is irreducible in R′. For, if

ϕ = g1(z1,..., zn+1) g2(z1,..., zn+1)

then gi(x1,..., xn+1) = 0 for i = 1 or 2, so that either g1 or g2 is in Y . Both cannot have a term in zn+1 with non zero coefficient. For then the degrees in zn+1 of g1 of g2 will both be less that of ϕ in zn+1 contradicting the definition of φ. So one g1, g2 say g1 is independent of zn+1. But this means that ϕ is not a primitive polynomial. Thus we have chosen in Y a polynomial ϕ which os irreducible, of the smallest degree in zn+1 and primitive in R′[zn+1]. Let ψ(z1,..., zn+1) be any other polynomial in Y . Since F = k(z1,..., zn) is a field, F[zn+1] is a Euclidean ring so that in F[zn+1] we have

ψ(z1,..., zn+1) = A(z1,..., zn+1) ϕ(z1,..., zn+1) + L(z1,..., zn+1)

60 where A and L are polynomials in zn+1over F. Here either L = o or degree of L in zn+1 is less than that of ϕ. If L , o, then we may multiply 1. F.K. Schmidt’s theorem 51

both sides of the above equation by a suitable polynomial in z1,..., zn over k so that

B(z1,..., zn)ψ = C(z1,..., zn+1)ϕ(z1,..., zn+1) + L1(z1,..., zn+1)

L1 having in zn+1 the same degree as L. Since ϕ and ψ are in Y , it follows that L Y . Because of degree of L , it follows that 1 ∈ 1 L1 = L = 0. Thus

B(z1,..., zn)ψ = A(z1,..., zn+1)ϕ(z1,..., zn+1) Since ϕ is a primitive polynomial, it follows that ϕ divides ψ and our theorem is proved. We call ϕ the irreducible polynomial of x1,... xn+1 over k. Note that since x1,... xn are algebraically independent over k, the polynomial ϕ1(zn+1) = ϕ(x1,..., xn, zn+1) over k(x1,..., xn) is irreducible in zn+1. Let x1,..., xn+1be of dimension n over k and ϕ the irreducible polynomial of x1,..., xn+1 over k. Let ϕ be a polynomial in z1,..., zi+1 but not in zi+2,..., zn+1, that is it does not involve zi+2,..., zn+1 in its expression. Consider the ﬁeld L = k(x1,..., xi+1). because x1,..., xi+1 are algebraically dependent (ϕ(x1,..., xi+1) = 0), 61 dim L i. k ≤ But k(x1,..., xn+1) = L(xi+2,..., xn+1) so that

dim k(x ,..., x + ) n i. L 1 n 1 ≤ − Since dimensions are additive, we have

n = dim L + dim k(x ,..., x + ) i + n i = n. k L 1 n 1 ≤ − Thus L has over k the dimension i. Since ϕ is a polynomial in z1,..., zi+1 every one of these variables occurring, with non zero co- eﬃcients, we get the 52 3. Algebraic function ﬁelds

Corollary. If x1,..., xn+1 be n + 1 elements of K/k and have dimension n, there exist among them i + 1 elements, i n, say x ,..., x + (in some ≤ 1 i 1 order) such that k(x1,... xi+1) has dimension i over k and every i of them are algebraically independent. Let K/k be a transcendental extension with a transcendence base B over k. Then K/k(B) is algebraic. We call K/k an algebraic function field if (1) B is a finite set (2) K/k(B) is finite algebraic.

Let dimk K = n. There exist x1,..., xn in K which form a transcendence base of K/k. If K is an algebraic function field then K/k(x1, ..., x ) is finite algebraic. Hence K = k(x ,..., x ), m n, is finitely n 1 m ≥ generated. This shows that algebraic function fields are identical with finitely generated extensions. 62 An algebraic function field K/k is said to be separably generated if there exists a transcendence base x1,..., xn of K/k such that K/k(x1,..., xn) is a separable algebraic extension of finite degree. x1,..., xn is then said to be a separating base. Clearly every purely transcendental extension is separably generated. Also, if k has characteristic zero and K is an algebraic function field, it is separably generated. In this case every transcendence base is a separating base. This is no longer true if k has characteristic p , o. For example, let K = k(x, y) be a function field of transcendence degree one and let x2 yp = o. − Let k have characteristic p , 2. Obviously x and y are both transcendental over k. But K/k(x) is a simple extension generated by y which is a root of zp x2 − over k(x)[z] and since k has characteristic p, K/k(x) is purely inseparable. On the other hand K/k(y) is separable since x satisfies over k(y) the polynomial x2 yp. − 1. F.K. Schmidt’s theorem 53

Thus y is separating but not x. An algebraic function field which is not separably generated is said to be inseparably generated. This means that for every base B of K/k. K/k(B) is inseparably algebraic. In algebraic geometry and in algebraic function theory, it is of im- 63 portance to know when an algebraic function field is separably generated. An important theorem in this regard is theorem 2 due to F.K. Schmidt. We shall first prove a

Lemma . Let k be a perfect ﬁeld of characteristic p , o and K = k(x1,..., xn+1) an extension ﬁeld of dimension n. Then K is separably generated.

Proof. Let ϕ be the irreducible polynomial of x1,..., xn+1 and let it be a polynomial in z1,..., zi+1 but in zi+2,..., zn+1. Then

ϕ(x1,..., zt,..., xi+1) is irreducible over k(x1,..., xt 1, xt+1,..., xi+1) for every t, 1 t i+1. − ≤ ≤ At least for one t,ϕ(z1,..., zt,..., zi+1) is a separable polynomial in zt over k(z1,..., zt 1,..., zi+1). For, if it is inseparable in every zt, then − p p ϕ(z ,..., z + ) k[z ,..., z ] 1 i 1 ∈ 1 i+1 and since k is perfect, this will mean that ϕ(z1,..., zi+1) is the pth power of a polynomial in k[z1,..., zi+1] which contradicts irreducibility of ϕ. So, for some zt, say z1, we have ϕ(z1, x2,..., xi+1) is a separable polynomial. Hence x1 is separable over k(x2,..., xi+1) and so over k(x2, ..., xn+1). But x2,..., xn+1 has dimension n and our lemma is proved.

Corollary. Under the conditions of the lemma, a separating base of n elements may be chosen from among x1,..., xn+1.

We are now ready to prove the theorem of F.K.Schmidt. 64

Theorem 2. Every algebraic function field K over a perfect field k is separably generated. 54 3. Algebraic function fields

Proof. Obviously, the theorem is true if k has characteristic zero. So let k have characteristic p , o. Let K = k(x1,..., xm) and let n be the dimension of K/k. Then m n. If m = n there is nothing to prove. ≥ Let m = n + q. If q = 1 then lemma 1 proves the theorem. So let us assume theorem proved for q 1 instead of q > 1. We may assume, − without loss in generality that x1,..., xn is a transcendence base of K/k. Consider the ﬁelds L = k(x1,..., xn, xn+1). It satisﬁes the conditions of the lemma. Hence there exist n elements among x1,..., xn+1 say x1,..., xt 1, xt+1,... xn+1 which from a separating base of L/k. Thus xt − is separable over k(x1,..., xt 1,..., xn+1) and hence over −

M = k(x1,..., xt 1, xt+1, xt+2,..., xm) − M/k now satisﬁes the induction hypothesis and so is separably generated. Since K/M is separable, it follows that K is separably generated.

We could prove even more if we assume as induction hypothesis the fact that among x1,..., xm there exists a separating base.

2 Derivations

Let R be a commutative ring with unit element e. A mapping D of R into itself is said to be a derivation of R if

65 (1) D(a + b) = Da + Db

(2) D(ab) = aDb + bDa

for a, b R. It is said to be a derivation over a subring R if for every ∈ ′ a R , Da = o. It then follows that for a R and x R ∈ ′ ∈ ′ ∈ Dax = aDx

The set R of a R with Da = o is a subring of R and contains e. o ∈ For, De = De2 = De. e + e. De = 2De ; 2. Derivations 55 so De = o. If Da = o, Db = o, then

D(a + b) =Da + Db = o D(ab) = aDb + bDa = o

Thus Ro is a subring. We call Ro the ring of constants of the derivation D. If R is a ﬁeld, then R also is a ﬁeld. For, if x R and x , o, then o ∈ o 1 1 1 o = De = Dx.x− = Dx.x− + x.Dx−

Thus 1 Dx− = o so that x 1 R . − ∈ o D is said to be a non-trivial derivation of R if there is an x R with ∈ Dx , o. It follows from above that

Theorem 3. A prime ﬁeld has no non-trivial derivations.

A Derivation D¯ of R is said to be an extension of a derivation D of a subring R of R if Da¯ = Da for a R . We now prove the 66 ′ ∈ ′ Theorem 4. If K is the quotient ﬁeld of an integrity domain R, then a derivation D of R can be uniquely extended to K. a Proof. Every element c in K can be expressed in the form c = , a, b b R. If an extension D¯ of D exists, then ∈ Da¯ = Da

But a = bc so that

Da = Da¯ = Dbc¯ = bDc¯ + cDb

Therefore Da cDb bDa aDb Dc¯ = − = − b b2 56 3. Algebraic function ﬁelds

a′ If c is expressed in the form , a′, b′ R then ab′ = ba′and so b′ ∈

Da.b′ + a.Db′ = Db.a′ + bDa′

or that Da cDb Da cDb ′ − ′ = − b′ b which proves that Dc¯ does not depend on the way c is expressed as the ratio of two elements from R. We have therefore only to prove the a existence of D¯ . In order to prove this, put for c = b bDa aDb Dc¯ = − ; b2 we should verify that it is a derivation, is independent of the way c is expressed as ratio of elements in R and that it coincides with D on R. These are very simple.

67 Let D1, D2 be two deviations of R. Deﬁne D = D1 + D2 by Da = D a + D a for a R. Then it is easy to verify that D is a derivation of 1 2 ∈ R. Furthermore if a R deﬁne aD by ∈ (aD)x = a.Dx

By this means, the derivations of R from an R module, Suppose D1,..., Dr from a basis of the module of derivations of R. Then every derivation D of R is of the form

D = a D , a R. i i i ∈ Xi

Let R be an integrity domain and D¯ 1,..., D¯ r the unique extensions of D1,..., Dr respectively to K, the quotient, ﬁeld of R. Then D¯ 1,..., D¯ r are linearly independent over K. For, if

1 D¯ = o, 1 K i i i ∈ Xi 2. Derivations 57

ai then write 1i = , ai, bi R. We get bi ∈ ai D¯ = o. bi i Xi

Multiplying throughout by b1,..., br which is not zero, we get ( λD¯ i)t = o for every t R. Therefore λiDi = o which implies i ∈ i = = thatP λi o or ai o. P Also, since every derivation D of K is an extension of a derivation of R, it follows that the derivations of K from an r-dimensional vector space over K. Let us now consider the case where R = k[x1,..., xn] is the ring of polynomials in n variables x1,..., xn. The n mappings ∂a Di : a , i = 1,..., n → ∂xi are clearly derivations of R over k. They form a base of derivations of 68 R which are trivial on k. For, if

a D = o, a R i i i ∈ Xi then, since Di(x j) = δi j, we get

o = ( aiDi)x j = a j, j = 1,..., n. Xi

Also, if D is any derivation of R which is trivial on k, then let Dxi = ai. Put D¯ = D a D . − i i Xi Then Dx¯ = Dx ( a D )x = o j j − i i j Xi which shows that since x1,..., xn generate R, D¯ = o. 58 3. Algebraic function ﬁelds

Suppose D is any derivation of k and let D¯ be an extension of D to R. Let Dx¯ i = ai. Let Do be another extension of D which has the property Do xi = ai, i = 1,..., n. Then D D is a derivation of R which is trivial on k. But since − o (D D )x = o, i = 1,..., n − o i

it follows that D = Do. This gives us the

Theorem 5. The derivations of K = k(x1,..., xn), the ﬁeld of rational functions of n variables over k which are trivial on k from a vector space of dimension n over K. A basis of this space of derivations is given by the n partial derivations

∂ Di = , i = 1,..., n. ∂xi

69 Furthermore if D is any derivation of k, there exists only one extension D¯ of D to K for which

Dx¯ i = ai, i = 1,..., n

where a1,..., an are any n quantities of K arbitrarily given.

We now consider derivations of algebraic function fields. Let K = k(x1,..., xm) be a finitely generated extension of k. Put T = k[x1,..., xm]. In order to determine all the derivations of K, it is enough to determine the derivations of T since K is the quotient field of T. Let D be a derivation of k; we wish to find extensions D¯ of D to K. Let R denote the ring of polynomials k[z1,..., zm] in m independent ∂ f variables. For any polynomial f (x1,..., xm) in T, denote by the ∂xi ∂ f¯ polynomial obtained by substituting zi = xi, i = 1,..., m in where ∂zi f¯ = f (z1,..., zm) is in R. If λ1 λm f = aλ ,...,λ x x 1 m 1 ··· m Xλ 2. Derivations 59 aλ ,...,λ k, put 1 m ∈

D λ1 λm f = (Daλ ,...,λ )x x . 1 m 1 · m Xλ Obviously f D is a polynomial in T. If D¯ is an extension of the derivation D, then clearly

m D ∂ f D¯ f = f + Dx¯ i = ∂xi Xi 1 for any f T. Also D¯ is determined uniquely by its values on x ,..., x 70 ∈ 1 m which generate T. Now Dx¯ i cannot be arbitrary elements of T. For, let Y be the ideal in R of the set x1,..., xm. Then for f (z1,..., zm) in Y ,

f (x1,..., xm) = o.

Therefore, since Do¯ = 0, the Dx¯ i would have to satisfy the inﬁnity of equations m D ∂ f 0 = f + Dx¯ i = ∂xi Xi 1 for every f in Y . Conversely, suppose u1,... um are m elements in T satisfying

m D ∂ f f + ui = 0 = ∂xi Xi 1 for every f in Y . For any ϕ in T deﬁne D¯ by

m D D ∂ϕ D¯ ϕ = ϕ + f + Dx¯ i = ∂xi Xi 1 where Dx¯ i = ui. Then clearly D¯ is a derivation of T and it coincides with D on k. Furthermore D¯ does not depend on the way ϕ is expressed 60 3. Algebraic function ﬁelds

as a polynomial in x1,..., xm. For, if ϕ = a(x1,..., xm) = b(x1,..., xm) then, since a b Y have − ∈ ∂a ∂b aD bD + u u = o − ∂x i − ∂x i i i i X which proves our contention. Hence

71 Theorem 6. Let K = k(x1,..., xm) and D a derivation of k. Let Y be ideal in k[z1,..., zm] of the set x1,..., xm. Let u1,..., um be any elements of K. There exists a derivation D¯ and only one satisfying

Dx¯ i = ui, i = 1,..., m

and extending the derivation D in k, if and only if, for every f Y ∈ m D ∂ f f + ui = o. = ∂xi Xi 1 and then for every ϕ in K,

m D ∂ϕ D¯ ϕ = ϕ + ui. = ∂xi Xi 1 The infinite number of conditions above can be reduced to a finite number in the following manner. Since R = k[z1,..., zm] is a noetherian ring, the ideal Y has a finite set f ,..., f of generators so that f Y 1 s ∈ may be written s f = Ai fi, Ai R = ∈ Xi 1 Suppose f1,..., fs satisfy the above conditions, then since D = D + D f (x) Ai fi fi Ai Xi Xi ∂ f ∂ f j ∂A j = A j + f j , ∂xi ∂xi ∂xi Xj Xj 2. Derivations 61 we get m D ∂ f f (x) + ui = o. = ∂xi Xi 1 We may therefore replace the above by the finitely many conditions

m D + ∂ f = = fi u j o, i 1,..., s. = ∂x j Xj 1 We now consider a few special cases. 72 Let K = k(x) be a simple extension of k. Let D be a derivation of k. We will study extensions D¯ of D into K.

(1) First let x be transcendental over k. The ideal of x in k[z] is zero. This means that we can prescribe Dx¯ arbitrarily. Thus for every u K there exists one and only extension D¯ with ∈ Dx¯ = u

(2) Let now x be algebraic over k. Suppose x is inseparable over k. Let f (z) be the minimum polynomial of x in k[z]. Then f (z) generates the ideal of x in k[z]. But x being inseparable, f ′(x) = o. This means that D has to satisfy

f D = o.

Also, D¯ is uniquely ﬁxed as soon as we assign a value u to Dx¯ . This can be done arbitrarily as can be easily seen. Thus there exist an inﬁnity of extensions D¯ .

(3) Finally, let X be separable. Then f (z), the irreducible polynomial of x over k is such that f ′(x) , o.

Since f (z) generates Y we must have

D f (x) + f ′(x)Dx¯ = o, 62 3. Algebraic function ﬁelds

or that Dx¯ is uniquely ﬁxed by D.

f D(x) Dx¯ = (*) − f ′(x) There is thus only one extension of D to K and it is given by ( ). ∗ 73 In particular, K has no derivations, except the trivial one, over k. We shall now prove

Theorem 7. In order that a ﬁnitely generated extension K = k(x1, ..., xn) be separably algebraic over k, it is necessary and suﬃcient that K have no non-trivial derivations over k.

Proof. If K/k is algebraically separable, then since K is ﬁnitely generated over k, it follows that K = k(x) for some x and the last of the considerations above shows that K has no nontrivial derivations over k.

Suppose now K/k has no-trivial derivations. In case n = 1, our considerations above show that K/k is separable. Let now n > 1 and assume that theorem is proved for n 1 instead of n. − Put K = K1(xn) K1 = k(x1,..., xn 1). − Then xn is separably algebraic over K1. For, if not, let x1 be inseparable over K1 or transcendental over K1. In both cases the zero derivation in K1 can be extended into a non-trivial deri-vation of K con- tradictions hypothesis over K. Thus xn is separable over K1. This implies, since K has no derivations over k that K1 and our theorem is proved. Note that in the theorem above, the fact that K/k is finitely generated is essential. For instance, if k is an imperfect field and K = kp−∞ then 74 K/k is infinite. Also if a k then a = bp for some b K. If D is a ∈ ∈ derivation of K, then

p p 1 Da = Db = pb − Db = o

This proves, in particular, that a perfect ﬁeld of characteristic p , o, has only the trivial derivation. 2. Derivations 63

If we take K to be the algebraic closure of the rational number ﬁled then K has only the trivial derivation. Let K = k(x, y) be an algebraic function ﬁeld of one variable. Let us assume that x is a separating variable and ϕ(X, Y) the irreducible polynomial of x, y over k. Then if D is a derivation of K over k, ∂ϕ ∂ϕ Dx + Dy = o ∂x ∂y ∂ϕ so that if we assume that y is separable over k(x), then , o and hence ∂y

∂ϕ Dy = − ∂x Dx ∂ϕ ∂y This shows that the ratio Dy/Dx is independent of D. Also, for any rational function ψ(x, y) of x, y ∂ψ ∂ψ Dψ = Dx + Dy ∂x ∂y which gives,if Dx , o

∂ϕ Dψ = ∂ψ ∂ψ ∂x Dx ∂x − ∂y ∂ϕ ∂y ! which is a well known formula in elementary calculus. We shall now obtain a generalisation of theorem 7 to algebraic function fields. Let K = k(x1,..., xm) be an algebraic function field of dimension n, so that o n m. Let f ,..., f be a system of generators of the 75 ≤ ≤ 1 s ideal Y of polynomials f (z1,..., zm) in k[x1,..., xm] which vanish for ∂ f x1,..., xm. Let for f in k[z1,..., zm] have the same meaning as ∂xi before. Denote by M the matrix ∂ f M = i = 1,..., m ∂x i ! 64 3. Algebraic function fields

j = 1,..., s

where i is the row index and j the column index. We denote by t the rank of the matrix M which is a matrix over K. Let VK(D) denote the vector space of derivations of K which are trivial on k. This is a vector space over K. Denote by l the dimension of VK(D) over K. We then have

Theorem 8. l+ t =m.

Proof. For any integer p, denote by Wp the vector space over K of dimension p, generated by p-tuples (β ,...,β ), β K. 1 p i ∈ Let σ denote the mapping

σD = (Dx1,..., Dxm)

of Vk(D) into Wm. This is clearly a homomorphism of Vk(D) into Wm. The kernel of the homomorphism is the set of D for which Dxi = o; i = 1,..., m. But since K is generated by x1,..., xm, this implies that D = o. Thus Vk(D) is isomorphic to the subspace of Wm formed the vectors

(Dx1,..., Dxm).

Consider now the vector space Ws and let τ be the mapping 76 (τ(α1,...,αm) = (α1,...,αm)M of Wm into Ws. Put

m ∂ fi β j = α j ; i = 1,..., s = ∂x j Xj 1 so that (β1,...,βs) = (α1,...,αm)M. The rank of the mapping τ is clearly y equal to the rank t of the matrix M. It is the dimension of the image by τ of Wm into Ws. The kernel of the mapping τ is the set of (α1,...,αm)with

βi = o; i = 1,..., s. 2. Derivations 65

which, by theorem 6 is clearly isomorphic to the subspace of Wm formed by vectors (Dx ,..., Dx ), D V (D). This, by previous considera- 1 m ∈ K tions, proves the theorem. We shall now prove the

Theorem 9. With the same notations as before, there exist ℓ elements, say x1,..., x1 of dimension n over k such that K/k(x1,..., x1) is a separably algebraic extension.

Proof. Since the matrix M has rank t, there exists a submatrix of M of t rows and which is non-singular. Choose notation in such a way, that this matrix is = + ∂ f j i m t 1,..., m P = , − ∂xi j = s t + 1,..., s ! − Note that t Min (s, m). Let L = k(x ,..., x ). Then ≤ 1 1 K = L(x1+p,..., xm). Let D be a derivation of K over L. Then since f j(x1,..., xm) = o, we must have, by theorem 6

m ∂ f j Dxi = o. = ∂xi Xi 1 But since D is zero on L, 77

Dxi = o, i = 1,..., l.

Thus m ∂ f j Dxi = o. = + ∂xi iXl 1 This means that Dxl+1 o P . = .  .   .  Dx  o  m        But since P , o, it follows that Dx l+1 = o ..., Dxm = o which | |     shows that D = o 66 3. Algebraic function ﬁelds

But K = L(xl+1,..., xm) is ﬁnitely generated over L. Using theorem 7, it follows that K/L is algebraic and separable. We get incidentally n 1. ≤ We now prove the important

Theorem 10. Let K = k(x1,..., xm) be of dimension n. Then K is separably generated over k, if and only if dim VK(D) = n. In that case there exists, among x1,..., xm, a separating base of n elements.

Proof. If VK(D) has dimension n then theorem 9 shows that there exist n elements x1,..., xn among x1,..., xm such K/k(x1,..., xn) is separably algebraic.

Suppose now that K/k is separably generated. Let y1,..., yn be a separating base so that K/k(y1,..., yn) is separably algebraic. k(y1, ..., yn) has n linearly independent derivations D1,..., Dn over k deﬁned by o, if i , j D y = i j = 1, if i j 78   Since K/k(y1,..., yn) is separably algebraic and ﬁnite, it follows that each of D1,..., Dn has a unique extension D¯ 1,..., D¯ n to K. Now D¯ 1,..., D¯ n are linearly independent over K. For, if

a D¯ = o, a K, i i i ∈ Xi then

aiD¯ i(y j) = o for all j. Xi Hence a j = o for j = 1,..., n. Let now D be a derivation of K/k and let Dyi = ai. Put D¯ = D a D¯ . − i i Xi 3. Rational function ﬁelds 67

Then Dy¯ i = o for all i. Therefore since K/k(y1,..., yn) is separable, D¯ = o.This proves that

dim VK(D) = n. and our theorem is completely established. Let K = k(x1,..., xn, y) where k is of characteristic p , o and k is an imperfect ﬁeld. Let y be algebraic over k and be a root of

zp t − t k. Then K/k is an inseparably generated extension and ∈

dim VK(D) = n + 1 where n is the dimension of K/k.

3 Rational function fields 79 Let us now consider the field K = k(x) of rational functions of one f (x) variable. Let y be any element of K. Hence y = where f and g are g(x) polynomials in x over k. Also K is the quotient field of the ring L = k[x] of polynomials in x. Assume that ( f (x), g(x)) = 1, that is that they have no factor in common. Let n be defined by

n = max(deg f (x), deg g(x)).

If n = o, then clearly f k, g k and so y k. Let us assume that ∈ ∈ ∈ n , o so that at least at least one of f and g is a non-constant polynomial. n is called the degree of y. Let F = k(y) be the ﬁeld generated over k by y. Then x satisﬁes over F the polynomial

ϕ(z) = f (z) yg(z). − 68 3. Algebraic function ﬁelds

ϕ(z) is not a constant polynomial over k(y). For, let

1 i f (z) = aiz i=o  X  ai, bi k. m  ∈ = i g(z) biz  i=o  X   Then n = max(1, m). The coeﬃcient of zn in ϕ(z) is

a1 if 1 > m = a1 yb1 if 1 m  −  ybm if 1 < m −  In every case, it follows that since y is not in k,ϕ(z) is a non-constant polynomial. Since ϕ(z) has degree n in z, it follows that

(K : F) n. ≤ 80 We assert that ϕ(z) is irreducible over F. For, if it is reducible over F[z], then since F = k(y), it will be reducible over k[y, z]. So let ϕ(z) = ψ1(y, z) ψ2(y, z) in k[y, z]. Since ϕ(z) is linear in y it follows that one of ψ1or ψ2 has to be independent of y. But then since ( f (z), g(z)) = 1, ϕ(z) is a primitive polynomial in y over k[z]. Therefore ϕ(z) is irreducible. This means that (K : F) = n It proves, in particular that y is transcendental over k. Hence

Theorem 11. k is algebraically closed in k(x).

We can extend it to the case where K = k(x1,..., xn) is a purely transcendental extension of dimension n. We use induction on n. Theo- rem 11 proves that n = 1, k is algebraically closed in k(x). Let, for n 1 − instead of n, instead of n, it be proved that k is algebraically closed in the purely transcendental extension k(x1,..., xn 1). Let K = k(x1,..., xn) be of dimension n. − 3. Rational function ﬁelds 69

Let z in K be algebraic over k. Then z K = K (x ), K = ∈ 1 n 1 k(x1,..., xn 1). Therefore by theorem 11 since z is algebraic over K1, z K . By induction− hypothesis, z k. ∈ 1 ∈ Thus

Corollary. If K = k(x1,..., xn) has dimension n over k, then k is algebraically closed in K It is easy to extend this to the case where K is a purely transcendental extension of any transcendence degree. Since K = k(x), we call x a generator of K over k. Let y also be a 81 generator so that K = k(y). Then (k(x) : k(y)) = 1 which shows by our considerations leading to theorem 11 that a(x) y = b(x) where a(x) and b(x) are coprime and have at most the degree 1 in x. Thus λx + µ y = νx + ρ where λ, µ, ν, ρ are in k and since y is transcendental over k, λρ µν , 0. − An automorphism of K which is identity on k, is uniquely fixed by its effect on x. If it takes x into y then x and y are related as above. If x and y are related as above, then the mapping which assigns to x the element y is an automorphism. If we consider the group of two rowed non-singular matrices, with elements in k, then each matrix gives rise to an automorphism of K/k. Obviously two matrices σ and τ give rise to the same automorphism if and only if σ = λτ for some λ , o in k. Hence Theorem 12. The group of automorphisms of K = k(x) over k is isomorphic to the factor group of the group of two rowed matrices over k modulo the group of matrices λE, λ , 0 k and E is the unit matrix of ∈ order 2. 70 3. Algebraic function fields

We shall call this group P2. 82 From theorem 11, it follows that if L is an intermediary ﬁeld between K and k then L is transcendental over k. But much more is true as in shown by the following theorem of Luroth.

Theorem 13. If K = k(x) is a simple transcendental extension of k and k L K,L = K(ω) for some ω K. ⊂ ⊂ ∈ Proof. We shall assume that L , k so that L is transcendental over k and contains an element t, transcendental over k andy by considerations leading to theorem 11, we have K/k(t) is ﬁnite algebraic. Since L k(t), ⊃ it follows that (K : L) < ∞ Let x satisfy over L the irreducible polynomial

n n 1 f (z) = z + a z − + + a 1 ··· n where a ,..., a L and n = (K : L). At least one a is not in k since x 1 n ∈ i is transcendental over k. The ai′ s are rational functions of x. Sowe may write b (x) f (z) = f (x, z) = b (x)zn + b (x)zn 1 + + b (x) where 0 0 1 − ··· n b0(x),..., bn(x) are polynomials in x and f (x, z) is a primitive polynomial in z over k[x]. Let m be the maximum of the degrees of b0(x), ... bn(x). Let ai be not in k. Then

bi(x) ai = b0(x) so that degree a m. Since a L and L k(a ), it follows that i ≤ i ∈ ⊃ i n m ≤ bi(x) Let us write w instead of ai. Then w = . b0(x) 83 Let w = h(x)/g(x) where (h(x), g(x)) = 1. Then L k(w). ⊃ Also x satisﬁes over k(w) the polynomial

h(z) wg(z) − 3. Rational function ﬁelds 71 so that since f (z) is irreducible, it follows that f (z) divides h(z) wg(z), − which is a polynomial of degree m in z. Let us therefore write ≤ f (x, z) ϕ(x, z)c (x) h(z) wg(z) = 1 − bo(x) co(x) where ϕ(x, z) is a primitive polynomial in z over k[x]. We therefore get h(x) on substituting the w = , g(x) g(x)c (x) h(z)g(x) g(z)h(x) = 1 . f (x, z)ϕ(x, z). − bo(x)co(x) The left hand side being a polynomial in x and z, f and ϕ being primitive polynomials in z over k[x], it follows that h(z)g(x) g(z)h(x) = f (x, z)ϕ (x, z) − 1 where ϕ (x, z) k[x, z]. We now compare degrees in x and z on both 1 ∈ sides of the above identity. On the right hand side the degree in x is m ≥ since one of bo(x),..., bn(x) has degree m. Therefore the left side has degree in x m. But the degree in x equals degree of w m. Thus ≥ ≤ degree of w = m. Since the left side is symmetrical in z and x,it follows that it has degree m in z. Therefore ϕ1 has to be independent of x. Hence h(z) wg(z) = f (z)ϕ(z), ϕ(z) being independent of x. This − can happen only if ϕ(z) is a constant. This proves that n = m.

Now (K : k(w)) = m = (K : L) and L k(w). Thus 84 ⊃ L = k(w) and our theorem is proved. The analogue of Luroth’s theorem for K = k(x1,..., xn) is not known for n > 1. Let K = k(x) and let G be a finite granite group of automorphism of K/k. If L is the fixed field of G, then K/L is a finite extension of degree equal to order of G. By Luroth’s theorem L = k(y) for some y. Thus degree y = order of G. 72 3. Algebraic function fields

For instance, let G be the ﬁnite group of automorphisms of K = k(x) deﬁned by

1 1 1 x x x, x 1 x, x , x 1 , x , x → → − → x → − x → 1 x → x 1 − − This is a group of order 6 and the ﬁxed ﬁeld will be k(y) where k(y) consists of all rational functions f (x) of x with

1 1 1 x f (x) = f (1 x) = f ( ) = f (1 ) = f ( ) = f ( ). − x − x 1 x 1 x − − We have only to ﬁnd a rational function of degree 6 which satisﬁes the above conditions. The function

(x2 x + 1)3 f (x) = − x2(x 1)2 − satisﬁes the above conditions and so y = f (x).

Theorem 14. If G is any ﬁnite subgroup of P2 of linear transformations

ax + b x → cx + d 85 a, b, c, d k, ad bc , 0 then there exists a rational function f (x) such ∈ − that every function ϕ(x) which is invariant under G is a rational function of f (x). f (x) is uniquely determined up to a linear transformation

λ f (x) + µ V f (x) + ρ λ, µ, νρ k, λρ µν , 0. ∈ −

We now consider the case of a rational function field K = k(x1, ... xn) of n variables. Let G be a finite group of automorphisms of K which are trivial on k and let L be the fixed field of G. Clearly L has transcendence degree n over k. It is not known, except in simple cases, whether L is a purely transcendental extension of k or not. We shall, however, consider the the case where G is the symmetric group on n 3. Rational function fields 73 symbols. So G S . Let G operate on K in the following manner. If σ ≃ n is an element of S n, then σ is a permutation 1, 2, 3,..., n σ = . σ ,σ ,σ ,...,σ 1 2 3 n! We define σ on K by = σxi xσi ; 1,..., n.

We then obtain a faithful representation of S n on K and we denote this group again by S n. An element of k which is ﬁxed under S n and which therefore is in L is called a symmetric function of x1,..., xn. Ob- viously (K : L) = n! and the galois group of K/L is S n. 86 Consider the polynomial f (z) = (z x ) ... (z x ). − 1 − n Since every permutation in S leaves f (z) ﬁxed, it follows that f (z) n ∈ L[z]. Let us write

n n 1 n 2 n f (z) = z s z − + s z − + (1) s − 1 2 −··· n where s = x x . i t1 ··· ti 1 t1

Theorem 15. Every rational symmetric function of x1,..., xn is a rational function over k of the elementary symmetric functions s1,..., sn. Incidentally since L/k has dimension n, the elementary symmetric functions s1,..., sn are algebraically independent over k. Chapter 4

Norm and Trace

1 Norm and trace 87 Let K/k be a ﬁnite extension and let ω1,...,ωn be a base of K/k so that every ω K may be written ∈

ω = aiωi Xi a k. By means of the regular representation i ∈

ω Aω → where Aω = (a ) is an n rowed square matrix with i j −

ωωi = ai jω j i = 1,..., n Xj the ﬁeld K becomes isomorphic to the subalgebra formed by Aω in the algebra mn(k) of n rowed matrices over k. We denote by NK/kω, S K/Kω the norm and trace respectively of ω K over k and they are deﬁned by ∈

N / ω = Aω K k | | S K/kω = trace Aω.

75 76 4. Norm and Trace

Deﬁned as such, it follows that

NK/kωω′ = NK/kω. NK/kω′

S K/k(ω + ω′) = S K/kω + S K/kω′

for ω, ω′ K. Obviously ω NK/kω is a homomorphism of K∗ into ∈ → + K∗ and similarly ω S K/kω is a homomorphism of K , the additive + → group, into k .

Let ω1′ ,...,ωn′ be any other basis of K/k. Then

ω1′ ω1 . = P .  .   .      ωn′  ωn         88 where P is a non-singular matrix  in mn(k). Since

ω1 ω1 ω . = A .  .  ω  .      ωn ωn         it follows that     ω1′ ω1′ ω . = PA P 1 .  .  ω′ −  .  ω  ω   n′   n′      which shows that by means of the new basis the matrix associated to ω is Bω where Bω = PAωP 1 − and then we have

Bω = Aω | | | | Trace Bω = Trace Aω.

This shows that NK/kω and S K/kω are invariantly deﬁned and do not depend on a basis of K/k. 1. Norm and trace 77

We write f / (x) = xE Aω K k | − | and call it the characteristic polynomial of ω. Obviously fK/k(0) = n ( 1) Aω so that − | | n n N / ω = ( 1) f / (0) = ( 1) a . (1) K k − K k − n We also see easily that

S / ω = a (2) K k − 1 where n n 1 f / (x) = x + a x − + + a K k 1 ··· n a ,..., a k. 1 n ∈ Let k L K be a tower of ﬁnite extensions. Let (K : L) = m ⊂ ⊂ and let Ω1,..., Ωm be a basis of K/L. Similarly let (L : k) = n and let 89 ω1,...,ωn be a base of L/k. Then (ω1Ω1,...,ωnΩm) is a base of K/k. Let ω L and consider the matrix of ω by the regular representation of ∈ K/k in terms of the base (ω1Ω1,...,ωnΩm). Call it A¯ω. Then it is trivial to see that

Aω 0 ¯ = Aω ········· ·········  0 A   ω    a matrix of mn rows and columns. Therefore (K:L) N / ω = A¯ω (N / ω) K k | | K k S K/kω = (K : L)S K/kω

Also the characteristic polynomials of ω as belonging to L and to K respectively are fL/k(x) and fK/k(x) and they are related by (K:L) fK/k(x) = ( fL/k(x)) , (3)

In particular let L = k(ω). Then fL/k(x) is the minimum polynomial of ω. We, therefore, have the 78 4. Norm and Trace

Theorem 1. If K/k is a ﬁnite extension and ω K, ϕ(x) its minimum ∈ polynomial over k and f (x) its characteristic polynomial, then

f (x) = (ϕ(x))r

where r = (K : k(ω)). From our formulae above, it follows that we can compute the norm and trace of ω in K from a knowledge of its minimum polynomial. 90 Let now K/k be a ﬁnite extension and ω an element of K. Let [K : k(ω)] = m′, [k(ω) : k] = n′ be the degrees of separability of K over k(ω) and k(ω) over k respectively. Then K/k has m′n′ distinct = k-isomorphisms in an algebraic closure Ω of k. Let σ , i 1,...,n′ , be i j j=1,...,m′ these isomorphisms and let notation be so chosen that σi1,...,σim, have ﬀ n o the same the same e ect on k(ω). Then we may take σ11,σ12,...,σn′1 as a complete system of distinct isomorphisms of k(ω)/k in Ω. By our considerations above fk(ω)/k(x) is the polynomial of ω as well as the characteristic polynomial of ω in k(ω)/k. If fK/k(x) is the characteristic polynomial of ω in k, then

(K:k(ω)) fK/k(x) = ( fk(ω)/k(x)) (4)

Now, because of the properties of the isomorphisms σi j

n′ m′ n′ m′ (x σi jω) = (x σi1ω) . (5) = = − = − Yi 1 Yj 1 Yi 1 But since fk(ω)/k(x) is the minimum polynomial of ω, we have

k(ω):k n′ { } f ω / (x) = (x σ ω) k( ) k − i1  i=1  Y  where k(ω) : k , as usual, denotes the degree of inseparability of k(ω) { }   over k. Using (5) we get

k:k m′ K:k { } n′ { } (x σi j(ω) = (x σi1ω) .  −  −  i, j   i=1  Y  Y          1. Norm and trace 79

But K : k = K : k(ω) k(ω) : k so that { } { }·{ } K:k { } m′ K:k(ω) (x σi jω) = fk(ω)/k(x) { }  −   i, j  Y    which proves that  91

K:k { } fK/k(x) = ((x σω) (6) σ − ( ) Q where σ runs through all the distinct isomorphisms of k(ω) in Ω. Using (1) and (2) we get K:k { } σ NK/kω = ω (7)  σ  Y  ωσ is a conjugate of ω. Similarly     σ S K/kω = K : k ω . (8) { } σ X If K/k is inseparable, then K : k = pt, t 1 so that for every ω K { } ≥ ∈

S K/kω = o.

On the other hand, suppose K/k is ﬁnite and separable. Let σ1, ...,σn be all the distinct isomorphisms of K/k in Ω, an algebraic closure of K. Then n = (K : k) and since σ1,...,σn are independent k-linear functions of K/k in Ω, it follows that

σ = σ + + σ 1 ··· n is a non - trivial k linear function of K/k in Ω. Therefore there exists a − ω K such that σω , 0. But by formula (8), ∈

σ1 σn σω = ω + ... + ω = S K/kω so that we have the 80 4. Norm and Trace

Theorem 2. A ﬁnite extension K/k is separable, if and only if there exist in K an element whose trace over k in not zero.

In case k has characteristic zero, or k has characteristic p ∤ n = (K : 92 k), the unit element 1 in k has trace , o. In order to obtain an element ω in K with S K/kω , 0, in every case we proceed thus: Let K/k be separable and K = k(α) for an element α. Let ϕ(x) be its irreducible polynomial over k and

ϕ(x) = (x α ) (x α ). − 1 ··· − n It follows then that

n 1 n 1 x α − 1 − = i . ϕ(x) ϕ′(αi) x αi Xi − n 1 Comparing coeﬃcients of x − on both sides we get

n 1 α − i = 1. ϕ′(αi) Xi n 1 α − If we put ω = and observe that ϕ′(α) K, we get ϕ′(α) ∈

S K/Kω = 1.

Using formula (6), it follows that if k L K is a tower of ﬁnite ⊂ ⊂ extensions and ω K, then ∈

NK/kω = NL/k(NK/Lω)

S K/kω = S L/k(S K/Lω)

We now give a simple application of formula (7) to finite fields. Let k be a finite field of q = pa elements so that p is the characteristic of k. Let K be a finite extension of k so that (K : k) = n. Then K has 2. Discriminant 81

qn elements. The galois group of K/k is cyclic of order n. Let σ be the Frobenius automorphism. Then for ω K, ∈ ωσ = ωq.

93 The norm of ω is

2 n 1 σ σ σ − N / ω = ω ω ω ω K k · · ··· qn 1 − = q 1 ω − .

Since K has qn 1 elements, ∗ − qn 1 α − = 1

for all α K . Since K is a cyclic group, the number of elements in K ∈ ∗ ∗ ∗ with qn 1 − q 1 = α − 1 is precisely qn 1/q 1, since q 1 divides qn 1. − − − − Now ω N / ω is a homomorphism of K into k and the kernel → K k ∗ ∗ of the homomorphism is the set of ω in K∗ with

qn 1 − = = q 1 1 NK/kω ω − .

By the ﬁrst homomorphism theorem we have, since k has only q 1 ∗ − elements, the

Theorem 3. If K/k is finite and k is a finite field, then every non-zero element of k is the norm of exactly (K∗ : k∗) elements of K.

It is clear that this theorem is not in general true if k is an inﬁnite ﬁeld. 82 4. Norm and Trace

2 Discriminant

Let K/k be a ﬁnite extension and ω1,...,ωn a basis of K/k. Suppose σ 94 is a k-linear map of K into k, that is

σ(ω) k ∈ σ(ω + ω′) = σω + σω′ σ(λ) = λσ(ω) where ω, ω K, λ K. Let Mσ denote the matrix ′ ∈ ∈ σ M = (σ(ωiω j)) of n rows and columns. We denote by DK/kσ(ω1,...,ωn) its determinant and call it the σ- discriminant of the basis ω1,...,ωn of K/k. If ω1′ ,...,ωn′ is another basis, then

ω1′ ω1 . = P .  .   .      ωn′  ωn         where P is an n rowed non-singular  matrix  with elements in k. If P = (pi j) then = ωi′ pi j ω j Xj so that = σ(ωa′ ωb′ ) pai pb jσ(ωiω j) Xi, j which proves that

σ 2 σ D (ω′ ,...,ω′ ) = P D (ω ,...,ω ). K/k 1 n | | K/k 1 n σ = Therefore DK/k 0 if it is zero for some basis. We now prove 2. Discriminant 83

Theorem 4. If ω1,...,ωn is a basis of K/k and σ a k-linear map of K into k, then σ = DK/k(ω1,...,ωn) 0 95 if and only if σ is the zero linear mapping. Proof. If σ is the zero linear map, that is, one that assigns to every σ = σ = element ω in K, the zero element, then DK/K 0. Now let DK/k 0. This means that the matrix Mσ with elements in k has determinant zero. Therefore there exist a1,..., an in k, not all zero, such that

a1 0 Mσ . = . .  .   . a  0  n       This means that     n σ(ωiω j)a j = 0; 1 = 1,..., n. = Xj 1 If we put z = a jω j. then we have j P σ(ωiz) = 0, i = 1,..., n.

Let ω be any element in K. Since a1,..., an are not all zero, z , 0 and so put ω = b ω + + b ω , b k. z 1 1 ··· n n i ∈ ω Then σ(ω) = σ z = biσ(ωiz) = o. This proves that σ is the z · i trivial map. P The mapping ω S / ω is also a k-linear map of K into k. For a → K K basis ω1,...,ωn of K/k we call

D / (ω ,...ω ) = (S / (ω ω )) K k 1 n | k K i j | the discriminant of the basis ω1,...ωn. Using theorem 2 and theorem 4, we get 84 4. Norm and Trace

Theorem 5. Discriminant of a base of K/k is not zero if and only if K/k is separable.

96 Let K/k be ﬁnite separable. Let σ1,...,σn be the distinct k-isomorphisms of K over k in Ω. Then

σi S K/kω = ω . Xi Therefore

σ1 σ2 σn 2 ω1 , ω1 , ..., ω1 = . DK/k(ω1,...,ωn) . . (9)

ωσ1 , ...... , ωσn 1 1

Since K/k is ﬁnite separable, K = k(ω) for some ω. Also 1,ω,ω2, n 1 ...,ω − form a base of K/k and we have 1, ..., 1 σ1 σ1 n 1 ω , ..., ω D(1,ω,...,ω − ) = ...... n 1 σ1 n 1 σn (ω − ) , ..., (ω − )

n 1 Also, D(1,ω,...,ω − ) D(ω 1,...,ωn) is the square of an element of k. The determinant 1, ..., 1 ωσ1 , ..., ωσn

...... n 1 σn n 1 σn (ω − ) , ..., (ω − )

n 1 is the called Van-der-Monde determinant. We call D(1,ω,...,ω − ) is the discriminant of ω and denote it by DK/k(ω). Let f (x) be the minimum polynomial of ω. Then f (x) = (x − ωσ1 ) ... (x ωσn ). Since f (ω) , 0 , we have − ′ σ1 σ2 σ1 σ3 σ1 σn f ′(ω) = (ω ω )(ω ω ).....(ω ω ) − − − and is an element of K. We call it the diﬀerent of ω and denote it dK/k(ω). Also the Vander - monde determinant shows that 2. Discriminant 85

n(n 1)2 D / (ω) = ( 1) N / (d / ω) K k − − K k K k

97 Suppose K/k is a ﬁnite galois extension. Let σ1,...,σn be the distinct automorphisms of K/k. We shall now prove

Theorem 6. If k contains suﬃciently many elements, then there exists in K an element ω such that ωσ1 ,...,ωσn form a basis of K over k.

Proof. It ω K such that ∈ D(ωσ1 ,...,ωσn ) , o

then ωσ1 ,...,ωσn form a base of K/k. For,if

a ωσi = o, a ,..., a k i 1 n ∈ Xi not all zero, then since σ1,...,σn form a group

σ jσi aiω = o, j = 1,..., n. Xi σ σ Therefore from the expression (9) for DK/k(ω 1 ,...,ω n ), it follows that a1 o . . S (ωσi ωσ j ) . = . K/k  .   .     an o     σ1 σn =     or that D(ω ,...,ω ) o which is a contradiction.   We have therefore to ﬁnd an ω with this property. Put

ωσi = x ωσi + + x ωσi , i = 1, 2,... 1 1 ··· n n

where x1,..., xn are indeterminates and ω1,...,ωn is a basis of K/k Then

σi σ j σ1σi σ1σ j σnσi σnσ j S / (ω ω ) = ω ω , + + ω ω K k ··· σ = σi j S K/k(ωa ωb )xa xb. Xa,b 86 4. Norm and Trace

σ σ Then DK/k(ω 1 ,...,ω n ) defined as the determinant of the matrix 98 σ σ (S K/k(ω i ω j )) is a polynomial in x1,..., xn with coefficients in k. σ σ In order to prove that DK/k(ω 1 ,...,ω n ) is non-zero polynomial, notice that by definition,

σ1 σ σ σ ω 1 1 1 x1 ω1 , ω2 , ..., ωn . = ......  .   .  σ  σn σn σn  ω n  ω1 , ω2 , ..., ωn  x       n       σ1   σi    so that if ω = 1 and ω = o for i > 1, then x1,..., xn are not all σ zero and for this set of values of x1,..., xn, the polynomial DK/k(ω 1 , ...,ωσn ) has a value , o, as can be see from the fact that

2 ωσ1 , ..., ωσn D (ωσ1 ,...,ωσn ) = ...... K/k ωσnσi , ..., ωσnσn

Therefore if k has sufficiently many elements, there exist values in k σ σ of x1,..., xn, not all zero, such that DK/k(ω 1 ,...,ω n ) , o. This proves the theorem In particular, if k is an infinite field, there exists a base of K/k consisting of an element and its conjugates. Such a base is said to be a normal base. The theorem is also true if k is a finite field. Chapter 5

Composite extensions

1 Kronecker product of Vector spaces 99 Let V and V be two vector spaces over a ﬁeld k and V V their 1 2 1 × 2 cartesian product. If W is any vector space over k, a bilinear function f (x, y) on V V is, by deﬁnition, a function on V V into W such 1 × 2 1 × 2 that for every x V the mapping λ : y f (x, y) is a linear function ∈ 1 x → on V into W and for every y V the function µ : x f (x, y) is a 2 ∈ 2 y → linear function on V1 to W. A Vector space T over k is said to be a Kronecker product or tensor product of V and V over k, if there exists a bilinear function θ on V V 1 2 1× 2 into T such that 1) T is generated by θ(V V ) 1 × 2 2) if V3 is any vector space over k, then for every bilinear function ϕ on V V into V there exists a linear function σ on T into V such that 1 × 2 3 3 σθ = ϕ. This is shown by the commutative diagram

ϕ / V1 V2 /; V3 × MMM vv; MMM vv θ MM& vv σ & T We shall now prove the

87 88 5. Composite extensions

Theorem 1. For any two vector spaces V1 and V2 over k there exists one and upto k-isomorphism only one Kronecker product T of V1 and V2 over k.

100 Proof. The uniqueness is easy to establish. For, let T1 and T2 be two vector spaces satisfying the conditions 1) and 2). Let T1 be generated by θ (V V ) and T by θ (V V ). Let σ be the linear map of T 1 1 × 2 2 2 1 × 2 1 into T such that σ θ = θ and τ the linear map of T into T such that 2 · 1 2 2 1 τ θ = θ . Since θ (V V ) generates T we see that τ σ is identity · 2 1 1 1 × 2 1 · on T . Similarly σ τ is identity on T . Thus σ and τ are isomorphisms 1 · 2 onto.

We now prove the existence of the space T. Let V be the vector space formed by ﬁnite linear combinations

axy(x, y) X a k and (x, y) V V . Every bilinear function f on V V into V xy ∈ ∈ 1 × 2 1 × 2 3 can be extended into a linear function f¯ of V into V3 by the prescription

f¯ axy(x, y) = ax,y f (x, y)

Let W be the subspace X of V generated X by elements of the type

(x + x1, y) (x, y) (x1, y) − − (x, y + y1) (x, y) (x, y1) − − (ax, y) a(x, y) − (x, by) b(x, y) − where x, x1 V ; y, y1 V and a, b k. W is independent of V . ∈ 1 ∈ 2 ∈ 3 Also if f is a bilinear function on V V , its extension f¯ on V vanishes 1 × 2 on W. Furthermore if f is any linear function on V vanishing on W, 101 its restriction to V V is a bilinear function. It is then clear that the 1 × 2 space of bilinear functions on V V is isomorphic to the space of linear 1 × 2 functions on V which vanish on W. If σ is any linear function on V/W into V and θ the natural homomorphism of V on V/W then σ θ is a 3 · linear function on V vanishing on W. Also σ σ θ is an isomorphism → · 1. Kronecker product of Vector spaces 89

of the space of linear functions on V/W into V3 on the space of linear functions on V vanishing on W. θ is clearly a bilinear function on V V 1× 2 into T = V/W and furthermore, by deﬁnition of V, V/W is generated by θ(V V ). Thus T is the required space. 1 × 2 By taking k itself as a vector space over k, we have Theorem 2. The space of bilinear functions on V V into k is isomor- 1 × 2 phic to the dual of the Kronecker product T. We denote by V x V the kronecker product space. When there is 1 k 2 no doubt about the ﬁeld over which the kronecker product is taken we will simply write V x V . 1 2 If T = V x V we denote by x x y the element in T which corre- 1 2 sponds to (x, y) by the bilinear function θ on V V into T. Since 1 × 2 θ(V V ) generates T, every element of T is of the form 1 × 2 a , (x x y) a k. x y xy ∈ Clearly X x x o = o x y = o x o

(x + x1) x y = x x y + x1 x y   1 1  x x (y + y ) = x x y + x x y   ax x y = a(x x y)   =  x x by b(x x y)  with obvious notations. 102  From our considerations it follows that in order to deﬁne a linear function on V x V it is enough to deﬁne it on elements of the type 1 2 x x y. Also for every such linear function, there is a bilinear function on

V V . 1 × 2 It is easy to see that the mapping x x y y x x of V x V to V x V → 1 2 2 1 is an isomorphism onto.

Suppose now that V1∗ and V2∗ are duals of V1 and V2 respectively over k. If σ V and τ V and we deﬁne for (x, y) V V the function ∈ 1∗ ∈ 2∗ ∈ 1 × 2 σ.τ(x, y) = σx.τy, then σ.τ is a bilinear function on V V into k. We shall now prove the 1 × 2 90 5. Composite extensions

Theorem 3. If xα is a base of V over k and yβ a base of V over k, { } 1 { } 2 then xα x yβ is a base of V x V over k. { } 1 2

Proof. We first prove that xα x yβ are linearly independent over k. For { } all aαβ(xα x yβ) = o for aαβ k, aαβ = o for all but a finite number of ∈ α, β. By the method of constructing the tensor product, it follows that P aαβ(xα, yβ) is an element of W. Let σ and τ be elements of V1∗ and V2∗ defined respectively by P

σ(xα) = 1 if α = γ = o otherwise

τ(yβ) = 1 if β = δ = o otherwise

103

Then σ τ is a bilinear function on V V and hence vanishes on · 1 × 2 W. Thus σ.τ aαβ(xα x yβ) = o.

X But the left side equals aγδ. Thus all the coeﬃcients aαβ vanish. Next any element of V x V is a linear combination of elements of 1 2 the type x x y, x V , y V . But then x = aα xα and y = bβyβ so ∈ 1 ∈ 2 that P P x x y = aα xα x bβyβ

which equals aαbβ(xα x yβ).X Our theorem X is proved.

We have incidentally the P

Corollary. If V1 and V2 are ﬁnite dimensional over k then

dim V x V = dim V dim V . 1 2 1 · 2

Also since the dual of V1∗ is isomorphic in a natural manner with V1 when dim V is ﬁnite, we get · 1 1. Kronecker product of Vector spaces 91

Corollary . If V and V are ﬁnite dimensional over k then V x V is 1 2 1 k 2 isomorphic to the dual of the space of bilinear functions on V V into 1 × 2 k.

Let now A1 and A2 be two associative algebras over a field k. We can form the Kronecker product A = A x A of the vector spaces A 1 k 2 1 and A2 over k. We shall now introduce a multiplication into A so as to make it into an associative algebra. In order to do this, observe that the multiplication defined has to be 104 a bilinear function on A A into A. Since A is generated by elements of × the type x x y, it is enough to define this bilinear function on elements of the type (z, z1) in A A where z = x x y and z1 = x1 x y1. Put × f (z, z1) = z z1 = x x1 x y y1. · · · Now since (x, y) xx1 x yy1 is a bilinear function on A A into → 1 × 2 A, by our previous considerations z z z1 is a linear function on A into → · A. Similarly z1 z z1 is a linear function on A. This proves that f is → · bilinear and that the multiplication so defined distributes addition. That the multiplication is associative is trivial to see. A is called the Kronecker product algebra. We obtain some very simple consequences from the definition. α) If e1 and e2 are respectively the unit elements of the algebras A1 and A then e x e is the unit element of A x A . 2 1 2 1 2 For, since A = A x A is generated by elements x x y, it is enough 1 2 to verify (x x y)(e x e ) = (e x e )(x x y). But this is trivial. 1 2 1 2 β) If A1 has x1,..., xm as a base over k and A2 has y2,..., yn as a base over k then (x x y ) is a base of A x A over k. Furthermore if the i j 1 2 multiplication tables for the bases are = (t) xi x j ai j xt t X = (t′) yiy j bi j yt′ Xt′ then 105 (λ) (µ) (x x y )(x x y ) = a b (xλ x yµ) p q r s pr qs Xλ,µ 92 5. Composite extensions

γ) If A1 and A2 have unit elements e1 and e2 respectively then the mappings

x x x e → 2 y e x y → 1

are isomorphisms of A1 and A2 into A. Thus A contains subalgebras isomorphic to A1 and A2. An important special case is the one where one of the algebras is a ﬁeld. Let A be an algebra over k and K an extension ﬁeld of k. Let A have unit element e1 and K the unit element e2. Form the Kronecker product A x K over k. Then A x K contains subalgebras A and K isomorphic to 1 1 A and K respectively. For any x x t in A x K we have

x x t = x x e .e x t = e x t.x x e 2 1 1 2 so that A and K commute. Also every element of A x K is of the form 1 1 aαβ(xα x tβ) where xα is a base of A over k and tβ a base of K over { } { } k. But this expression can be written P

aαβ(e1 x tβ) (xα x e2). α β X X This shows that A x K is an algebra over K with the base xα x e . 1 { 2} If we identify A with A and K with K then A x K can be considered as 1 1 an algebra over K, a basis of A over k serving as a base of A x K over

K. A x K is then called the algebra got from A by extending the ground

106 ﬁeld k to K. We shall denote it by AK. It is clear that AK is commutative if and only if A is a commutative algebra.

Note. Even if A is a field over k, AK need not be a field over K. Let Γ be the rational number field and Γo = Γ( √d) the quadratic field over Γ. Let Γ be the real number field and consider the Kronecker product A = Γ x Γ¯ over Γ. Let e , e be a basis of Γ over Γ with the o 1 2 o multiplication table, 2 = = = 2 = e1 e1, e1e2 e2e1 e2, e2 de1. 2. Composite fields 93

The elements of Γo are of the form ae1 +be2, a, b in Γ. The elements of A are of the form ae + be with a, b Γ¯. 1 2 ∈ Let ﬁrst d > o. Then Γ x Γ¯ is not an integrity domain. For, o ( √de + e )( √de e ) = o. 1 2 1 − 2 It can however be seen that A is then the direct sum of the two ﬁelds λΓ¯ and µΓ¯ where

1 e2 1 e2 λ = (e1 + ), µ = (e1 ). 2 √d 2 − √d 2 2 Let d < o. Then A is a ﬁeld. For if ae1 + be2 , o, then a db , o. a b − Put f = , g = − . Then a2 db2 a2 db2 − − (ae1 + be2)( f e1 + ge2) = e1.

2 Composite ﬁelds

Let K1 and K2 be two extension fields of k. Suppose K1 and K2 are both contained in an extension field Ω of k. Then the composite of K1 and K2 is the field generated over k by K1 and K2. In general, given 107 two fields K1 and K2 which are extensions of k, there does not exist an extension field Ω containing both. Suppose, however, there is a field Ω σ τ /k which contains k-isomorphic images K1 , K2 of K1 and K2, then a composite of K and K is defined to be the fields k(Kσ Kτ). A 1 2 1 ∪ 2 composite extension of K1 and K2 is therefore given by a triplet (Ω, σ, τ) consisting of 1) and extension field Ω of k and 2) isomorphisms σ, τ of K1 and K2 respectively in Ω which are identity on k. The composite extension is then k(Kσ Kτ). We wish to study these various composites 1 ∪ 2 of K1 and K2. If Ω′ is another extension of k and σ′, τ′ two k-isomorphisms of τ K and K respectively in Ω then k(Kσ′ K ′ ) is another composite 1 2 ′ 1 ∪ 2 extension. We say that these two composite extensions are equivalent if τ τ there exists a k-isomorphism µ of k(Kσ K ) on k(Kσ′ K ′ ) such that 1 ∪ 2 1 ∪ 2 µσ = σ′,µτ = τ′. 94 5. Composite extensions

Obviously this is an equivalence relation and we can talk of a class of composite extensions. σ τ σ τ If k(K1 K2) is a composite extension we denote by R(K1 K2) ∪ σ τ Ω ∪ the ring generated over k by K1 and K2 in . This ring is, in general, different from k(Kσ Kτ). 1 ∪ 2 Let now K¯ = K x K be the Kronecker product of K and K . K¯ 1 2 1 2 contains subfields isomorphic to K1 and K2. We shall identify these subfields with K and K respectively. Let now k(Kσ Kτ) be a com- 1 2 1 ∪ 2 108 posite extension and R(Kσ Kτ) the ring of the composite extension. 1 ∪ 2 Define the mapping ϕ of K¯ into R(Kσ Kτ) by 1 ∪ 2 ϕ a (x x y) = a xσyτ, xy xy where a k. Then Xϕ coincides withXσ on K and with τ on K . xy ∈ 1 2 Since σ and τ are isomorphisms, it follows that ϕ is a k-homomorphism of K¯ on R(Kσ Kτ). Since Ω is a field, it follows that kernel of the 1 ∪ 2 homomorphism is a prime ideal Y of K¯ . Thus

K¯ /G R(Kσ Kτ). ≃ 1 ∪ 2 If k(Kσ′ Kτ′ ) is another composite extension, then µ defined ear- 1 ∪ 2 lier, is an isomorphism of R(Kσ Kτ) on R(Kσ′ Kτ′ ). Consider the 1 ∪ 2 1 ∪ 2 homomorphism ϕ defined above. Defineϕ ¯ on K¯ byϕ ¯ = µ ϕ. We have · ϕ¯( a x x y) = µ a xσyτ = a xσ′ yτ′ . xy xy xy X X X Thenϕ ¯ is a homomorphism of K¯ on R(Kσ′ Kτ′ ). But, since µ is an 1 ∪ 2 isomorphism, it follows thatϕ ¯ has G as the kernel. Thus the prime ideal G is the same for a class of composite extensions. Conversely, if two composite extensions correspond to the same prime ideal of K¯ , it can be seen that they are equivalent. We have, now, only to prove the existence of a composite extension associated with a prime ideal of K¯ . Let G be a prime ideal of K¯ and G , K¯ . Since K¯ has a unit element, a prime ideal G , K¯ always exists. Let A be the integrity domain

A = K¯ /G 2. Composite ﬁelds 95

109 ¯ ϕ ϕ Let ϕ be the natural homomorphism of K on A. Then K1 and K2 are subfields of A. Since K¯ is generated by elements of the type x x y, ϕ ϕ ϕ ϕ K and K , are different from zero. Clearly A = R(K K ). Hence 1 2 1 ∪ 2 the quotient field of A is a composite extension. Hence for every prime ideal G , K¯ , there exists a composite extension. We have hence proved the

Theorem 4. The classes of composite extensions of K1 and K2 stand in (1,1) correspondence with the prime ideal G , K¯ of the Kronecker product K¯ ofK1 and K2 over k.

σ Consider now the special case where K2/k is algebraic. Let k(K1 τ ∪ K2) be a composite extension. Then

k(Kσ Kτ) R(Kσ Kτ) Kσ. 1 ∪ 2 ⊃ 1 ∪ 2 ⊃ 1 Since every element of K is algebraic over k, k(Kσ Kτ) is algebraic 2 1 ∪ 2 over Kσ. This means that R(Kσ Kτ) is a ﬁeld and so coincides with 1 1 ∪ 2 k(Kσ Kτ). Thus 1 ∪ 2 Theorem 5. If K/k is algebraic, then every prime ideal G , K¯ of K is a maximal ideal.

Let K/k be an algebraic extension and L/k any extension. The Kro- necker product K¯ = K x L is the extended algebra (K) of K by extend- k L ing k to L. K¯ is thus an algebra (commutative) over L. If G is a prime of K¯ , then by above, it is a maximal ideal and K¯ /G gives a composite extension. Since K¯ is an algebra over L we may regard K¯ /G as an extension ﬁeld of L. Let now G1,..., Gm be m distinct maximal ideals of K¯ , none of them 110 equal to K¯ . Let Li = K¯ /Gi be a composite extension. Form the direct sum algebra

Li Xi as a commutative algebra over L. We shall now prove 96 5. Composite extensions

Theorem 6. i K¯ /Gi K¯ / Gi ≃ i P T Proof. We shall construct a homomorphism ϕ of K¯ on i Li and show that the kernel is Gi. i P T In order to do this let us denote by σi the natural homomorphism of K¯ on L (this is identity on L), i = 1,..., m. If x K¯ , then σ x L . i ∈ i ∈ i Deﬁne ϕ on K¯ by ϕ(x) = σi x. Xi That this is a homomorphism on K¯ is easily seen; for, if x, y K¯ ∈

ϕ(x + y) = σi(x + y) = σi x + σiy = ϕx + ϕy. Xi Xi Xi ϕ(xy) = σi(xy) = (σi x)(σiy) = ( σi x)( σiy) = ϕxϕy. Xi Xi Xi Xi The kernel of the homomorphism is set of x such that ϕx = o. Thus σ x = o so that x G for all i. Hence x G . But every y G has i ∈ i ∈ ∩i i ∈ ∩i i the property ϕy = o. Thus the kernel is precisely Gi. i We have only to prove that the homomorphismT is onto. 111 To this end, notice that for each i, i = 1,..., m, there is a b K¯ with i ∈ G , j , i b ∈ j i G < i   For, since Gi and G j, j , i are distinct, there is a j G j which is ∈ not in Gi. Put bi = a j. Then bi satisﬁes above conditions since Gi is i, j maximal. Q Let now ci be an element in the direct sum, ci Li. By deﬁnition i ∈ of bi P = o if j , i σ b j i = , o if j i   3. Applications 97

Since Li is a ﬁeld, there exists xi , o in Li such that

xiσibi = oi.

σ being a homomorphism of K¯ on L , let y K¯ with σ y = x . Put i i i ∈ i i i

c = biyi Xi Then ϕ(c) = σ j(biyi) = ci Xi Xj Xi which proves the theorem completely. Suppose in particular K/k is ﬁnite. Then KL has over L the degree (K : k). Since, for a maximal ideal, G , K¯ , K¯ /G has over L at most the degree (K : k), we get

1 (K¯ /G : L) (K : k) i = 1,..., m. ≤ i ≤ This means that K¯ has only ﬁnitely many maximal ideals and

K¯ / G K¯ /G G ≃ G T X the summations running through all maximal ideals of K¯ . 112 Thus K and L have over k only ﬁnitely many inequivalent composite extensions.

3 Applications

Throughout this section Ω will denote an algebraically closed extension of k and K and L will be two intermediary ﬁelds between Ω and k.A composite of K and L in Ω will be the ﬁeld generated over k by K and L. It will be denoted by KL. Let K¯ be the Kronecker product over k of K and L. There is, then, a homomorphism ϕ of K¯ on R(K L) given by ∪ ϕ a (x x y) = a x y. xy xy · X X 98 5. Composite extensions

Suppose that this homomorphism is an isomorphism of K¯ onto R(K ∪ L). This means that a (x x y) = o a x y = o xy ⇐⇒ xy · or that every set ofX elements of K whichX are linearly independent over k are also so over L. Incidentally, this gives K L = k. ∩ Conversely, suppose K and L have the property that every set of elements of K which are linearly independent over k are also so over L. Then the mapping ϕ is an isomorphism of K¯ on R(K L). For, if ∪ axy xy = o we express x and y in terms of a base of L/k and a base of K/k giving P bαβ xαyβ = o

113 But this means all bαβ areX zero. Therefore ϕ is an isomorphism. It shows that every set of elements of L which are linearly independent over k are also so over K. We call two such fields L and K linearly disjoint over k. Note that L K = k. We deduce immediately ∩ 1) If L and K are are linearly disjoint over k then any intermediary field of K/k and any intermediary field of L/k are also linearly disjoint. Suppose now that K/k is algebraic. Then every prime ideals of K¯ is maximal. Let, in addition, K and L be linearly disjoint over k. Since K¯ is isomorphic to R(K L), it follows that (o) is a maximal ideal. Hence ∪ K¯ is a field. Thus 2) If K/k is algebraic and K and L are linearly disjoint over k, there exists but one class of composite extensions of K and L over k. Let K/k be a finite extension. Then, for some maximal ideal G , K¯ /G is isomorphic to KL. Since K¯ /G may be considered as an extension field of L, we get (K¯ /G : L) (K : k), that is ≤ (KL : L) (K : k) ≤ Clearly if G = (o), equality exists, and then K and L are linearly disjoint over k. The converse is true, by above considerations. Hence, in particular, 3. Applications 99

3) If (K : k) = m and (L : k) = n, then

(KL : k) mn; ≤ equality occurs if and only if K and L are linearly disjoint over k. 114 We now consider the important case, K/k galois. By the considerations above, it follows that KL/k is algebraic over L. Since Ω is a algebraically closed, it contains the algebraic closure of KL. Let σ be an isomorphism of KL in Ω, which is identity on L. Its restriction to K is an isomorphism of K in Ω. But Ω contains the algebraic closure of K and hence σK = K. Since KL is generated by K and L, it follows that σKL KL. Hence KL/L is a galois extension. (KL/L is already ⊂ separable since elements of K are separable over k).

Ω

KL L rr LL rr LL rrr LL K MM r L MMM rrr MM rrr K L ∩ k

We now prove the

Theorem 7. If K/k is a galois extension and L, any extension of k then

G(KL/L) G(K/K L) ≃ ∩ Proof. In the ﬁrst place, we shall prove that K and L are linearly disjoint over K L. For this, it is therefore enough to prove that every ﬁnite set ∩ y ,..., y of elements of L which are linearly independent over K L, 1 m ∩ are also so over K.

If possible, let y1,..., ym be dependent over K so that xiyi = o, xi K. Let us assume that y1,..., ym are dependent over K but no ∈ P 100 5. Composite extensions

proper subset of them is linearly dependent over K. Therefore, all the xi are diﬀerent from zero. We may assume x1 = 1. Let σ be an element of G(KL/L). Then 115

0 = σ( x y ) = σx σy i i i · i Xi Xi By subtraction we get, since σ1 = 1,

0 = (xi σxi)yi , − Xi 1

This means that xi = σxi, i = 2,..., m. But, since σ is arbitrary in G(KL/L), it follows that x L. But x K. Thus y ,..., y are linearly i ∈ i ∈ 1 m dependent on K L which is a contradiction. Hence K and L are linearly ∩ disjoint over K L. ∩ Therefore, KL is isomorphic to the Kronecker product of K and L over K L. ∩ Suppose σ is any L-automorphism of KL. Its restriction to K is an automorphism of K and leaves K L ﬁxed. Consider the mapping ∩ σ σ¯ of G(KL/L) into G(K/K L). This is clearly a homomorphism. → ∩ Ifσ ¯ is identity element of G(K/K L), then σ is identity on K. Since it is ∩ already identity on L, it is identity on KL. Thus, G(KL/L) is isomorphic to a subgroup of G(K/K L). To see that this isomorphism is onto ∩ G(K/K L), let τεG(K/K L). Any element of KL may be written ∩ ∩ (since K and L are linearly disjoint) in the from,

xαyα α X where xα are linearly independent elements of K over K L and yα L. ∩ ∈ This expression is unique. Extend τ toτ ¯ in G(KL/L) by deﬁning

τ¯ xαyα = yατ(xα).  α  α X  X   116 This is well deﬁned; for, if  yατ(xα) = 0, then, since xα are α { } linearly independent over K L, τ(xα) are also linearly independent ∩ P{ } 3. Applications 101 over K L and since K and L are linearly disjoint over K L, all yα are ∩ ∩ zero. Thusτ ¯ is an automorphism of KL/L. Our theorem is thus proved. In particular, if K and L are both galois extensions of k and K L = k ∩ then, by above,

G(KL/L) G(K/k) ≃ G(KL/K) G(L/k) ≃ Also, since KL/k is algebraic, let σ be an isomorphism (trivial on k) of KL in Ω. Since its restrictions on K and L are auto morphisms, it follows that KL/k is a galois extension. We now prove

Theorem 8. G(KL/k) is isomorphic to the direct product of G(K/k) and G(L/k).

Proof. G(K/k) and G(KL/L) are isomorphic and for every element σ in G(K/k), by the previous theorem, we have the extensionσ ¯ , an element of G(KL/L) determined uniquely by σ. Similarly, if τ G(L/k),τ ¯ ∈ denotes its unique extension into an element of G(KL/k).

KL tt tt tt K

L tt ttt ttt k

We now consider the mapping

(σ, τ) σ¯ τ¯ → of the direct product G(K/k) G(L/k) into G(KL/k). · Suppose λ is an element of G(KL/k). Its restriction λ to K is an 117 ¯ ¯ 1 element of G(K/k). Consider λ1. Now λ1− λ is identity on K. By the 102 5. Composite extensions

isomorphism of G(KL/k) and G(L/k), this deﬁnes a unique element µ1 of G(L/k). Hence λ = λ¯ 1µ¯ 1. Thus the mapping above is a mapping of the direct product G(K/k) · G(L/k) onto G(KL/k). We have only to prove that is a homomorphism to obtain the theorem. It is clearly seen that λ is identity if and only if λ1 and µ1 are identity. Let σ, σ′ be two elements of G(K/k) and τ, τ′ in G(L/k). Let λ and µ be the unique elements in G(KL/k) deﬁned by

= = ¯ ¯ λ σ¯ τ,¯ µ σ′, τ′.

Consider to element λµ. Its restriction to K is σσ′ and its restriction to L is ττ′. Thus λµ = σσ ττ′ ′ · which proves that the mapping is a homomorphism. The theorem is now completely proved. Chapter 6

Special algebraic extensions

1 Roots of unity 118 Consider the polynomial xm 1 in k[k], where k is a ﬁeld. Let k have − characteristic p. If p = o, then xm 1 is a separable polynomial over − k, whereas if p , o, the derivative of xm 1 is mxm 1 which is zero, − − if p divides m. If p ∤ m, then xm 1 is a separable polynomial over − k. Therefore, let m > o be an arbitrary positive integer, if k has characteristic zero and m, an integer prime to p, if k has characteristic p , 0. Then xm 1 is a separable polynomial over k and it has m roots in Ω, an − algebraic closure of k. We call these m roots, the mth roots of unity. If ρ and τ are mth roots of unity then ρm = 1 = τm. Therefore m m m 1 m m 1 (ρτ) = ρ τ = 1, (ρ− ) = (ρ )− = 1 which shows that the mth roots of unity from a group. This group Gm is abelian and of order m. Let now d/m. Any root of xd 1 in Ω is also a root of xm 1.If ρ is − − an mth root of unity such that ρd = 1, then ρ is a root of xd 1. Since − xd 1 has exactly d roots in Ω, it follows that G has the property that − m for every divisor d of m, the order of Gm, there exist exactly d elements of Gm whose orders divide d. Such a group Gm is clearly a cyclic group. Hence The mth roots of unity form a cyclic group Gm of order m. 119 There exists, therefore, a generator ρ of Gm. ρ is called a primitive mth root of unity. Clearly, the number of primitive mth roots of unity is

103 104 6. Special algebraic extensions

ϕ(m). All the primitive mth roots of unity are given by ρa with 1 a < ≤ m, (a, m) = 1, ρ a ﬁxed primitive mth roots of unity. Let Ω be an algebraically closed ﬁeld. Let m and n be two positive integers which are arbitrary, if Ω has characteristic zero and, prime to p, if Ω has characteristic p , o. If ρ is an mth root of unity and τ an nth root of unity, then (ρτ)mn = ρmn τmn = 1, · so that ρτ is an mn th root of unity. This shows that the roots of unity in Ω form a group H(Ω). We now determine the structure of H.

Theorem 1. If Ω has characteristic zero, then H is isomorphic to the additive group of rational numbers mod 1, whereas, if Ω has characteristic p , o, H is isomorphic to the additive group of rational numbers a , (a, b) = 1, p ∤ b, mod 1. b

Proof. Let R denote the group (additive) of rational numbers and ν1 < ν < ν the sequence of natural numbers, if Ω has characteristic 2 3 ··· zero, whereas, if Ω has characteristic p , o, let R denotes the rational a numbers , (b, a) = 1, p ∤ b and ν < ν < ν the sequence of b 1 2 3 ··· natural numbers prime to p. Put

µ = ν ν . n 1 ··· n

120 Denote, by Hn, the group of µn the roots of unity in Ω. Since every integer m(p ∤ m, if Ω has characteristic p) divides some µn, if follows that

H = Hn. n [ Since Hn is cyclic, we can choose a generator ρn of Hn is such a way that = νn+1 ρn ρ′n+1 . 2. Cyclotomic extensions 105

a Any x in R may be written as where a is an integer. Deﬁne the µn mapping σ as follows = a σx ρn, so that σ is a function on R with values in H. The mapping is well b deﬁned: for if x = , then µma = µnb. Suppose m > n; then µm

b = aν + ν n 1 ··· m b = aνn+1 νm = a so that ρm ρm ··· ρn by choice of ρn. We now verify that σ is a a b homomorphism of R on H. If x , y = are in R and m n, then µn µm ≥

+ + aνn 1 νm b aνn+1 νm+b σ(x + y) = σ( ··· ) = ρm ··· µm which equals ρa ρb = σx σy. Also, since any root of unity is in some n · m · a = a = a Hn, it is of the form ρn so that, for x , σx ρn. We have, therefore, µn to determine the kernel of the homomorphism. It is the set of x in R such that σx = 1. = a = = a If x , then 1 σx ρn so that µn a and so x is an integer. The 121 µn | | converse being trivial, it follows that the kernel is precisely the additive group of integers and our theorem is established.

2 Cyclotomic extensions

Let k be a field and xm 1 a separable polynomial in k[x]. This implies, − in case k has characteristic p , o, that p does not divide m. Let ρ be a primitive mth root of unity in Ω, an algebraic closure of k. Then K = k(ρ) is the splitting field of xm 1 in Ω. Therefore, K/k is a − separable, normal extension. Let G be its galois group. If σ G, σ is ∈ determined by its effect on ρ. Since ρ is a primitive mth root of unity, so is σρ. For, (σρ)m = σ(ρm) = 1; 106 6. Special algebraic extensions

so σρ is a root of unity. Also, if (σρ)t = 1, then σ(ρt) = 1. Since σ is an automorphism, it follows that ρt = 1 or m/t. Thus

σρ = ρν, (ν, m) = 1.

If σ, τ are in G, let σρ = ρν, (ν, m) = 1 and τρ = ρµ (µ, m) = 1. Then στ(ρ) = σ(τρ) = σ(ρµ) = ρµν which shows that στ = τσ or that G is abelian. Consider now the mapping

g : σ ν → where σρ = ρν, (ν, m) = 1. This is clearly a homomorphism of G into the multiplicative group prime residue classes mod m. The kernel of the mapping g is set of σ with σρ = ρ. 122 But then σt = t for all t K, so that by, galois theory, σ is the ∈ identity. Let us call extension K = k(ρ) a cyclotomic extension of k. We then have proved the Theorem 2. The Cyclotomic extension k(ρ)/k is an abelian extension whose galois is isomorphic to a subgroup of the group of prime residue classes mod m where ρm = 1 and ρ is a primitive mth root of unity. Let Γ be the prime ﬁeld contained in k. Then Γ(ρ) is a subﬁeld of k(ρ) = K. Let G be the galois group K/k.

K = k(ρ) qq qqq qqq k

Γ(ρ) p ppp ppp Γ p

Let σ be in G andσ ¯ the restriction of σ to Γ(ρ). Since σ is identity on k and σρ is again a primitive root of unity, it follows thatσ ¯ is an 2. Cyclotomic extensions 107 automorphism of Γ(ρ)/Γ. It is easy to see that the mapping σ σ¯ is an → isomorphism of G into the galois group of Γ(ρ)/Γ. We shall therefore confine ourselves to studying the galois group of Γ(ρ)/Γ. First, let Γ be the rational number field and ρ a primitive mth root of unity. Let Γ(ρ) be the cyclotomic extension. Let f (x) be the primitive integral polynomial which is irreducible in Γ[x] an which ρ satisfies. m 1 Then f (x) is a monic polynomial. For, since f (x) divides x − ,

xm 1 = f (x)ψ(x). − a ψ(x) has rational coeﬃcients and so ψ(x) = ψ (x) where ψ (x) is a 123 b 1 1 primitive integral polynomial and a and b are integers. From the the theorem of Gauss on primitive polynomials it follows that f (x) is monic. Let p be a prime not dividing m. Let ϕ(x) be the minimum polynomial (which is monic and integral) of ρp. We assert that f (x) = ϕ(x). For, if not, f (x) and ϕ(x) are coprime and so

xm 1 = f (x) ϕ(x) h(x). − · · for some monic integral polynomials h(x). Consider the polynomial ϕ(xp). It has ϕ as a root and so f (x) divides ϕ(xp). Hence ϕ(xp) = f (x)g(x), g(x), again, a monic and integral polynomial. Considering the above mod p we get

f (x)g(x) ϕ(xp) (ϕ(x))p(mod p) ≡ ≡ so that f (x) divides (ϕ(x))p( mod p). If t(x) is a common factor of f (x) and (ϕ(p)( mod p) then (t(x))2 divides xm 1 mod p, which is − impossible, since p ∤ m and xm 1 does not have x as a factor. Thus our − assumption f (x) , ϕ(x) is false. This means that, for every prime p ∤ m, ρp is a root of f (x). If (ν, m) = 1, then ν = p p p where p ,..., p are primes not dividing 1 2 ··· 1 1 1 108 6. Special algebraic extensions

m. By using the above fact successively, we see that, for every ν, (ν, m) = 1, ρν is a root of f (x). Therefore

ν ρm(x) = (x ρ ) (1) = − (ν,Ym) 1 124 divides f (x). But ϕm(x) is ﬁxed under all automorphisms of Γ(ρ)/Γ so that f (x) = ϕm(x). We have proved the Theorem 3. If Γ is the ﬁeld of rational numbers and ρ is a primitive mth root of unity, then the galois group of the cyclotomic extension Γ(ρ)/Γ is isomorphic to the group of prime residue classes mod m. The irreducible polynomial ϕm(x) of ρ is given by (1).

ϕm(x) is called the cyclotomic polynomial of order m. Its degree is ϕ(m). In order to be able to obtain an expression for ϕm(x) in terms of polynomials over Γ, we proceed thus. We introduce the Mobius function deﬁned as follows: It is a function µ(n) deﬁned for all positive integers n such that 1) µ(1) = 1 2) µ(p p ) = ( 1)t where p ,..., p are distinct primes. 1 ··· t − 1 t 3) µ(m) = o if p2/m, p being a prime. From this, one deduces easily 4) µ(m) µ(n) = µ(mn) for (m, n) = 1. · For, one has to verify it only for m = p1 pt, n = q1 q1 where ··· t ··· 1 p′ s and q′ s are all distinct primes. Then µ(m) = ( 1) , µ(n) = ( 1) + − − and µ(mn) = ( 1)t 1. − We now prove the following simple formula 5) µ(d) = o if m > 1 d m X| = 1 if m = 1

125 the summation running through all divisors d of m. 2. Cyclotomic extensions 109

If m = 1, the formula reduces to (1). So, let m > 1. Let m = pa1 pat be the prime factor decomposition of m. Any divisor d of m 1 ··· t is of the form pb1 pbt where o b a , i = 1,..., t. In view of (3), 1 ··· t ≤ i ≤ i it is enough to consider divisors d of m for which o b 1. In that ≤ i ≤ case, t µ(d) = (t)( 1)i = o. i − d m i=o X| X Let now f (n) be a function deﬁned on positive integers, with values in a multiplicative abelian group. Let g(n) also be such a function. We then have the M¨obius inversion formula,

f (d1) µ(d); n d1 n d Y| X| d1 using formula (5), we obtain the inversion formula. The converse follows in the same way. Consider now the integers mod m. Divide them into classes in the following manner. Two integers a, b are in the same class if and only if (a, m) = (b, m).

Let d/m and Cd, the class of integers a ( mod m) with (a, m) = d. 126 m m Then a is of the form dλ where (λ, ) = 1. Thus C has ϕ( ) elements. d d d The classes C , for d m, exhaust the set of integers mod m. If ρ is a d | primitive mth root of unity, then m xm 1 = (x ρt). − = − Yt 1 110 6. Special algebraic extensions

In view of the above remarks, we can write

xm 1 = (x ρt) . − − d m t Cd Y| Y∈ m m But if t C , ρt = ρdλ, (λ, ) = 1 and so ρt is a primitive th root ∈ d d d of unity. Using the deﬁnition of ϕm(x), if follows that

xm 1 = ϕ (x) − d d m Y| By the inversion formula we get

d µ( m ) ϕm(x) = (x 1) d d m − | Q Comparison of degrees on both sides gives the formula

m µ(d) ϕ(m) = dµ( ) = m . d d d m d m X| X|

We may compute ϕm(x) for a few special values of m. Let m = p, a prime number. Then

xp 1 ϕ (x) − = xp 1 + + x + 1. p x 1 − ··· − If m = pq, the product of two distinct primes, then

(xpq 1)(x 1) ϕ (x) = − − . pq (xp 1)(xq 1) − − Let us now consider the case when Γ is the prime ﬁled of p elements. 127 Obviously, Γ(ρ)/Γ is a cyclic extension. If we deﬁne the cyclotomic polynomial as before, it is no longer irreducible over Γ[x]. For instance, let p = 5 and m = 12. Then

12 2 (x 1)(x 1) 4 2 ϕ12(x) = − − = x x + 1. (x6 1)(x4 1) − − − 2. Cyclotomic extensions 111

Also

x4 x2 + 1 = (x2 2x 1)(x2 + 2x 1) (mod 5). − − − − Therefore, Γ(ρ)/Γ has degree < ϕ(m). It is obvious since Γ(ρ)/Γ is cyclic, that, for ϕm(x) to be irreducible, the group of prime residue classes mod m should be cyclic. We shall prove Theorem 4. If Γ is the prime ﬁled of characteristic p , o and p ∤ m,the cyclotomic polynomial ϕm(x) is irreducible, if and only if the group of prime residue classes mod m is cyclic and p is a generator of this cyclic group.

Proof. We already know that Γ(ρ)/Γ is cyclic. If ϕm(x) is irreducible, then Γ(ρ)/Γ has order ϕ(m). Let σ be the Frobenius automorphism of Γ(ρ)/Γ. Then ρσ = ρp, p being the number of elements in Γ. The ϕ(m) automorphisms 2 ϕ(m) 1 1,σ,σ ,...,σ − are distinct. Hence

2 ϕ(m) 1 ρ, ρp, ρp ,...,ρp −

2 ϕ(m) 1 are all distinct, which means 1, p, p ,..., p − are distinct mod m. Therefore, p is a generator of the multiplicative group of prime residue classes mod m.

The converse is trivial. The theorem is true, if, instead of Γ being the prime filed of p ele- 128 ments, Γ is a finite filed of q elements. Then q has to be a generator of the group of prime residue classes mod m. If Rm denotes the group of prime residue classes mod m, then Rm is cyclic, if and only if

m = 2a, a = 1, 2 or m = qb or 2qb, where q is a prime. Thus m has necessarily to have one of these forms. We now use the irreducibility of ϕm(x) over the rational number ﬁeld to prove a theorem of Wedderburn. 112 6. Special algebraic extensions

Theorem 5. A division ring with a finite number of elements is a filed. Proof. Let D be the division ring and k its centre. Then k is a field. D being finite, let k have q elements. If D is of rank n over k, then D has qn elements. We shall prove that n = 1.

Let D1 be a subalgebra of D over k. Let D1 have rank m over k. Then D has qm elements. But D is a sub group of D so that qm 1 1 1∗ ∗ − divides qn 1. This means that m n. For, if n = tm + µ, o µ< m; then − | ≤ qn 1 = qµ(qtm 1) + (qµ 1) − − − so that qm 1 qµ 1 which cannot happen unless µ = 0. Thus every − | − subalgebra of D has rank d dividing n. 129 Let x D. Consider the y D such that xy = yx. They form a ∈ ∈ subalgebra over k. Therefore, the number of y in D∗ which commute with x form a group of order qd 1, for some d dividing n. This group − is the normaliser of x. Hence, the number of distinct conjugates (in the sense of group theory) of x in D∗ is qn 1 qd 1. − | − For the ﬁnite group D∗, we have = + D∗ k∗ D∗x, x X where D∗x is the set of all conjugates of x. Comparing number of elements on both sides qn 1 = q 1 + qn 1/qd 1 − − − − Xd for some divisors d of n. Since ϕ (x) xn 1, we see that ϕ (q), which is an integer, divides n | − n qn 1. Also ϕn(x) divides xn 1 xd 1 for any d n, d , n. Therefore, − − | − | ϕ (q) divides q 1. But n − ϕ (q) > (q 1)ϕ(n) n − which shows that n ≯ 1. This proof is due to Ernst Witt. 3. Cohomology 113

3 Cohomology

Let G be a ﬁnite group and A an abelian group on which G acts as a group of left operators. Let A be a multiplicative group. We denote elements of G by σ,τ,ρ,..., and elements of A by a, b, c,.... We denotes by aσ the eﬀect of σ on a. Then

(ab)σ = aσbσ (aτ)σ = aστ

130 Let 1 be the unit element of A. Denote by Gn, n 1 the Cartesian ≥ product of G with itself n times. A function on Gn with values in A is said to be an n dimensional Cochain or, simply, an n cochain. This function f (x1,..., xn) has values in A. We denote by a σ1,...,σn the element in A which is the value taken by the n cochain for values σ1,...,σn of its variables. We denote the function also, by aσ1,...,σn. If aσ1,...,σn and bσ1,...,σn are two functions, we deﬁne their product by

oσ ,...,σ = aσ ,...,σ bσ ,...,σ . 1 n 1 n · 1 n Similarly = 1 oσ1,...,σn (aσ1,...,σn )− is called the inverse of aσ1,...,σn . With these deﬁnitions, the n cochains form a group Cn(G, A) or simply Cn. We deﬁne Co(G, A), the zero dimensional cochains, to be the group of functions with constant values, that is functions aσ on G such that

aσ = a is the same for all σ G. It is then clear that Co is isomorphic to A. ∈ We now introduce a coboundary operator ∂ in the following manner. n n+1 ∂(= ∂n) is a homomorphism of C into C deﬁned by = aσ1,...,σn+1 ∂aσ1,...,σn 114 6. Special algebraic extensions

n i n+1 = σ1 ( 1) ( 1) aσ2,...,σn+1 (aσ1,...,σi 1,σiσi+1,...,σn+1 ) − (aσ1,...,σn ) − . − n=1 Y 131 For n o, we deﬁne the groups Zn(G, A) and Bn+1(G, A) in ≥ the following manner. Zn(G, A) is the kernel of the homomorphism ∂n Cn Cn+1 and Bn+1(G, A) is the homomorphic image. The elements −→ of Zn(G, A) are called n dimensional cocycles or n-cocycles and the elements of Bn+1(G, A) are n + 1-dimensional coboundaries or n + 1 coboundaries. The homomorphism ∂ has the property ∂∂ = identity (2) which proves that Bn+1(G, A) is a subgroup of Zn+1(G, A) and we can form for every n > o the factor group Hn(G, A) the n-dimensional cohomology group. We verify (2) only in case n = o and 1 which are the ones of use in our work. The coboundary of a zero cochain is a one cochain given by aσ ∂a = a = . (3) σ a Its coboundary, by deﬁnition is σ aσ aτ ∂∂a = aσ,τ = · . aστ

Substituting from (3), we get aσ,τ = 1 which veriﬁes (2) for n = o. We deﬁne the zero coboundary, that is the elements in Bo(G, A), to be the function with value 1 on all σ G. Thus Bo(G, A) consists only ∈ of the identity. An element in Zo(G, A) will be the constant function a with aσ = a for all σ. Hence 132 Ho(G, A) is isomorphic with the set of a A with the property aσ = ∈ a for all σ G. ∈ A one dimensional cocycle is a function aσ for which ∂aσ = 1. But σ aσ aτ ∂aσ = · . aστ 3. Cohomology 115

Therefore, the elements in Z′(G, A) are functions aσ with σ = aσaτ aστ.

σ A one coboundary is, already, a function aσ of the form a /a. It is, certainly, a one cocycle. For our purposes, we shall need also the additive cohomology, instead of the multiplicative cohomology above. We regard now A+ as an additive, instead of a multiplicative, group. Then Cn(G, A+) is an additive abelian group. We deﬁne the coboundary operator as = aσ1,...,σn+1 ∂aσ1,...,σn n + = aσ1 + ( 1)ia + ( 1)n 1a . σ2,...,σn+1 σi,...,σi 1,σi,τσi+1,...,σn+1 σi,...,σn = − − − Xi 1 As before, a zero cochain is a constant function on G and its coboundary aσ is σ ∂a = aσ = a a. − A one cochain aσ is a cocycle if

∂aσ = 0 which is the same thing as + σ = aσ aτ aστ. Exactly as before, we see that Ho(G, A), the zero dimensional addi- + tive cohomology group is isomorphic to the set of elements a in A with 133 aσ = a for all σ. We now consider the case when G is a cyclic group. Let σ be a gen- 2 n 1 erator of G so that 1,σ,σ ,...,σ − are all the elements of G. Taking multiplicative cohomology, if aσ is a one cocycle, then = τ aτµ aτaµ

τ = n 1 or aµ aτµ/aτ. Substituting for τ successively 1,σ,...,σ − we get

+ + + n 1 1 σ σ − = aµ ··· 1. 116 6. Special algebraic extensions

n 1 1+σ+ +σ − We denote aµ ··· by Naµ and call it the norm of aµ. Thus, if aµ is a cocycle, then, Naµ = 1. Conversely, suppose a is an element in A with

+ + + n 1= Na = a1 σ .. σ − 1 = 1.

ν We can deﬁne a cocycle aµ such that aσ = a. For let µ = σ , for some ν. Put ν 1 1+σ+ +σ − aµ = a ··· . ν Obviously, aσ = a. Also, aσ is a cocycle. For, if τ = σ ′ .

ν 1 ν 1 ν µ 1+σ+ +σ − 1+σ+ +σ ′− σ aµaτ = a ··· (a ··· ) + 1+σ+ +σν 1+σν+ +σν ν′ 1 = a ··· − ··· −

= aµτ. = + τ In a similar manner, for additive cohomology, we have aτµ aτ aµ n 1 134 where aµ is a cocycle. If G is cyclic, on substituting 1,σ,...,σ − for τ, we get n 1 σ σ − S aµ = aµ + a + + a = o. µ ··· µ We call S aµ the spur or trace of aµ. If a is any element of A, with n 1 trace S a = a + aσ + + aσ − = o, then the cocycle ··· ν 1 σ σ − aµ = a + a + + a ··· where µ = σν, has the property

aσ = a. We now apply the considerations above, in the following situation. K/k is a ﬁnite galois extension with galois group G. Then G acts on + K∗ and also the additive group K as a group of operators. We might, o 1 therefore, consider the cohomology groups H (G, K∗), H (G, K∗),... o + 1 + o o + and H (G, K ), H (G, K ) ... etc. As before, H (G, K∗) and H (G, K ) + σ are isomorphic to subgroups of K∗ and K with a = a for all σ. This, by galois theory, shows 3. Cohomology 117

o o + Theorem 6. H (G, K∗) is isomorphic to k∗ and H (G, K ) is isomorphic to k+.

But what we are interested in, is the following important theorem due to Artin.

1 1 + Theorem 7. The group H (G, K∗) and H (G, K ) are trivial.

Proof. Let us ﬁrst consider multiplicative cohomology. If aσ is a cocycle, we have to prove that it is a coboundary. The elements σ,τ,... of G are independent k-linear mappings of K into Ω, the algebraic closure. Hence, if (aσ) are elements of K∗,

aσ σ σ · X is a non-trivial k-linear map of K into Ω. Therefore, there exists a θ , o 135 in K such that σ aσθ , o. σ X

1 σ τ Put b− = aσθ = aτθ . Then σ τ P P 1 σ = σ στ (b− ) aτ θ . τ X Therefore aσ = σ στ σ aσaτ θ . b τ X Since (aσ) is a cocycle, we get

aσ = στ = τ = 1 σ aστθ aτθ b− . b τ τ X X Thus bσ a = = bσ 1 σ b − which is a coboundary. 118 6. Special algebraic extensions

Consider now the additive cohomology. K/k being ﬁnite and separable, there exists aθ K such that ∈ σ θ = S K/kθ = 1. σ X Put now σ b = aσθ − σ X aσ being an additive cocycle. Then σ = σ στ b aτ θ . − τ X τ But aσ = aσ 1 = aσ θ so that · τ · σ = + σ στ = στ = aσ b P (aσ aτ )θ aστθ b − τ τ − X X σ which proves that aσ = b b is a coboundary. − Our theorem is completely proved. 136 We apply the theorem in the special case where G is cyclic. Let σ be a generator of G. If aσ is a cocycle then, in multiplicative theory

+ + + n 1 1 σ σ − = aσ ··· 1

or NK/kaσ = 1. Similarly in additive cohomology,

n 1 σ σ − aσ + a + + a = o σ ··· σ or S K/kaσ = o. Using theorem 7, we obtain ’theorem 90’ of Hilbert. Theorem 8. If K/k is a ﬁnite cyclic extension, σ, a generator of the galois group of K/k and a and b two elements of K with NK/ka = 1 and S K/kb = 0 respectively, then

1 σ a = c − b = dσ d − for two elements c, d in K. 4. Cyclic extensions 119

4 Cyclic extensions

Let K/k be a cyclic extension of degree m. Put m = npa, (n, p) = 1 if p is the characteristic of k; otherwise, let m = n. Let G be the galois group of K/k. It has only one sub-group of order pa. Let L be its fixed field. Then K/L is cyclic of degree pa and L/k is cyclic of degree n prime to p. Let ρ be a primitive nth root of unity and k(ρ) the cyclotomic extension. The composite F = Lk(ρ) is cyclic over k(ρ) and of degree prime to p. We shall see that K over L and F over k(ρ) can be described in a simple manner. We shall, therefore, consider the following case, first. K/k is a cyclic extension of degree m and p ∤ m, if k has character- 137 istic p , o; otherwise, m is an arbitrary integer. Also, k contains all the mth roots of unity. We then have the theorem of Lagrange. Theorem 9. K = k(w) where wm k. ∈ Proof. Let ρ be a primitive mth root of unity. ρ is in k.

Hence, since K/k has degree m,

m NK/k ρ = ρ = 1.

By Hilbert’s theorem, therefore, if σ is a generator of the galois group of K/k, then there is an ω K such that ∈ 1 σ ω − = ρ.

Since ρ is a primitive mth root of unity, ω,ωσ,ωσ2 ,... are all distinct and are conjugates of ω. Hence our theorem. ω satisﬁes a polynomial xm a, a k. If K = k(ω ), where ω also − ∈ ′ ′ satisﬁes a polynomial xm b, b k,then − ∈ σ + ω′ = ω′ ρ , · where ρt is a primitive mth root of unity and so (t, m) = 1. Now ω σ ω ρt ω ′ = ′ · = ′ ωt ωt ρt ωt ! · 120 6. Special algebraic extensions

which shows that t ω′ = ω c, c k. · ∈ We shall call xm a a normed polynomial. We have, then, the − Corollary. If k(ω) = K = k(ω′), where ω and ω′ are roots of normed polynomials, then t ω′ = ω c, · 138 where (t, m) = 1 and c k. ∈ We consider the special case, m = q, a prime number. Let K/k be a cyclic extension of degree q. Let K = k(α) and let α1 = (α),α2,...,αq be the irreducible polynomial of α over k. Suppose σ is a generator of the galois group of K/k. Let notation be so chosen that σ = σ = σ = σ = α1 α2,α2 α3,...,αq 1 αq,αq α1. − Since k contains the qth roots of unity and every qth root of unity ρ , 1 is primitive, we construct the Lagrange Resolvent.

q 1 ω = ω(α, ρ) = α + ρα + + ρ − α . 1 2 ··· q Then σ q 2 q 1 ω = α + ρα + + ρ − α + ρ − α 2 3 ··· q 1 σ 1 which shows that ω = ρ− ω. Hence K = k(ω). Also,

(ωq)σ = ωq

which proves that ωq k and ω satisﬁes a normed polynomial. ∈ In particular, if k has characteristic , 2, and K/k has degree 2, then

K = k(√d)

for d k. ∈ The polynomial xq a for a k is, thus, either irreducible and, then, − ∈ a root of it generates a cyclic extension, or else, xq a is a product of − linear factors in k. 4. Cyclic extensions 121

We study the corresponding situation when K has characteristic p , o and K/k is a cyclic extension of degree pt, t 1. ≥ 139 We ﬁrst consider cyclic extensions of degree p. Let K/k be a cyclic extension of degree p and σ, a generator of the galois group of K/k. Let µ be a generic element of the prime ﬁeld Γ contained in k. Introduce the operator P x = xp x. Then − P(x + µ) = P x.

Also, the only α in k satisfying Pα = o are the elements of Γ and these are all the roots of P x = o. The element 1 in k has the property

S K/k 1 = o,

so that by Hilbert’s theorem, there is an ω K such that ∈ 1 = ωσ ω. − Therefore, since K/k has degree p,

K = k(ω).

In order to determine the polynomial of which ω is a root, consider Pω. (Pω)σ = Pωσ = P(ω + 1) = Pω which shows that Pω k. If we put Pω = α, then ω is a root the ∈ irreducible polynomial xp x α − − in k[x]. ω is a root of the polynomial and ωσ = ω + 1. Since σ is a generator of the galois group of K/k, the roots of xp x α are ω,ω + − − 1,...,ω + p 1. − A polynomial of the type xp x α, α k is called a normed − − ∈ polynomial. Suppose K = k(ω′) where ω′ also satisﬁes a normed polynomial. 140 122 6. Special algebraic extensions

Then ω ,ω + 1,...,ω + p 1 are the roots of this normed polynomial. ′ ′ ′ − So σ ω′ = ω′ + h, o < h p 1. ≤ − Now ω hω satisfies ′ − σ σ σ (ω′ hω) = ω′ hω = ω′ + h hω h = ω′ hω − − − − − which shows that ω hω k. We have, hence, the ′ − ∈ Theorem 10. If K/k is cyclic of degree p and σ, a generator of the galois group of K/k, then K = k(ω) where ω satisfies a normed polynomial xp x α in k[x] and ωσ = ω + 1. IfK = k(ω ) and ω also satisfies a − − ′ ′ normed polynomial, then

ω′ = µω + a,

µ Γ and a k. ∈ ∈ In order to be able to construct ω, we proceed thus. Let α be an element in K with S K/kα = 1.

Let α1(= α),...,αp be the roots of the irreducible polynomial which α satisﬁes over k. Construct the resolvent

ω = α1 + 2α2 + + (p 1)αp 1 + pαp. − ··· − − Let notation be so chosen that σ = σ = σ = σ = α1 α2,α2 α3,...,αp 1 αp,αp α1. − Then ωσ = α + 2α + + (p 1)α − 2 3 ··· − p and so ωσ ω = α + α + + α = 1. Therefore ωp ω = t for some − 1 2 ··· p − t in k. This gives the normed polynomial. 141 It should be noticed that, here, we use additive cohomology whereas, in case K/k has degree m prime to p, we used multiplicative cohomology. Furthermore, in the second case, when K/k is of degree p, the 4. Cyclic extensions 123 elements of Γ serve the same purpose as the roots of unity in the ﬁrst case. If k has characteristic 2 and K/k is a separable extension of degree 2, then K = k(ω) where ω2 ω k. − ∈ Observe, also, that any polynomial xp x α, for α k, is either − − ∈ irreducible over k and so generates a cyclic extension over k, or splits completely into linear factors in k. For, if ω is a root of xp x α then, − − for µ Γ, ω + µ is also a root. ∈ Just as we denote a root of xm α by √m α, we shall denote, for α k, α − α ∈ p α ω by P , a root of x x . It is obvious that P is p valued and if is α − − one value of P , all the values are

ω,ω + 1,...,ω + p 1. − We now study the case of cyclic extension of degree pn, n 1. ≥ Let K/k be a cyclic extension of degree pn and σ, a generator of the galois group G of K/k. Since G is cyclic of order pn, it has only one subgroup of order p and hence, there exists a unique subﬁeld L of K n 1 such that K/L is cyclic of degree p and L/k is cyclic of degree p − = m. Let us assume that n 2. ≥ σm is of order p and hence is a generator of the galois group of K/L. 142 Thus K = L(ω) where σmω = ω + 1 and ω satisﬁes Pω = α L. ∈ Put σω ω = β. Then σmβ = σ(σmω) σmω = β which shows that − − β L. Also, ∈ m 1 σ σ − S / β = β + β + + β . L k ··· Substituting the value of β, we get

2 m m 1 S / β = (σω ω) + (σ ω σω) + + (σ ω σ − ω) L k − − ··· − = σmω ω = 1. − 124 6. Special algebraic extensions

Hence β has the property

S L/kβ = 1.

It is easy to see that α is not k. For,

Pβ = σ(Pω) Pω = σα α − −

and α in k would mean Pβ = o or β Γ. This means that S / β = o. ∈ L k We now proceed in the opposite direction. Let L be cyclic of degree n 1 p − > 1 over k. We shall construct an extension K which is cyclic over k and contains L as a proper subﬁeld. Let σ be a generator of the galois group of L/k. Let us choose β L, such that ∈

S L/kβ = 1.

Now S L/k(Pβ) = P(S L/kβ) = o which shows that

Pβ = σα α − 143 for some α in L. Also, α is not in k. We claim that for λ k, ∈ xp x α λ (4) − − − is irreducible over L[x]. For, if it is not irreducible, it is completely reducible in L. Let ω be a root of xp x α λ in L. Then − − − ωp ω = α + λ. − This means that

P(σω ω) = σ(Pω) Pω = σα α = Pβ. − − − Thus P(σω ω β) = o − − or σω ω β = µ. Since S / (σω ω) = o and S / µ = o, it follows that − − L k − L k S / β = o, which is a contradiction. Thus for λ k, (4) is irreducible L k ∈ 4. Cyclic extensions 125 in L[x]. Let ω be a root of this irreducible polynomial for some λ. Then K = L(ω)/L is cyclic of degree p. Letσ ¯ denote an extension of σ to an isomorphism of K/k in Ω, the algebraic closure of k. Thenσ ¯ is not identity on L. Since Pω = α + λ and λ k, we get ∈ σ¯ (Pω) = σα + λ.

Now

P(σω ¯ ω) = σ¯ (Pω) Pω = σα α = Pβ − − − and, again, we have P(σω ¯ ω β) = o − − or thatσω ¯ = ω + β + µ which shows thatσ ¯ is an automorphism of K/k. If t is any integer,

t 1 σ¯ tω = ω + β + βσ + + βσ − + tµ. ···

144 n 1 This shows that 1,σ, ¯ σ¯ 2,..., σ¯ p − have all diﬀerent eﬀects on ω, so that they are distinct automorphisms of K/k. But, since K/k has degree pn, it follows that K/k is cyclic of degree pn. We have, hence, the important

Theorem 11. If k is a ﬁeld of characteristic p and admits a cyclic extension K of k containing L and of degree pm,m > n, for every m.

In fact, if k is an infinite field, we may very λ over k and obtain an infinity of extensions K over k with the said property. It follows theorem 10.

Corollary. If k is a ﬁeld of characteristic p and admits a cyclic extension of degree pn for some n 1, then its algebraic closure is of inﬁnite ≥ degree over it. 126 6. Special algebraic extensions

5 Artin-Schreier theorem

We had obtained, in the previous section, a sufficient condition on a field so that its algebraic closure may be of infinite degree over it. We would like to know if there are fields K which are such that their algebraic closures are finite over them. The complete answer to this question is given by the following beautiful theorem due to Artin and Schreier. Theorem 12. If Ω it an algebraically closed field and K is a subfield such that 1 < (Ω : K) < ∞ 145 then K has characteristic zero and Ω= K(i), where i is a root of x2 + 1. Proof. The proof is as follows:-

1) K is a perfect field. For if not K p−∞ Ω and K p−∞ is of infinite ⊂ degree over K. This shows that Ω/K is a finite separable extension. Since it is is trivially normal, it is galois extension. Let n be the order of the galois group G of Ω/K. 2) If p is the characteristic of K, then p ∤ n (if p , o). For, if p/n, then G has a subgroup of order p generated by an element σ. Let L be the fixed field of this subgroup. Then Ω/L is a cyclic extension of degree p. By corollary of theorem 11 it follows that Ω/L has infinite degree. This is a contradiction. Therefore, the order n of G is prime to the characteristic of K, if different from zero. 3) Let q be a prime dividing n. Then q , p. Let L now be the fixed field of a cyclic subgroup of G, of order q. Then Ω/L is cyclic, of order q. Now L contains the primitive qth root ρ of unity. For, if not since ρ satisfies an irreducible polynomial of degree q 1, it follows that L(ρ)/L has degree q 1. But − − (L(ρ) : L) (Ω : L) | which means that L(ρ) = L. By theorem 9, therefore

Ω= L(ω1) 5. Artin-Schreier theorem 127

where ωq = ω L. 1 ∈ 4) Any irreducible polynomial over L is either linear or of degree q. 146 For, if t is its degree and α, a root of it then t = (L(α) : L) divides q. Thus, every polynomial in L[x] splits in L into product of linear factors and of factors of degree q. Consider, in particular, the polynomial xq2 ω. In Ω, we can write −

q2 q2 x ω = πµ(x µ √ω) (5) − − where µ runs through all q2 th roots of unity. Since (Ω : L) > 1, the polynomial xq2 ω has, in L[x], an irreducible factor of degree q. − Since this factor is formed by q of the linear factors on the right of (5), this factor is of the form

xq + + ε√ε ω, ··· ε being a q2th root of unity. We assert that this is a primitive q2th root of unity. For if not, it is either 1 or a primitive qth root of unity. Since ε√q ω L, it would then follow that √q ω L. Therefore we get ∈ ∈ Ω= L(ε).

5) Let Γ be the prime field contained in L. Consider the field Γ(ε). Let Γ(ε) contain the qνth but not qν+1 roots of unity. This integer certainly exists. For, if L has characteristic zero ν = 2. If L has characteristic p (and this is different from q, from (2)), then Γ(ε) contains p f elements where f is the smallest positive integer with p f 1( mod q2). If qν is the largest power of q dividing p f 1 then ≡ + − Γ(ε) contains the qνth, but not the qν 1 the roots of unity. 147 Let ρ be a primitive qv+1th root of unity. Then ρ satisfies a polynomial of degree q(irreducible) over L. Thus Γ(ρ) is of degree q over Γ(ρ) L. Now, ρ being a primitive qν+1th root of unity ρq is a qνth ∩ root of unity and so is contained in Γ(ε). But all the roots of xq ρq + − are qv 1th roots of unity. Thus xq ρq is the irreducible polynomial − 128 6. Special algebraic extensions

of ρ over Γ(ε). Hence

Γ(ρ) OO sss OOO sss OO Γ(ε) Γ(ρ) L LL o ∩ LLL ooo LLL ooo Γ oo

and (Γ(ρ) : Γ(ε)) = q = (Γ(ρ) : (Γ(ρ) L). ∩ If Γ is a finite field, then (Γ(ρ)/Γ is cyclic and it cannot have two distinct subfields Γ(ε) and Γ(ρ) L over which it has the same degree. ∩ Since ε < L, it follows that Γ(ε) and Γ(ρ) L are distinct and so, Γ ∩ has characteristic zero. In this case, if q is odd, Γ(ρ)/Γ is cyclic and the same thing holds. Thus q = 2. We know, then that ν = 2. If i denotes a fourth root of unity, then

Ω= L(i).

6) Form above, it follows that n = 2t for some t and K does not contain the 4th root of unity. Now t = 1, for if not, let t > 1. Let M be a subﬁeld of L such that (L : M) = 2. Then (Ω : M(i)) = 2 and, by above considerations, M(i) cannot contain 1 which is a contradiction. So K(i) =Ω.

148 We shall make use of this theorem in the next chapter.

6 Kummer extensions

We now study the structure of finite abelian extensions of a field k. We use the following notation:- k is a field of characteristic p, not necessarily > o. n is a positive integer not divisible by p if p , o, otherwise arbitrary. α, the group of non-zero elements of k. A generic element of α will also be denoted, following Hasse and Witt, by α. 6. Kummer extensions 129

αn is the group of nth powers of elements of α. ω, a subgroup of α containing αn and such that

(ω : αn) < . (6) ∞ Let k contain the nth roots of unity. We shall establish a(1, 1) correspondence between abelian extensions of k of exponent dividing n and subgroups ω of α satisfying (6). Let K be an extension ﬁeld obtained from k by adjoining to k the nth roots of all the elements of ω. We shall obtain some properties of K. n Since ω/α is a ﬁnite group, let λ1,...,λt in ω, form a system of generators of ω mod αn. Then any ω ω is of the form ∈ = a1 ai n ω λ1 ,...,λt α Λ = 1/n = Λ where a1,..., at are integers. Put i λi , i 1,..., t. Then i is uniquely determined up to multiplication by an nth root of unity, which is already in k. This means that

K = k(Λ1,..., Λt).

149 Therefore

1) K/k is a ﬁnite extension. Each Λ is a root of a polynomial of the form xn λ . This polyno- i − i mial, by the condition on n, is separable over k. Also, xn λ splits − i completely in K. Thus

2) K/k is a ﬁnite galois extension and is the splitting ﬁeld of the polynomial t n f (x) = (x λi) = − Yi 1 over k[x].

Let us denote by Λ the group generated by Λ1,..., Λt and α. Let G denote the galois group of K/k. In the ﬁrst place, 130 6. Special algebraic extensions

3) Λ/α ω/αn. (These are isomorphic groups). ≃ Consider the homomorphism Λ Λn of Λ into itself. It takes α into → αn. The kernels in Λ and α are both the same. Since Λ is taken to ω by this homomorphism, it follows that Λ/α ω/αn. Incidentally, ≃ therefore, Λ/α is a ﬁnite group. We shall now prove the important property, 4) G is isomorphic to ω/αn. (This proves that K/k is a ﬁnite abelian extension.) In order to prove this, consider the ’pairing’, (τ, Λ) of G and Λ, given by 1 σ (σ, Λ) =Λ − , (7)

150 σ G, Λ Λ. Because of deﬁnition of Λ, ∈ ∈ Λn (σ, Λ)n = = 1. (Λn)σ Thus (σ, Λ) is an nth root of unity. Also, Λ Λ Λ σ Λ Λ (στ, Λ) = = = = = (σ, Λ) (τ, Λ). Λστ Λσ Λτ Λσ · Λτ · ! Furthermore, Λ Λ Λ Λ (σ, ΛΛ ) = ′ ′ = , ′ = (σ, Λ)(σ, Λ ). ′ Λ Λ σ Λσ Λσ ′ ( ′ ) Thus (σ, Λ) deﬁned by (7) is a pairing. Let Go be the subgroup of G consisting of all σ with (σ, Λ) = 1, for all Λ. Since Λ is generated by Λ1, Λ2,..., Λt,α we get Λ =Λσ i i . But Λ ,..., Λ generate K. Therefore,β = βσ for all β K. 1 t ∈ By galois theory, σ = 1. Thus Go = (1). Let Λo be the subgroup of all Λ with (σ, Λ) = 1, for all σ. For the same reason as before, Λo = α. Thus, by theorem on pairing, G is isomorphic to Λ/α. (4) is thus proved. 6. Kummer extensions 131

Observe, now, that since Λn α, it follows the G is an abelian group ⊂ whose exponent divides n. We will denote K symbolically by K = k(√τ ω). We shall now prove

Theorem 13. Let K/k be a ﬁnite extension with abelian galois group G of exponent dividing n, then K = k(√n ω) for a sub group ω α with ⊂ ω/αn ﬁnite.

Proof. Let Λ¯ denote the group of non-zero elements of K with the prop- 151 erty Λn α for Λ Λ¯ . Then, by considering the homomorphism ∈ ∈ Λ Λn, it follows that → Λ¯/α ω/αn, ≃ where ω = Λ¯ n. We shall prove that K = k(√n ω¯ ). In order to prove this, it suﬃces to prove that G ω/αn. (8) ≃

For, in that case, construct the ﬁeld k(√n ω). Because of the properties of K, it follows that K k(√n ω¯ ). But, by the previous results, (k(√n ω) : ⊃ k) = order of ω/αn = order of G. Hence K = k(√n ω). We shall, therefore, prove (8). Let G denote the dual of G. For every Λ in Λ¯ , deﬁne the function ∗ 1 σ χΛ (σ) =Λ − on G into Λ¯ . Since Λn α, it follows that χΛ(σ) is an nth root of unity. ∈ Also, Λ Λ Λ σ Λ Λ χΛ(στ) = = = = χΛ(σ) χΛ(τ) Λστ Λσ Λτ Λσ · Λτ · ! so that χΛ is a character of G. 132 6. Special algebraic extensions

Let χ be any character of G. Since the exponent of G divides n, χn(σ) = 1 for all σ. Therefore, χ(σ) is an nth root of unity and hence is an element of k. Also, since χ is a character of G,

χ(στ) = χ(σ)χ(τ)

152 which equals χ(σ) (χ(τ))σ. Thus χ(σ) is a one cocycle and, by theo- · rem 7, 1 σ χ(σ) = β − for β K. But (χ(σ))n = 1 so that ∈ βn = (βn)σ

for all σ or βn α. This means, by deﬁnition of Λ¯ that β Λ¯ . ∈ ∈ Consider the mapping ω χΛ of Λ¯ into G . This is, trivially a → ∗ homomorphism. By above, it is a homomorphism onto. The kernel of the homomorphism is set of Λ for which χΛ(σ) = 1 for all σ. That is

Λ=Λσ

for all σ. By galois theory Λ α. But every element in α satisﬁes this ∈ condition. Thus α is precisely the kernel, and

Λ¯ /α G∗. ≃ Since G is a ﬁnite abelian group, G G and our theorem is com- ≃ ∗ pletely proved. We call a ﬁnite abelian extension K/k, a Kummer extension if

1) k contains the nth roots of unity, p ∤ n, if p(, o) is characteristic of k,

2) K/k has a galois Galois group of exponent dividing n.What we have then proved is

Theorem 14. The Kummer extensions of k stand in (1, 1) correspondence with subgroups ω of α with ω/αn ﬁnite. 7. Abelian extensions of exponent p 133

7 Abelian extensions of exponent p 153 We had introduced earlier the operator P which is deﬁned by

P x = xp x. − Let k be a ﬁeld of characteristic p and denote by k+ the additive group of k. Then α Pα is a homomorphism of k+ into itself. The → kernel of the homomorphism is precisely the set of elements in Λ, the prime ﬁeld of characteristic p, contained in k. α p If α k, we denote by P , a root of the polynomial x x α. ∈ α − − Obviously P is p valued. Also, + α β = α + β P P P . (9)

Let us denote by Pk+, the subgroup of k+ formed by elements Pα, α k+. Let ω be a subgroup of k+ with the properties, ∈ + + k ω Pk ⊃ ⊃

ω/Pk+ is ﬁnite. Let K be the extension ﬁeld of k, formed by adjoining to k all the α elements , fro α ω. We denote P ∈

= ω K k(P ).

We then obtain in exactly the same way as before

1) K/k is a ﬁnite abelian extension.

2) The galois group G of K/k is isomorphic with ω/Pk+ .

3) G is an abelian group of exponent p. For the proofs, we have to use additive, instead of multiplicative, pairing and the property (9). 134 6. Special algebraic extensions

Suppose, now, on the other hand, K/k is a ﬁnite abelian extension of 154 exponent p, where p is the characteristic of k. Then ω = , K k(P ) where ω is a subgroup of k+ with ω/Pk+ ﬁnite. For the proof of this, we have to use additive instead of multiplicative, cohomology. For, let G be the galois group of K/k and G∗ its character group. Denote, by Λ¯ , the additive subgroup of K+ formed by elements Λ with PΛ k+. ∈ Denote, by ω, the group PΛ¯ . Then

Λ¯ / + ω/P + k ≃ k Deﬁne χΛ(σ) by σ χΛ(σ) =Λ Λ . − Then σ P(χΛ(σ)) = PΛ (PΛ) ). − + But, PΛ k by deﬁnition. Hence, P(χΛ(σ)) = o. ∈ Therefore, χΛ(σ) is an element of Γ. The rest of the proof goes through in the same way and we have the

Theorem 15. If k has characteristic p , o, the ﬁnite abelian extensions K/k of exponent p stand in a(1, 1) correspondence with subgroups ω of + + ω k such that ω/Pk is ﬁnite, and then K = k( P ).

8 Solvable extensions

We propose to study now the main problem of the theory of algebraic equations, namely the ‘solution’ of algebraic equations by radicals 155 k will denote a field of characteristic p,(p = o or p , o), Ω will be its algebraic closure and n, n1, n2,... integers > o which are prime to p, if k has characteristic p , o, otherwise arbitrary. An element ω Ω will be ∈ said to be a simple radical over k if ωn k, for some integer n. k(ω)/k ∈ is then said to be a simple radical extension. k(ω)/k is clearly separable. If ω is a root of xp x a, for a k,ω is called a simple pseudo radical − − ∈ 8. Solvable extensions 135 over k.k(ω)/k is a pseudo radical extension and is separable. In fact, it is a cyclic extension over k. This situation occurs, only if p , o. An extension field K/k is said to be a generalized radical extension if it is a finite tower k = K K K K = K o ⊂ 1 ⊂ 2 ⊂···⊂ m where Ki/Ki 1 is either a simple radical or a simple pseudo radical extension. Every− element it K is called a generalized radical. A typical element would be

n a 1 + n2 + 3 + a1 a2 P .. s r Clearly, K/k is a separable extension. A generalized radical extension is called a radical extension if Ki/Ki 1 is a simple radical extension for every i. A typical element of K would,− then be

n 1 a + n√2 a + . 1 2 ··· Let f (x) be a polynomialq in k[x] and let it be separable. Let K be its splitting field. Then K/k is a galois extension. The galois group G of K/k is called the group of the polynomial f (x). The polynomial f (x) is said to be solvable by generalized radicals, if K is a subfield 156 of a generalized radical extension. It is, then, clear that the roots of f (x) are generalized radicals. In order to prove the main theorem about solvability of a polynomial by generalized radicals, we first prove some lemmas. Lemma 1. Every generalized radical extension is a subfield of a generalized radical extension which is galois, with a solvable galois group. Proof. Let K/k be a generalized radical extension so that k = K K K = K, o ⊂ 1 ⊂···⊂ m where Ki/Ki 1 is a simple radical or simple pseudo radical extension. − n Let Ki = Ki 1(ωi). Then either ω Ki 1 for some ni or Pωi Ki 1. − i ∈ − ∈ − Let (K : k) = n. Put N = n, if k has characteristic zero; otherwise, let n = N pa, · 136 6. Special algebraic extensions

a o, (p, N) = 1. Let ρ be a primitive Nth root of unity. ≥ Let L1 = Ko(ρ). The L1/Ko is a simple radical extension. Since K1 = Ko(ω1), put L2 = Ko(ρ, ω1). Then L2 is the splitting ﬁeld of the polynomial (xN 1)(xn1 a ) or (xN 1)(P x a ) depending on whether − − 1 − − 1 ω1 is a simple radical or a simple pseudo radical. In any case, L2/Kois a galois extension. Furthermore

L2 = L1(ω1)

so that L2/L1 is cyclic, since, when ω1 is a simple radical, L1 contains 157 the requisite roots of unity. Let σ1,...,σℓ be the distinct automorphisms of L2/Ko. Put ℓ = n2 σi f (x) (x a2 ), = − Yi 1 n2 = if ω2 is a simple radical with ω2 a2, and

ℓ = P σi f (x) ( x a2 ), = − Yi 1 if ω2 is a simple pseudo radical with Pω2 = a2. Then f (x) is a polynomial in k[x]. Let L3 be its splitting ﬁeld. Then L3 is galois over k. Also, L3 is splitting ﬁeld of f (x) over L3 so that L3/L2 is either a Kummer extension or else, an abelian extension of exponent p. In this way, one constructs a galois extension T of k such that Lo = k L1 L2 L3 Lm 1 T = Lm, ⊂ ⊂ ⊂ ⊂···⊂ − ⊂ where Li/Li 1 is either a kummer extension or an abelian extension of − exponent p. Clearly, Li/Li 1 and, hence, T/k is a generalized radical − extension. Let Gi, i = o, 1, 2,..., m be the galois group of T/Li. Then

G = G G ...... G = (e) o ⊃ 1 ⊃ ⊃ m

is a normal series. Further, by our construction, Gi 1/Gi is the galois − group of Li/Li 1 and hence, abelian. Thus, G is a solvable group. The lemma is thus− proved. 8. Solvable extensions 137

Lemma 2. If K/k is a ﬁnite solvable extension, then K is a subﬁeld of a generalized extension over k.

Proof. Let G be the galois group of K/k and G solvable. Let n be the order of G and put n = paN with the same connotation, as before. Let ρ be a primitive Nth root of 158 unity and L = k(ρ). Then L/k is a simple radical extension. Let M be a composite of K and L. Then M/L is a galois extension with a galois group which is isomorphic to a subgroup of G and hence, solvable. Let Go be the galois group of M/L and let it have a composition series

G G G = (e). o ⊃ 1 ⊃···⊃ m

Then Gi/Gi+1 is a cyclic group of prime degree. Let Lo = L, L1,... Lm = M be the ﬁxed ﬁelds of Go, G1,..., Gm respectively. Then Li/Li 1 is a cyclic extension of prime degree. Since Li 1 contains the − − requisite roots of unity, Li is a simple radical or a simple pseudo radical extension of Li 1. Then M/k is a generalized radical extension and our lemma is proved.− We are, now, ready to prove

Theorem 16. A separable polynomial f (x) k[x] is solvable by gener- ∈ alized radicals if and only if its group is solvable.

Proof. Let K be the splitting ﬁeld of f (x) and G the galois group of K/k. Suppose f (x) is solvable by generalized radicals. Then K L ⊂ where L/k is galois and by lemma 1, has solvable galois group H. Let Go be the galois group of L/K Then H/Go is isomorphic to G and so G is solvable.

Let, conversely, G be solvable. Then, by lemma 2, K is contained in a generalized radical extension and so f (x) is solvable by generalized radicals. We can easily prove 159 138 6. Special algebraic extensions

Corollary. A separable polynomial f (x) k[x] is solvable by radicals ∈ if and only if its splitting ﬁeld has a solvable galois group of order prime to the characteristic of k, if diﬀerent from zero.

Let k be a ﬁeld. The polynomial

m m 1 m 2 m f (x) = x x x − + x x − + ( 1) x , − 1 2 ··· − m

where x1,..., xm are algebraically independent over k, is said to be the general polynomial of the mth degree over k. It is so called, because every monic polynomial of degree m over k is obtained by specializing the values of x1,..., xm to be in k. Let L = k(x1,..., xm). Let y1,..., ym be roots of f (x) over L. Then

f (x) = (x y ) (x y ) − 1 ··· − m

and y1,..., ym are distinct. The splitting ﬁeld k(y1,..., ym) of f (x) over L is a galois extension whose galois group is isomorphic to S m. Hence, the general polynomial of the mth degree over k has a group isomorphic to the symmetric group on m symbols. But,S m is not solvable, if m > 4, so that in virtue of theorem 16, we have the theorem of Abel.

Theorem 17. The group of the general polynomial f (x) of the mth degree is isomorphic to the symmetric group S m on m symbols and hence, for m > 4, f (x) is not solvable by generalized radicals.

We shall now explicitly show how to obtain the roots of a polynomial of degree 4 in terms of generalized radicals. ≤ 160 Let f (x) be a general polynomial of degree m over k and let K be the splitting field. K/k has the group S m. Let y1,..., ym be the roots of f (x). Put 2 D = (yi y j) . < − Yi j Then D is fixed under all permutations in S and, hence, D k. m ∈ If we assume that the characteristic of k is , 2, then k( √D) is a galois extension of k and K/k( √D) has the galois group isomorphic to 8. Solvable extensions 139 the alternating group on m symbols. D called the discriminant of the polynomial f (x). Let us, first, consider the general polynomial of the second degree

f (x) = (x y )(x y ) = x2 x x + x . − 1 − 2 − 1 2 Then y1 + y2 = x1, y1y2 = x2. Suppose, now, that k has characteristic , 2. Then

D = (y y )2 = (y + y )2 4y y = x2 4x . 1 − 2 1 2 − 1 2 1 − 2 Also, y + y = x , y y = √D so that 1 2 1 1 − 2 ± x + √D x √D y = 1 , y = 1 − 1 2 2 2 and K = k( √D) is the splitting ﬁeld of f (x) and is a radical extension. Let, now, k have characteristic 2. Then x1 , o, since f (x) is separa- x y ble. Put x x instead of x. Then the polynomial x2 x + 2 has roots 1 1 2 x − x1 1 y2 x2 and . But this is a normal polynomial so that if λ = , then 161 x1 x12

= λ = λ + y1 x1 P , y2 x1 P x1 λ and thus k(P ) is a pseudo radical extension and is splitting ﬁeld of f (x). We shall now study cubic and biquadratic polynomials. Let, ﬁrst, k have characteristic , 2 or 3. Let m = 3 or 4 and

m m 1 m f (x) = x x x − + + ( 1) x − 1 ··· − m be the polynomial of the mth degree whose roots are y1,..., ym. x If we put x + 1 instead of x, we get a polynomial whose roots are m x x y 1 ,..., y 1 and which lacks the terms in xm 1. 1 − m m − m − 140 6. Special algebraic extensions

We shall, therefore, take the polynomial f (x) in the form

m m 2 m f (x) = x + x x − + ... + ( 1) x . 2 − m

If y1,..., ym are the roots, then

y1 + y2 + ... + ym = o.

2 Also , Dm = (yi y j) . A simple computation shown that i< j − Q D = 4x3 27x2 3 − 2 − 3 and

D = 16x4 x 4x3 x2 128x2 x2 + 144x x2 x 4 2 4 − 2 3 − 2 4 2 3 4 27x4 + 256x3. − 3 4

If K is the splitting field of f (x) over L = k(x1,..., xm) then L( √D) is the fixed field of the alternating group Am and L( √D)/L is a radical extension. In order to study the extension K/L( √D), let us, first, take the case m = 3. The symmetric group on 3 symbols, S 3, has the composition series S A (e). 3 ⊃ 3 ⊃ 162 K/L( √D) is thus a cyclic extension of degree 3. Let ρ be a primitive cube root of unity and let M = L( √D, ρ). Let N = KM be a composite of K and M. Then KM/M has degree 1 or 3, according as ρ is in K or not. In the first case, M = K. In the second case, KM/M is a cyclic extension of degree 3 over K and M contains the cube roots of unity. Thus KM = M(√3 ω), for some ω M. In order of determine this ω, we ∈ use Lagrange’s method. KM is the splitting field over M of the polynomial x3 + x x x . Let 2 − 3 y1, y2, y3 be roots of this polynomial. Put

2 ω = y1 + ρy2 + ρ y3 2 ω′ = y1 + ρ y2 + ρy3. 8. Solvable extensions 141

27 Then ω3 = x + 3 √D(ρ 1 ). Changing ρ into ρ2 we get ω 3. 2 3 − 2 ′ Hence a 3 27 1 ω = ρ x3 + 3 √D(ρ ) r 2 − 2 and we have a similar expression for ω . Here a o. In order to de- ′ ≥ termine a, we use the fact that ωω′ = 3x2 and so, choosing the root a − of unity ρ for ω arbitrarily, the root of unity in the expression for ω′ is uniquely determine. Now

y1 + y2 + y3 = o 2 y1 + ρy2 + ρ y3 = ω 2 y1 + ρ y2 + ρy3 = ω′ and since the matrix 1 1 1 1 ρ ρ2 1 ρ2 ρ      is non-singular, the values of y1, y2, y3 are uniquely determined K M is 163   · a radical of L and contains K. Consider, now, the polynomial of the fourth degree

f (x) = x4 + x x2 x x + x 2 − 3 4 whose roots are y1, y2, y3, y4 with y1 + y2 + y3 + y4 = o. The galois group of the splitting ﬁeld K/k is S 4. This has the composition series

S A B C (e). 4 ⊃ 4 ⊃ 4 ⊃ 4 ⊃

Let K1, Kb, Kc be the ﬁxed ﬁelds of A4, B4 and C4 respectively. Now A4 is the alternating group, B4 the group consisting of the permutations

(1), (12)(34), (13)(24), (14)(23) and C4 is the group of order 2 formed by

(1), (12)(34). 142 6. Special algebraic extensions

Ka = k( √D) and is of degree 2 over k. Now Kb/Ka is of degree 3 and is cyclic. Hence K = K (θ) where θ K is an element ﬁxed by B b a ∈ 4 but not by A4. Such as elements, for instance, is

θ1 = (y1 + y2)(y3 + y4).

θ1 has 3 conjugates θ1, θ2, θ3 obtained from θ1 by operating on θ1, by representatives of cosets of A4/B4. Thus

θ2 = (y1 + y3)(y2 + y4)

θ3 = (y1 + y4)(y2 + y3).

164 Consider the polynomial

ϕ(x) = (x θ )(x θ )(x θ ). − 1 − 2 − 3 It is ﬁxed under A4 and so, its coeﬃcients are in Ka. A simple computation shows that

ϕ(x) = x3 2x x2 + (x2 4x )x + x2. − 2 2 − 4 3 ϕ(x) is called the reducing cubic or the cubic resolvent. By the method adopted for the solution of the cubic, if ρ is a primitive cube root of unity and M = Ka(ρ), then Kb M, the composite, is a radical extension of k in which θ1, θ2, θ3 lie. Kc/Kb is of degree 2 and so Kc = Kb(α) where α is ﬁxed under B4, but not by C4. such as element is

α = y1 + y2. Now α2 = (y + y )2 = (y + y )(y + y ) = θ . Thus K = 1 2 − 1 2 3 4 − 1 c K ( √ θ ). Hence, if K M = K , then K (α) is a radical extension of k b − 1 b d d containing α. Put now T = Kd(α, β) where β = y +y = √ θ . Then T is a radical extension of k containing 1 3 − 3 K. The indeterminacy signs in taking the square roots of θ and θ − 1 − 3 can be ﬁxed by observing that

(y1 + y2)(y1 + y3)(y1 + y4) = x3. 8. Solvable extensions 143 we now have

y + y = θ , y + y = θ 1 2 − 1 3 4 − − 1 y + y = p θ , y + y = p θ 1 3 − 2 2 4 − − 2 y1 + y4 = p θ3, y2 + y3 = p θ3. − − − p p for which y1, y2, y3, y4 can be obtained. We have, hence, proved 165 Theorem 18. If K has characteristic , 2 or 3, then the cubic and biquadratic polynomials over k can be radicals.

3 2 Let us, now, assume that k has characteristic 3. Let x + x1 x + x2 x+ x3 be a cubic polynomial and K, its splitting field. K/k has the galois group S 3. Let L be the fixed field. K/k has the galois group S 3. Let L be the fixed field of A3. Then

L = k( √D) where D = (y y )2(y y )2(y y )2. 1 − 2 2 − 3 3 − 1 K/L is now a cyclic extension of degree 3 and since k has characteristic 3, K = L(ω), where

ω3 ω α = o − − for some α L. Thus K is a generalized radical extension. We shall ∈ now determine ω and α and therefrom, y1, y2 and y3. In order to do this, we have to consider two cases, x1 = o, and x1 , o. Let, ﬁrst, x1 = o. Then y1 + y2 + y3 = o. Let σ be a generator of the galois group fo K/L and let notation be so chosen that σ = σ = σ = y1 y2, y2 y3, y3 y1.

= + σ + σ2 = Since S K/Ly1 y1 y1 y1 o, by Hilbert’s theorem, there is a λ in K such that y = λσ λ. 1 − 144 6. Special algebraic extensions

Also 166

σ σ2 σ σ σ2 2 x2 = y1y2 + y2y3 + y3y1 = (λ λ)(λ λ ) = (λ + λ + λ ) . σ − − − X Since x2 , o (otherwise, polynomials is not separable), we see that x = t2, for some t L. The polynomial, therefore, has the form 2 − ∈ x3 t2 x + x . − 3 y Put now tx for x. Then the polynomial x3 x + x /t3 has roots 1 , 3 t y y − 2 , 3 . Let ω = x /t3. Then t t 3 = ω = ω + = ω + y1 t P , y2 t P t, y3 t P 2t and thus the roots are all obtained. The indeterminacy in the sign of t does not cause any diﬃculty; for, ω ω if we use ( t) instead of t, then, observing that = − , we see that − −P P ω ω ω y = t − , y = t − t, y = t − + t 1 − P 3 − P − 2 − P

so that y2 and y3 get interchanged. We now consider the case x1 , o. Put x + a instead of x. Then the new polynomial is

3 2 2 3 x + x1 x + x(x2 + 2x1a) + x3 + x2a + a + a .

Choose a so that x2 + 2x1a = o. For this value of a, µ = x3 + x2a + a2 + a3 , o; for, otherwise, the polynomial will be reducible and the roots are o, o, x . The roots of this new polynomial are y a, y a, − 1 1 − 2 − y3 a. − 1 Let now be written for x, then the polynomial is reduced to x3 + x x 1 1 1 1 x+ . We are in the previous case. The roots, now, are , , µ µ y1 a y2 a 1 x1 1 ν − − 167 . If t2 = , t L and ν = , then K = L( ) and the roots y a − µ ∈ µt3 P 3 − 8. Solvable extensions 145

y1, y2, y3 are given by

= + 1 = + 1 = + 1 y1 a ν , y2 a + ν , y3 a + ν t P t t P 2t t P

Since the characteristic is 3, the biquadratic polynomial can be taken in the form 4 2 x + x2 x + x3 x + x4.

The proof is similar to the old one except that Kb/Ka is a cyclic ω = ω extension of degree 3 and so Kb Ka P for a suitable in Ka. To find ω, we use the foregoing method. One finds that K is a generalized radical extension. We now consider the case where k has characteristic 2. In this case, 3 the cubic polynomial cab be taken in the form, x + x2 x + x3. Let y1, y2, y3 be the roots. Then y1 + y2 + y3 = o 2 + 2 = 2 and therefore y1 y2 y3 and so on. Put y y y ω = 1 + 2 + 3 , y2 y3 y1 y2 y3 y1 ω′ = + + . y1 y2 y3 Then 2 + 2 + 2 + 2 + 2 + 2 y1(y2 y3) y2(y3 y1) y2(y1 y2) ω + ω′ = = 1, y1y2y3 which shows that ω < k, because ,then, it will be symmetric and equal to ω′. Thus k(ω) is a quadratic subfield of K. A simple computation shows that ω and ω′ = ω + 1 are roots of

x2 x + x3. − 2 K/k(ω) is cyclic of degree 3 and one uses the method of Lagrange 168 to obtain a generalized radical extension. 146 6. Special algebraic extensions

Suppose now that f (x) is a polynomial of degree 4. Let f (x) = 4 3 2 x + x1 x + x2 x + x3 x + x4. We have to consider two cases. Let, ﬁrst, x1 = o. Then the roots y1, y2, y3, y4 satisfy

y1 + y2 + y3 + y4 = o.

As before, put

θ = (y + y )(y + y ), θ = ...,θ = 1 1 2 3 4 2 3 ··· The reducing cubic is then = 3 + 2 + ϕ(x) x x2 x x3.

Furthermore θ1 + θ2 + θ3 = o. Put now θ θ θ ω = 1 + 2 + 3 . θ2 θ3 θ1

Then ω is ﬁxed under A4 but not under S 4. Also Ka = k(ω) is a simple pseudo radical extension. Now Kb/Ka is a cyclic extension of degree 3 and has to be solved by Lagrange’s methods. Further, Kc/Kb is of degree 2. Put now = y3 = y4 ω1 , ω1′ . y1y2y4 y1y2y3 Then y2 + y2 + + + = 3 4 = (y1 y2)(y3 y4) = θ1 ω1 ω1′ . y1y2y3y4 y1y2y3y4 x4

We assume x3 , o. Then θ1 , o. If we put x ω = x4ω1 = 4 1′ ω2 , ω1′ , θ1 θ1 + = = 169 then ω2 ω2′ 1 and Kc Kb(ω2). In similar manner, K = Kc(ω3) where

x4 y1 ω3 = , . θ2 y2y3y4 8. Solvable extensions 147

Suppose, now, that x1 , o. Then, as before, we construct the ﬁeld Kb. In order to exhibit Kc as a pseudo radical extension of Kb, observe that y1 + y2 is ﬁxed under C4 but not under B4. Also

2 (y1 + y2) = (y1 + y2)(y3 + y4 + x1) = θ1 + x1(y1 + y2) which shows that y + y K = K 1 2 . c b x 1 ! y1 + y3 Similarly, K = Kc . x1 Our contentions are completely! established.

Chapter 7

Formally real ﬁelds

1 Ordered rings 170 A commutative ring R is said to be ordered if there is an ordering relation > (greater than) such that

(1) for every a R, a > o, a = o or a > o. ∈ − (1) a, b R, a > o, b > o a + b > o, ab > o. ∈ ⇒ We may then deﬁne a > b by a b > o. If a > b, then, for any c in − R, a + c > b + c and if c > o, ac > bc. If a > o, we shall say a is negative and if a > o, a said to be − positive. We denote “a is negative” by a < o. Let us denote, by P, the set of elements a R with a o. Then, ∈ ≥ from the deﬁnition or ordered ring, we have

A )P + P P 1 ⊂ A )P P P 2 · ⊂ A )P ( P) = (o) 3 ∩ − A )P ( P) = R, 4 ∪ − where P + P denotes the set of elements of R of the form a + b, a, b P; ∈ P denotes the set of elements a, a P P shall be called the set of − − ∈ · 149 150 7. Formally real ﬁelds

non-negative elements of R. The only element which is both positive and negative is zero. Clearly, if R is a any subset P of R satisfying the four conditions above again determine an order on R. Two elements a, b R, a , o , b are said to have the same or ∈ opposite signs according as ab > o or ab < o. 171 Let now R be an ordered ring with unit element 1. Let a R, a , o. ∈ Then a2 = a a = ( a) ( a), so that a2 > o for a , o in R. More · − · − generally, every ﬁnite sum of squares of elements of R is positive. These elements will be contained in the set of non-negative elements in every order of R. Since R has a unit element 1 and 12 = 1, we have 1 > o. Also, n 1 = 1 + 1 + 1, n times so n > o. This proves · ··· 1) An ordered ring with unit element has characteristic o. Let a , o, b , o be elements of R. Then a or a is positive. − Similarly b or b is positive. Hence ab or ab is positive which − − proves that ab , o. Therefore

2) An order ring is an integrity domain. We deﬁne an ordered ﬁeld to be an ordered ring whose non-zero elements form a multiplicative commutative group. We have

3) If k is an ordered ﬁeld, its positive elements form a multiplicative group. 1 1 1 For, if x k, x > o; then xx− > o. If x− > o, then xx− > o ∈ 1 −1 − which contradicts x− x > o. Thus x− > o and (3) is proved. Suppose R and R are two rings, R R . If R is ordered, clearly ′ ⊂ ′ ′ R is ordered by means of the induced order. If however, R is ordered, it may not be possible, in all cases, to extend this order to R′. However, in case, it is always possible, namely 4) If K is the quotient ﬁeld of an ordered ring R then the order in R can be uniquely extended to K.

172 Proof. Let R have an order. It is an integrity domain. Any elements a x K is of the form x = , a, b R, b , o. Let x , o. Then a , o. ∈ b ∈ 1. Ordered rings 151 define x > o by ab > o. Then this defines an order in K. In the first place, the definition does not depend on the way x is expressed in the a a′ form . Suppose x = . Then ab′ = a′b. Since b′ , o, multiplying b b′ both sides of this equality by bb′, we have

2 2 ab b′ = a′b′ b . · · a Since ab > o, b 2 > o, b2 > o, it follows that a b > o, that is ′ > o. ′ ′ ′ b a ′ In order to prove that A ,..., A are satisfies, let x = > o and 1 4 b a′ y = > o. Then ab > o and a′b′ > o. b′ ab + a b x + y = ′ ′ bb′ 2 2 Now (ab′ + a′b)bb′ = ab b′ + a′b′ b . Since ab > o, a′b′ > o, 2 2 · · b > o, b′ > o, it follows that (ab′ + a′b)bb′ > o or x + y > o. In similar manner, xy > o. Suppose x o and y o and x + y = o; then x = o, y = o. For, if a c ≥ ≥ ad + bc x = , y = , then ab o, cd o and x + y = = o, so that b d ≥ ≥ bd ad + bc = o. Thus abd2 + cdb2 = o, which means that since all elements are in R, ab = o, cd = o, i.e. a = o = c. ab If x R, than x = . If x = o, then ba2 = o or b = o, so that the ∈ a order coincides on R with the given order in R. That the extension is unique can be, trivially, seen. Since every ordered field has characteristic zero, it contains a sub- 173 field isomorphic to Γ, the rational number field Γ has thus induced order. We shall now prove (5) Γ can be ordered in one way only. For, if Z denotes the set of integers, then Γ is the quotient field of Z. On Z, there is only one order since 1 > o and hence n = 1+1+ +1 > o. ··· Thus, all natural integers have to be positive. 152 7. Formally real fields

2 Extensions of orders

Hereafter, we consider ordered fields k. Our main task will be the study of extensions of orders in k to extension fields K of k. For this purpose, we introduce the notion of a positive form on k. m 2 Let k be an ordered field. A polynomial ai xi , ai k, is said to be i=1 ∈ = an m-ary form over k. It is said to be positivePif ai > o, i 1,... m. An m-ary form is said to represent a k, if there exist α ,...,α in k such ∈ 1 m that 2 = aiαi a. Xi Clearly a positive form represents zero, only if α1,...,αm = o. Let k be an ordered field and K/k, an extension field. We shall prove Theorem 1. K has an order extending the order in k, if and only if, every positive form over k is still positive in K.

174 Proof. We have only to prove the suﬃcient. To this end, consider the family M of subsets Mα of K having the following properties. Denote, { } by S , the set of elements in K of the form

2 aiαi , Xi a k and a > o and α K. (α can be all zero also). i ∈ i i ∈ i Then

1) Mα S 2) Mα + Mα Mα ⊃ ⊂ 3) Mα Mα Mα 4) Mα ( Mα) = (o). ⊂ ∩ − This family is not empty, since S satisfies this condition. In the usual way, we make M a partially ordered set and obtain a maximal set P. We have now to show that P ( P) = K ∪ − and then P will determine an order. Let x , o be an elements of K which is not in P. Define Q = P xP − 2. Extensions of orders 153 as the set of elements of the form a xb, a, b P. Obviously Q satisfies − ∈ (1). To see that Q satisfies (2), observe that if a xb, c xd are in Q − − then (a xb) + (c xd) = (a + c) x(b + d); − − − but a, b, c, d being in P which is an element of M, a + c, b + d are in P. In a similar way Q satisfies (3). That Q satisfies (4) can be seen as follows: Let a xb and c xd be in Q with a, b, c, d, in P. Let − − (a + c) x(b + d) = o. Then b + d = o. For, if b + d , o, then − a + c 1 x = = (a + c)(b + d) b + d (b + d)2 and so is an element of P, which is a contradiction. Hence b + d = 0 175 and, therefore, a + c = 0. Since a, b, c, d are all in P, it follows that a = b = c = d = 0. Thus Q M. But Q P so that, by maximality of ∈ ⊃ P, Q = P. This means that x P or x P. The theorem is therefore − ∈ ∈− proved. If Γ is the field of rational numbers and n is a positive integer, then a n = 1 + 1 + + 1. If r = is a positive rational number, then ··· b a ab 1 + 1 + 1 = = ··· , b b2 b2 so that every positive rational number is a sum of squares. This shows Γ 2 + + 2 that every positive form over can be put in the form x1 xn. 2 = ··· = If k is an ordered field then αi 0, αi k implies that αi 0, i ∈ 2 i = 1,.... On the other hand, if kPhas the property that α = 0, αi k i ∈ implies α = 0, then n = 1+1+ +1 , 0 so that k has characteristic zero. i ··· Since every positive form over Γ is essentially of the typeP x2 + + x2, 1 ··· n we have the 2 = = = Theorem 2. k is ordered if and only if αi 0, αi k implies αi 0, i = i 1, 2,.... P 2 = It is obvious that, if, in a field k, αi 0, with α1,α2,... not all i zero, then P 1 = β 2, − i Xi 154 7. Formally real fields

176 β k. A field in which 1 is not a sum of squares is called a for- i ∈ − mally real field. From theorem 2, it follows that formally real fields are identical with ordered fields. We shall, now, prove the following application of theorem 1.

Theorem 3. Let k be a formally real ﬁeld with a given order and f (x), an irreducible polynomial in k[x]. Let a, b be two elements in k such that f (a) f (b) < 0. Suppose α is a root of f (x) in Ω, an algebraic closure of k. Then K = k(α) is ordered with an order which is an extension of the given order in k.

Proof. Let f (x) be of degree n. Then every element in k(α) is a polynomial in α of degree n 1 with coeﬃcients in k. If k(α) is not ordered − with an order extending that in k, then there is a positive form in k which represents 1. That is, − 1 = a ϕ (α) 2, − i{ i } Xi

ai > 0 in k. This means that, in k[x],

1 + a ϕ (x) 2 = f (x)ψ(x). i{ i } Xi Since f (x) has degree n and left side has degree 2n 2, ψ(x) has, ≤ − at most, the degree n 2. − We now use induction on n. If n = 1, these is nothing to prove. Assume theorem proved for n 1 instead of n. Let g(x) be in irreducible − factor of ψ(x). Then g(x) has degree n 2. Now ≤ − 0 < 1 + a ϕ (a) 2 = f (a)ψ(a) i{ i } Xi 0 < 1 + a ϕ (b) 2 = f (b)ψ(b). i{ i } Xi 177 Since f (a) f (b) < 0, it follows that ψ(a)ψ(b) < 0. Therefore, at least one irreducible factor, say g(x), of ψ(x) has the property g(a)g(b) < 0. 2. Extensions of orders 155

If β is a root of g(x) in Ω, then g(β) = f (β) = 0. But by induction hypothesis, k(β) has an order extending the given order in k. Hence

0 = 1 + a ϕ (β) 2 > 0, i{ i } Xi which is a contradiction. Thus our theorem is completely proved. Suppose k is an ordered ﬁeld, let us denote, by a , the absolute value | | of a k by ∈ 0 if a = 0 = a a if a > 0 | |   a if a < 0. −  It is then easy to prove that

ab = a b , | | | | | | a + b a + b . | | ≤ | | | | Let f (x) = xn + a xn 1 + + a be a polynomial in k[x]. Put 1 − ··· n M = max(1, a + + a ). If t , 0 k and t > M, then | 1| ··· | n| ∈ | | n 1 n t− f (t) = 1 + a t− + + a t− > 0 1 ··· n which shows that tn and f (t) have the same sign. Suppose now f (x) is irreducible and of odd degree. Then, if M is deﬁned as above, f (M) f ( M) < 0. − Therefore, by theorem 3, if β is a root of f (x) in an algebraic closure of k, then k(β) has an order extending that in k. If a k and a > 0, then the polynomial x2 a changes sign in k. 178 ∈ − For, (a + 1)2 > a and a < 0. Thus − ((a + 1)2 a)(0 a) < 0. − − Therefore, k( √a) has an order extending the given order in k. 156 7. Formally real ﬁelds

3 Real closed ﬁelds

We had seen above that, under certain circumstances, an order in a field k can be extended to a finite algebraic extension of k. We shall consider, now, a class of fields called real closed fields defined as follows: k is said to be real closed, if

1) k is ordered

2) k has no proper algebraic extension K with an order extending that in k.

Before we establish the existence of such ﬁelds, we shall obtain some of their properties. We ﬁrst prove

Theorem 4. For a formally real ﬁeld k, the following properties are equivalent:

1) k(i) is algebraically closed, i being a root of x2 + 1.

2) k is real closed.

3) Every polynomial of odd degree over k has a root in k and every positive element of k is a square in k.

Proof. 1 2. k(i) being of degree 2 over k, there are no intermediary ⇒ ﬁelds, so that, k being ordered, and k(i) being algebraically closed, k has no ordered algebraic extension. 179 2 3. Suppose f (x) is a polynomial of odd degree. Then it ⇒ changes sign in k. Hence an irreducible factor of f (x), also of odd degree, changes sign in k. If α is a root of this irreducible factor, then k(α) is ordered with an order extending that in k. Hence α k. If a > 0 in k, ∈ then x2 a changes sign in k. Thus √a k. − ∈ 3 1. The poof of this part consists of three steps. Firstly, every ⇒ element of K = k(i) is a square. For, let a + bi be an element of K, a, b k. Put ∈ a + ib = (c + id)2 3. Real closed ﬁelds 157 where c and d have to be determined in k. Since 1, i form a base of K/k we get c2 d2 = a, 2cd = b. − Therefore, (c2 + d2)2 = a2 + b2. But a2 + b2 > 0 in k and, since every positive element is a square, there is a λ> 0 in k such that

c2 + d2 = λ.

Solving for c2 and d2, we have

λ + a λ a c2 = , d2 = − . 2 2

But, since λ2 = a2 + b2, it follows that λ a , λ b . Therefore λ + a λ a ≥ | | ≥ | | 0, − 0. Therefore 2 ≥ 2 ≥

λ + a λ a c = , d = − ± r 2 ± r 2 exist in k. The arbitrariness in the signs of c and d can be ﬁxed from the fact that 2 cd = b.

This proves that every quadratic polynomial over k has a root in 180 K. For, if ax2 + bx + c k[x], then, in an algebraic closure of K, ∈ b √b2 4ac − ± − are its roots (a , 0). But √b2 4ac K. 2a − ∈ The second step consists in showing that every polynomial in k[x] has a root in K. Let f (x) be in k[x] and let N be its degree N = 2n q, q · odd. We shall use induction on n. If n = 0, then N = q, and so whatever odd number q be, f (x) has a root already in k. Let us, therefore, assume n 1 proved that every polynomial of degree 2 − q′, where q′ is odd, with coeﬃcients in k has a root in K. Let f (x) be of degree N = 2n q, q odd. · Let α1,...,αt be the distinct roots of f (x) in an algebraic closure of K. Let µ k be an element to be suitably chosen later. Put ∈

λi j(µ) = λi j = αi + α j + µαiα ji, j = 1,..., t, i , j. 158 7. Formally real ﬁelds

Consider now the polynomial

ϕµ(x) = (x λi j). , − Yi j N(N 1) This has degree − = 2n 1 q , q odd. Also, by every per- 2 − · ′ ′ mutation of the symbols 1,..., t, the polynomial goes over into itself. Thus ϕµ(x) k[x]. Since its degree satisfies the induction hypothesis, ∈ for every µ k, there is an i and a j such that λ (µ) K. Since k is ∈ i j ∈ an infinite field, there exist µ, µ′, µ , µ′ and both in k such that λi j(µ) and λi j(µ′) for two integers i and j are in K. This means that αiα j and, 181 hence, α + α are in K. The polynomial x2 x(α + α ) + α α is a i j − i j i j polynomial in K[x]. By what we proved above, both its roots are in K. Thus our contention is proved. The third step consists in proving that every polynomial in K[x], has a root in K. For, let f (x) be a polynomial in K[x]. Let σ be the generating automorphism of K/k. It is of order 2. Denote by f σ(x) the polynomial obtained from f by applying σ on the coefficients of f . Then ϕ = f (x) f σ(x) is a polynomial in k[x]. The second step shows that ϕ has a root α in K. Furthermore if α is a root of ϕ, ασ is also a root of ϕ, so that either α is a root of f (x) or ασ is a root of f (x). We have thus proved theorem 4 completely. We deduce from this an important corollary due to Artin and Schreier.

Corollary 1. If Ω is an algebraically closed ﬁeld and K a subﬁeld such that 1 < (Ω : K) < , then K is real closed. ∞ Proof. We had already proved that K(i) =Ω and that K has characteristic zero. By virtue of theorem 4, it is enough to prove that K is formally real. Every element of Ω is of the form a + ib, a, b K. Also ∈ a + ib = (c + id)2,

for c, d in K, since Ω is algebraically closed. Thus

a2 + b2 = (c2 + d2)2. 3. Real closed ﬁelds 159

Hence every sum of two squares and, hence, of any number of squares is a square. Therefore

1 = a2 − i Xi is impossible in K. By theorem 2, therefore, K is formally real. 182

This proves that the real closed fields are those and only those which are such that their algebraic closures are finite over them. We have again Corollary 2. A real closed field has only one order. For, the set of positive elements coincides with the set of squares of the elements of the field. This shows that, if on ordered field has two distinct orders, it has algebraic extensions which are ordered. It must be remembered that if a field has only one order, it is not necessarily real closed. The rational number field, for example, has only one order. Suppose k is a real closed field. Then every irreducible polynomial in k[x] is of degree one or two. Suppose f (x) is a polynomial in k[x] and a, b in k such that f (a) f (b) < 0. Then one of the irreducible factors ϕ(x) of f (x) must have the property that ϕ(a)ϕ(b) < 0. If α is a root of ϕ(x), then k(α) is ordered. But, k being real closed, α k. ϕ(x) must be a linear polynomial. Thus ∈ ϕ(x) = x α and − (a α)(b α) < 0 − − which means that α lies between a and b. Hence the Theorem 5. If k is a real closed field, f (x) a polynomial in k[x], a, b in k such that f (a) f (b) < 0, then there is a root α of f (x) in k between a and b.

Furthermore, we had seen that there is an M in k depending only 183 on the coeﬃcients of f (x) such that f (a) has the same sign as an, n = deg f (x), if a > M. This shows | | 160 7. Formally real ﬁelds

All the roots of f (x) that lie in k lie between M. ± Let k be a real closed ﬁeld and f (x) a polynomial in k[x]. Let f ′(x) be its derivative. Put ϕ0 = f , ϕ1 = f ′. Since k[x] is a Euclidean ring, deﬁne, by the Euclidean algorithm, the polynomials

ϕ = A ϕ ϕ 0 1 1 −− 2 ϕ = A ϕ ϕ 1 2 2 −− 3 ......

ϕr 1 = Ar ϕr. −

It is then well-known that ϕr(x) is the greatest common divisor of ϕ0 and ϕ1. The sequence ϕ0,ϕ1,...,ϕr of polynomials in k[x], is known as the Sturmian polynomial sequence. Let a k be such that ϕ (a) , 0. Then φ (a) , 0. Consider the ∈ 0 r set of elements ϕ0(a), ϕ1(a),...,ϕr(a) in k. The non-zero ones among them have a sign. Denote, by ω(a), the number of changes of sign in the sequence of elements,

ϕ0(a),ϕ1(a),...,ϕr(a),

in this order, taking only the non-zero elements. A very important theorem due to Sturm is

Theorem 6. Let b and c be two elements of k, b < c and ϕ0(b) , 0, 184 ϕ0(c) , 0. Let ω(b) and ω(c) denote the number of changes of sign in the Sturmian sequence for the values b and c. Then f (x) has precisely ω(b) ω(c) distinct roots in k between b and c. −

Proof. Since ϕr(b) , 0, ϕr(c) , 0, we may divide all the Sturmian polynomials by ϕr(x) and obtain the sequenceϕ ¯ 0(x),ϕ ¯ 1(x),... ϕ¯r 1(x), − ϕ¯ r(x)(= 1). Nowϕ ¯0(x) has no multiple roots. For, if α is a root of f (x) of multiplicity t, then

ϕ (x) = (x α)tψ (x), ψ (α) , 0 0 − 1 1 t 1 t ϕ (x) = t(x α) − ψ (x) + (x α) ψ′ (x) 1 − 1 − 1 3. Real closed ﬁelds 161 so that (x α)t 1 is the highest power of of x α, that divides ϕ (x). − − − 1 Hence

ϕ¯ (x) = (x α)ψ (x) 0 − 2 ϕ¯ (x) = tψ (x) + (x α)ψ (x), 1 2 − 3

ψ2(x) and ψ3(x) being polynomials over k. Note thatϕ ¯1(x) is not the derivative ofϕ ¯ 0(x). We shall drop the ‘bars’ on the ϕ′ s and write them as ϕ0,ϕ1,...,ϕr 1,ϕr = 1 and ϕ0 having no multiple roots. Note that ω(b) or ω(c) is not− altered by doing the above.

The finite number of polynomials ϕ0,ϕ1,...,ϕr 1 have only finitely many roots between b and c. By means of these roots,− we shall split the interval (b, c) into finitely many subintervals, the end points of which are these roots. We shall study how the function ω(a) changes as a runs from b to c.

1) No two consecutive functions of the Sturmian series ϕ0(x), 185 ϕ1(x),...,ϕr 1(x) can vanish at one and the same point, inside the − interval (b, c). For, suppose b < a < c and ϕi(a) = 0 = ϕi+1(a), 0 < i + 1 < r. Then

ϕ (x) = A ϕ + (x) ϕ + (x) i i i 1 − i 2

so that ϕi+2(a) = 0, and, so on, ﬁnally ϕr(a) = 0. But ϕr(a) = 1. 2) Inside any one of the intervals, each function keeps a constant sign; for, if any function changed sign then, by theorem 5, there would be a zero inside this interval. Let d denote an end point of an interval and L and R the intervals to the left of d and to the right of d, having d as a common end point.

3) Suppose d is a zero of ϕ1 for 0 < 1 < r. Then

ϕl 1 = Alϕl ϕl+1, − −

so that ϕl 1(d) = ϕl+1(d) and none of them is zero by (1). Because − − of (2), ϕl 1 has in L the same sign as at d. Similarly in R. The same − 162 7. Formally real ﬁelds

is true of ϕl+1. Thus, whatever sign ϕl might have in L and R, the function ω(a) remains constant when a goes from L to R crossing a zero of ϕl, 0 < 1 < r.

4) Let now d be a zero of ϕ0. Then d is not a zero of ϕ1. We have

ϕ (x) = (x d)ψ(x) 0 − ϕ (x) = mψ(x) + (x d)ψ (x), 1 − 1

where m is an integer > 0, ψ(x) and ψ1(x) are polynomials over k, and ψ(d) , 0. At d, ϕ1(d) has the same sign as ψ(d).

186 In L, ϕ has the sign of ϕ (a) = (a d)ψ(a). But a d < 0, so that 0 0 − − ϕ has the sign of ψ(a). In L, ϕ (a) has the same sign as ψ(d). Also 0 − 1 ψ(x) has no zero in L. Hence ϕ1(a) has the same sign as ψ(a). Therefore in L, ϕ0 and ϕ1 have opposite signs. In R, exactly the opposite happens, namely ϕ (a) = (a d). ψ(d), a d > 0. Hence ϕ and ϕ have the 0 − − 0 1 same sign in R. Thus ω(a) is lessened by 1, whenever a crosses a zero of ϕ0(x) and remains constant in all other cases. Our theorem is, thus, completely proved. We make the following remark.

Remark. Suppose k is a formally real ﬁeld and

n n 1 f (x) = x + a x − + + a , 1 ··· n a polynomial in k[x]. Let, as before, M = max(1, a + + a ). | 1| ··· | n| Suppose there exists a real closed algebraic extension K of k with an order which is an extension of the given order in k. f (x) can, then, be considered as a polynomial in K[x] and, as seen earlier, f (x) has no roots in K outside the interval ( M, M). The number of these distinct − roots is thus independent of K. We now prove

Theorem 7. Let k be an ordered field, Ω its algebraic closure. Suppose there exist two real closed subfields K, K∗ of Ω/k with orders extending the given order in k. Then K and K∗ are k isomorphic. 3. Real closed fields 163

Proof. 1) Let f (x) be a polynomial in k[x] and α1,...,αt the distinct 187 roots of f (x) in K. Let α1∗,...,αt∗ be the distinct roots of f (x) in = = K∗. Put L k(α1,...,αt) and L∗ k(α1∗,...,αt∗). Since L/k is ﬁnite, L = k(ξ) for some ξ in K. Suppose ϕ(x) is the minimum polynomial of ξ in k[x]. Let ξ∗ be a root of ϕ(x) in K∗. Then k(ξ) and k(ξ∗) are k-isomorphic. If ρ is this isomorphism, then ρξ = ξ∗. But k(ξ) = L = k(α1,...,αt). Hence ρα1,...,ραt will be distinct roots of f (x) in K∗. Thus L∗ = k(ξ∗). Hence

L L∗. ≃

2) Suppose ϕ(x) is any polynomial in k[x], β1,...,βs its distinct roots in K. Let β1∗,...,β∗s be the corresponding roots in K∗. Consider all the positive quantities among β β , i , j. Their square roots exist i − j in K. Let ψ(x) be a polynomial in k[x], among whose roots are these

square roots and let δ1,...,δg be the roots of ψ(x) in K, δ1∗,...,δg∗ the corresponding roots in K∗. Then, from above, F = k(β ,...,β , δ ,...,δ ) 1 s 1 g ≃ k(β∗,...,β∗, δ∗,...,δ∗) = F∗. ≃ 1 s 1 g Let τ be this isomorphism. Let notation be such that = τ(βi) βi∗ = τ(δi) δi∗.

Suppose β > β . Then β β > 0 so that β β = δ2, for some t. i j i − j i − j t Also, τ(β β ) = β β . But i − j i∗ − ∗j 2 2 τ(β β ) = τ(δ ) = δ∗ . i − j t t Hence β β = δ 2 > 0, which proves that i∗ − ∗j t∗

βi∗ > β∗j.

The isomorphism τ between F and F∗ preserves order between the 188 roots of ϕ(x) in K. 164 7. Formally real ﬁelds

3) In order, now, to construct the isomorphism between K and K∗, we remark that any such isomorphism has to preserve order. For, if σ is this isomorphism and α > β in K, then α β = δ2 so that − σ(α β) = σ(δ2) = (σ(δ))2 > 0 − so that σα > σβ. We shall, therefore, construct an order preserving map which we shall show to be an isomorphism.

4) Let α be an element in K, h(x) its minimum polynomial over k. Let α1,...,αt be the distinct roots of h(x) in K and let notation be so = chosen that α,< α2 <...<αt. Let α αi. Let α1∗,...,αt∗ be the distinct roots of h(x) in K∗ and, again, let the notation be such that

α1∗ <α2∗ <...<αt∗.

Deﬁne, now, σ on K by = σα αi∗. Let α and β be two elements of K and let f (x) be a polynomial in k[x]n whose roots in K are α, β, α + β, αβ,.... Construct the ﬁelds F and F∗ and the k-isomorphism τ of F on F∗. Since τ preserves order of roots of f (x), it preserves order of roots of the factor of f (x) which has α as its root. Similarly of the factor having β as a root. Hence

τ(τ) = σ(α), τ(β) = σ(β), τ(α + β) = σ(α + β), τ(αβ) = σ(αβ).

189 Hence σ(α + β) = σα + σβ, σ(αβ) = σα σβ. Thus σ is an isomor- · phism of K into K∗. We have similarly an isomorphism σ∗ of K∗ into K. Thus σ σ∗ is identity on K∗. Hence K and K∗ are k-isomorphic. ·

Corollary. The only automorphism of K over k is the identity. We shall now prove the theorem regarding the existence of real closed ﬁelds, namely 3. Real closed ﬁelds 165

Theorem 8. If k is an ordered ﬁeld with a given order and Ω, its algebraic closure, there exists in Ω, upto k-isomorphism, only one real closed ﬁeld K with an order extending the given order in k.

Proof. Let V be the family of formally real subfields of Ω/k which have an order extending that in k. V is not empty, since k V. We partially ∈ order V by inclusion. Let kα be a totally ordered subfamily of V. Let { } K0 = kα. Then K0 is a field which is contained in V. This can be α easilyS seen. By Zorn’s lemma, there exists a maximal element K in V. K has an order extending the order in k. To prove that K is real closed, let f (x) be a polynomial of odd degree over K. It changes sign. Therefore there is an irreducible factor, also of odd degree, which changes sign in K. This factor must have a root in K. Else, by theorem 3 there exists an algebraic extension with an order extending that in k. This will contradict maximality of K. In a similar way, every positive element of K is a square. By theorem 4, K is real closed. If K and K∗ are two real closed subfields with orders extending that in k, then, by theorem 7, they 190 are k-isomorphic.

Suppose now that k is a perfect field and Ω, its algebraic closure. Let G be the galois group of Ω/K. If G has elements of finite order (not equal to identity), let K be the fixed field of the cyclic group generated by one of them. Then (Ω : K) is finite and by Artin-Schreier theorem this order has to be two. Thus any element of finite order in G has to have the order two. Furthermore, in this case, k is an ordered field. On the other hand, if k is an ordered field and Ω, its algebraic closure, there exists, then by theorem 8, a real closed subfield of Ω/k, say K. This means that G has an element of order 2. Moreover no two elements of order 2 commute. For if α, β are of order 2 and commute, then 1, α, β, αβ is a group order 4 which must have a fixed field L such that (Ω : L) = 4. This is impossible, by Artin-Schreier theorem. Hence the

Theorem 9. If k is a perfect field, Ω its algebraic closure and G the galois group of Ω/k then G has elements of finite order if and only if k is formally real. Also, then, all these elements have order 2 and no two of them commute. 166 7. Formally real fields

4 Completion under an order

This implies that a b a b . | | − | | ≤ | − | 191 Also a a is a homomorphism of k into the set of positive ele- → | | ∗ ments of k. The function defines a metric on the field k. We define | | a Cauchy sequence in k to be a sequence (a1,..., an,.) of elements of k such that for every ε> 0 in k, there exists n0, an integer such that

a a < ε, n, m > n . | n − m | 0

Obviously, if m > n0 + 1,

a = a a a <ε + a , | m| | m − no − no| | n0 |

so that all elements of the sequence from n0 onwards have a value less than a certain positive element of k. A Cauchy sequence is said to be a null sequence if, for every ε > 0 in k, there is an integer n0 = n0(ε) such that

a < ε, n > n . | n | 0 The sum and product of two Cauchy sequence is deﬁned as follows:-

(a1, a2,...) + (b1, b2,...) = (a1 + b1, a2 + b2,...) (a , a ,...) (b , b ,...) = (a + b , a + b ,...) 1 2 − 1 2 1 1 2 2

and it is easy to verify that the Cauchy sequences in k form a ring R and the null sequences, an ideal Y of R. We assert that Y is a maximal 4. Completion under an order 167

ideal. For, let (a1,...) be a Cauchy sequence in k which is not a null sequence. Then there exists a λ> 0 in k and an integer n such that

a > λ, m > n. (1) | m | For, if not, for every ε > 0 and integer n, there exist an inﬁnity of 192 m > n for which a < ε. Since (a ,...) is a Cauchy sequence, there | m | 1 exists n0 = n0(ε) such that

a a < ε, m , m > n . | m1 − m2 | 1 2 0 Let m > n such that a <ε. Then, for all m > m , 0 0 | m0 | 0 a a a + a < 2ε | m| ≤ | m − m0 | | m0 which proves that (a1,...) is a null sequence, contradicting our assumption. Let 1 be the unit element of k. Then the sequence (1, 1,...) is a Cauchy sequence and is the unit element in R. Let c = (a1,...) be in Y = 1 R but not in . Let m be deﬁned as in (1). Then c1 (0, 0,... 0, am− , 1 am−+1,...) is also a Cauchy sequence. For, let ε > 0 and n, an integer so that a a < ε, n , n > n . | n1 − n2 | 1 2 0 Then 1 1 1 ε = a a < , a − a a a n2 − n2 λ2 n1 n2 n1 n2 | | | | if n , n > max( m, n ). Also cc = (0, 0, 0,..., 1, 1,...) which proves 1 2 0 1 that cc (1, 1, 1,... (mod Y ). This proves that Y is a maximal ideal 1 ≡ and, therefore,

1) R/Y is a ﬁeld k¯.

k¯ is called the completion of k under the given order. For every a k, consider the Cauchy sequencea ¯ = (a, a,...). Then a a¯ is a ∈ → non-trivial homomorphism of k into k¯. Since k¯ is a field, this is an isomorphism. k¯ thus contains a subfield 193 isomorphic to k. We shall identify it with k itself. 168 7. Formally real fields

(2) k¯ is an extension ﬁeld of k.

We shall now make k¯ an ordered ﬁeld with an order which is the extension of the order in k. To this end, deﬁne a sequence c = (a1,..., ) of R to be positive if there is an ε> 0 and an n0 = n0(ε) such that

an > ε, n > n0.

If b = (b1,...) is a null sequence, then

b < ε/ , n > m = m(ε). | n| 2

If p > max(m, n0), then ap + bp > ε/2 which shows that a + b is also a positive sequence. The deﬁnition of positive sequence, therefore, depends only on the residue class mod Y . Let P denote the set of residue classes containing the positive sequences and the null sequence. We shall show that P determines an order in k¯ First, let c1 and c2 be two positive sequences. There exist ε1 > 0, ε2 > 0 and two integers n1 and n2 such that

ap > E1, p > n1,

bp > E2, p > n2;

if p > max(n1, n2), then

ap + bp >ε1 + ε2,

194 so that c + c is a positive sequence or P + P P. In a similar manner, 1 2 ⊂ PP P. ⊂ Let now c = (a ,...) be not a null sequence. Then c and c cannot 1 − both be positive. For then, there exist ε,ε′ both positive and integers n, n′ such that ap > ε, p > n a >ε , p > n. − p ′   4. Completion under an order 169

If p > max(n, n1), then

0 = a a >ε + ε′ > 0, p − p which is absurd. Thus P ( P) = (0). ∩ − Suppose now that c = (a1,...) is not a null sequence. Suppose it is not positive. Then c = ( a , a ,...) is a positive sequence. For, − − 1 − 2 otherwise, for every ε and every n,

a < ε, p > n. − p But, c being not positive, we have

ap < ε, p > n1.

c being a Cauchy sequence, there exists n0 with

a a < ε, p, q > n . | p − q| 0

Hence, if p > max(n0, n1),

a a a + a < 2ε | p| ≤ | p − q| | q| which means that c is a null-sequence. This contradiction proves that

P ( P) = k¯. ∪ − We therefore see that 195

(3) k¯ is an ordered ﬁeld with an order extending the order in k.

We shall denote the element (a1,...) in k¯ bya ¯ and write

a¯ = lim an. n →∞ Then, clearly, given any ε> 0, there exists n0 such that the element a¯ (a ) in k has all its elements, from some index on, less than ε in − n0 absolute value. This justiﬁes our notation. One clearly has

lim an + lim bn = lim an + bn. 170 7. Formally real ﬁelds

lim an. lim bn = lim an + bn. lim a a n = lim n , lim bn bn

if (b1, b2,...) is not a null sequence. If Γ is the rational number field, it is ordered and the completion Γ¯ under this unique order is called the real number field. This method of construction of the real number field goes back to Cantor.

5 Archimedian ordered ﬁelds

Ordered ﬁelds can be put into two classes. A ﬁeld k is said to be archimedian ordered if, for every two elements a, b in k, a > 0, b > 0 there exists an integer n such that

nb > a

196 and, similarly, there is an integer m with ma > b. We may state equivalently that, for every a > 0, there is an integer n with n > a. A ﬁeld k is said to be non-archimedian ordered if there exists a > 0 such that a > n for every integer n. Γ, the rational number ﬁeld is archimedian ordered. Consider, now, the ring Γ [x] of polynomials. For

n n 1 f (x) = a x + a x − + + a , a , 0 0 1 ··· n 0

deﬁne f (x) > 0, if a0 > 0 as a rational number. That this is an order can be veriﬁed easily. Also the order is non-archimedian because

x2 + 1 > n, 5. Archimedian ordered ﬁelds 171 for every integer n. This order can be extended to Γ(x). This examples shows that an archimedian order in k can be extended into a non-archimedian order in an extension ﬁeld which is transcendental over k. That this is not possible in algebraic extensions is shown by Theorem 10. If k is archimedian ordered and K/k is algebraic with an order extending the order in k, the extended order in K is again archimedian.

n n 1 Proof. Let α be > 0 in the extended order in K. Let f (x) = x +a1 x − + a be the minimum polynomial of α in k[x]. Consider the quantities 197 ··· n 1 a , i = 1,..., n. They are in k and since k and is archimedian ordered, − i there exists an integer t > 0 such that

t > 1 a , i = 1,..., n. − i We now assert that α< t. For if α t, then ≥ n n 1 0 = f (α) = α + a1α − + ... + an n n 1 α + (α − + ... +)(1 t) ≥ − n n 1 α + (α − + ... + 1)(1 α) = 1, ≥ − which is absurd. Our theorem is proved. We had already introduced the real number ﬁeld. It is clearly archimedian ordered. In fact, the completion of an archimedian ordered ﬁeld is archimedian. We can prove even more, as shown by

Theorem 11. Every complete archimedian ordered field is isomorphic to the field of real numbers. Proof. Let K be an archimedian ordered field. It has a subfield isomorphic to Γ, the field of rational numbers. We shall identify it with Γ it self. Let k¯ be the completion of k. Then

k¯ Γ¯. ⊃ 172 7. Formally real ﬁelds

In order to prove the inequality the other way round, let R be the ring of Cauchy sequences in k and Y , the maximal ideal formed by null sequences. We shall show that every residue class of R/Y can be represented by a Cauchy sequence of rational numbers. Therefore, let c = (a1, a2,...) be a positive Cauchy sequence in k. Since k is archime- 198 dian ordered, there exists, for every n greater than a certain m, an integer λn such that λn < nan < 1 + λn which means that λ 1 a n < . n − n n

λm λm+1 Let d = (0, 0, 0,... , ,... ) be a sequence of rational num- m m + 1 bers. Let ε > 0 be any positive quantity in k. There exists, then, an integer t such that εt > 1, since k is archimedian ordered. Then, for n > t and m, λ a n <ε n − n which shows that c d Y , which proves that k¯ Γ¯. − ∈ ⊂ We shall now prove the important Theorem 12. The real number field Γ¯ is real closed. Proof. We shall show that every polynomial which changes sign in Γ¯ has a root in Γ¯. Let b < c be two elements of Γ¯ and f (x) a polynomial in Γ¯[x] with f (b) > 0 and f (c) < 0. We define two sequences of rational numbers b0, b1, b2, ... and c0, c1, c2,... in the following way. Define the integers λ0, λ1, λ2,... inductively in the following manner. λ0 = 0, and having defined λ0, λ1,...,λn 1, λn is defined as the largest integer such that − (c b)λ f (b + − n ) 0. 2n ≥ (c b)λ We shall put b = b + − n and since we want b to be an n 2n n increasing sequence, we shall, in addition, require that

λn λn 1 + 1. 2 ≤ − 5. Archimedian ordered ﬁelds 173

199 That such a sequence can be found is easily seen. Put now c b c = b + − . n n 2n

Then c0, c1,... is a decreasing sequence

b b b + c + c c. ≤ n ≤ n 1 ≤ n 1 ≤ n ≤

Also (bn) and (cn) are Cauchy sequences; for, c b b + b < − . | n 1 − n | 2n+1 Furthermore, the sequence (c b ) is a null sequence. For, n − n c b c = b = − n n 2n There is, therefore, an α between b and c such that

α = lim bn = lim cn.

By deﬁnition of λ , f (c ) < 0 and f (b ) o. n n n ≥ Therefore

f (α) = lim f (b ) 0, f (α) = lim f (c ) < 0. n ≥ n This shows that f (α) = 0. Hence Γ¯ is real closed. We have the

Corollary. Γ¯(i) is algebraically closed.

Γ¯(i) is the complex number ﬁeld. We have thus proved the ‘fundamental theorem of algebra’. The algebraic closure of Γ in Γ¯ is called the ﬁeld of real algebraic numbers and is real closed.

Chapter 8

Valuated ﬁelds

1 = 12 = 1 2= 1 1 | | | | | | | | · | | so that 1 = 1. If ζ is a root of unity, say an n say an n th root of unity, | | then 1 = ρn = ρ n | | | | and so ρ = 1. Thus means that 1 = 1 and so, for a k, | | |− | ∈ a = a . |− | | | 175 176 8. Valuated ﬁelds

Also, since a = a b + b, we get − a b a b . | | − | |≤ − | A valuation is said to be trivial if a = 1 for all a , 0. | | Two valuations and are said to be equivalent if for every a , 0 | |1 | |2 in k,

a < 1 a < 1, | |1 ⇒| |2 a = 1 a = 1. | |1 ⇒| |2 201 It is obvious that the above relation between valuations is an equivalence relation. All valuations equivalent to a given valuation form an equivalence class of valuations. If is a valuation then, for 0 c 1, 1 c is also a valuation. We | | ≤ ≤ | | shall now prove

Theorem 1. If and are equivalent valuations, there exists a real ||1 ||2 number c > 0 such that a = a c | |1 | |2 for all a k. ∈ Proof. Let us assume that is non-trivial. Then is also non-trivial. ||1 ||2 Also, there exists a b k such that b > 1, b > 1. Let a k, a , 0. ∈ | |1 | |2 ∈ Then a and b being positive real numbers. | |1 | |1 a = b λ | |1 | |1 log a where λ = | |1 . log b | |1 We approximate to the real number λ from below and from above m by means of rational numbers. Let < λ. Then n

a > b m/n | |1 | |1 2. Classiﬁcation of valuations 177

an which means that > 1. Since and are equivalent, this means | bm |1 ||1 ||2 that a > b m/n . | |2 | |2 In a similar manner, if p/q > λ, then

a < b p/q . | |2 | |2 202 m p This means that if λ and λ,then n → q →

A valuation is said to be archimedian if for every a k, there exists an ∈ integer n = n(a) (that is ne, if e is the unit element of k) such that

a < n . | | | | (Compare this with archimedian axiom in ordered fields). A valuation of k which is not archimedian is said to be non-archimedian. We shall deduce a few simple consequences of these defini- tions. 1) is archimedian there exists an integer n in k such that n > 1. || ⇐⇒ | | Proof. If is archimedian, there exists a k with a > 1 and an || ∈ | | integer n with n > a . This means that | | | | n > 1. | | 178 8. Valuated fields

Let be a valuation and n, a rational integer so that n > 1. Let a any || | | element of k. If a 1, then clearly a < n . Let a > 1. Since | |≤ | | | | | | archimedian axiom holds in real number ﬁelds, we have an integer m with a < n m. | | | |· 203 If m = 1, there is nothing to prove. So let m > 1. Then m = n λ log m | | (λ = > 0). Let µ be a positive integer greater than λ. Then log n m < n µ=| n|µ . Therefore | | | | a < n m < n µ+1 | | | |· | | and our assertion is proved. We deduce

2) non-archimedian n 1, for every integer n in k. || ⇐⇒| |≤ This shows at once that

3) All the valuations of a ﬁeld of characteristic p are non-archimedian. We now prove the important property

4) is an non-archimedian valuation if and only if for every a, b in k || a + b Max( a , b ). | |≤ | | | | Proof. If is a non-archimedian valuation, then for every integer || n, n 1. Let m be any positive integer. Then | |≤ + m = m + m m 1 + + m (a b) a (1 )a − b ... b

so that

Taking m th roots and making m we get →∞ a + b Max( a , b ). | |≤ | | | | The converse is trivial, since n = 1 + + 1 1. | | | ··· |≤

We deduce easily 204

5) If is non-archimedian and a , b , then || | | | | a + b = Max( a , b ). | | | | | | Proof. Let, for instance, a > b . Then | | | | a + b Max( a , b ) = a | |≤ | | | | | | Also a = a + b b, so that − a Max( a + b , b ). | |≤ | | | | But, since a > b , a + b b . Thus a a + b and our | | | | | |≥| | | |≤| | contention is proved.

More generally, we have, if a > a , j , 1, then | 1 | | j | a + a + ... a = a . | 1 2 n | | 1 | In the case of non-archimedian valuations, many times, the so-called exponential valuation is used. It is defined thus: If is an non-archi- ||0 median valuation, define the function by || a = log a , a , 0. | | − | |0 This has a meaning since a > 0 for a , 0. We introduce a quantity | |0 which has the property ∞ + = ∞ ∞ ∞ + a = ∞ ∞ 180 8. Valuated fields

for any real number a and a = 0 ∞ for any real number a. Then satisﬁes || 1’) 0 = | | ∞ 2’) a is a real number | | 3’) ab = a + b | | | | | | 4’) a + b Min ( a , b ). | |≥ | | | | 205 Then 1 = 0, ρ = 0 for a root of unity ρ and the valuation is trivial | | | | if a = 0 for a , 0. For two valuations and which are equivalent | | ||1 ||2 a = c a , | |1 | |2 c 0 being a real number. i

3 Examples

First, let Γ be the finite field of q elements. Every element of Γ∗ satisfies the polynomial xq 1 1. Therefore, any valuation on Γ is trivial. − − | | Let now Γ be the field of rational numbers. Let be an archimedian valuation on Γ. It is enough to determine its | | effect on the set of integers in Γ. There is an integer n such that n > 1. | | Let m be any positive integer. Then

m = mC | | where c > 0 is a constant. Obviously, the valuation is determined by its effect on positive integers. From the definition of equivalence, is || equivalent to the ordinary absolute value induced by the unique order in Γ. Let now be a non-trivial non-archimedian valuation. It is enough || to determine its effect on Z, the ring of integers. Since is non-trivial, consider the set Y of a Z with a < 1. Y | | ∈ | | is an ideal. For,

a < 1, b < 1 a + b Max( a , b ) < 1. | | | | ⇒ | |≤ | | | | 182 8. Valuated ﬁelds

Also, if a < 1 and b Z, then | | ∈ ab = a b < 1. | | | | | | Furthermore, if ab Y , then ab < 1. But ab = a b and a 1, ∈ | | | | | || | | | ≤ b 1, since valuation is non-archimedian. Hence a < 1 which means | |≤ | | that Y is a prime ideal. Thus g = (p) generated by a prime p. Since, by deﬁnition of Y , M < 1 p/n, we have, if n = pλ n , (n , p) = 1, ⇔ · 1 1 n = p λ. | | | | 207 a If we denote p by c, 0 < c < 1, then for any rational number , the | | b value is a = cλ b

a λ a′ where = p where (a′, p) = 1 = (b′, p) and λ a rational integer. b b This valuations is′ called the p-adic valuation. Thus with every non-archimedian valuation, there is associated a prime number. Conversely, let p be any prime number and let n be any integer, n = pλ n , λ 0, where (n , p) = 1. Put · 1 ≥ 1 n = p λ , 0 < p < 1. | | | | | | Then determines a non-archimedian valuation on Γ. Further, if p | | and q are distinct primes, then the associated valuations are inequivalent. For, if and are the valuations, then | |p | |q q = 1, q < 1. | |p | |q We shall denote the valuation associated with a prime p, by . Then ||p we have the

Theorem 2. The ordinary absolute valuation and the p-adic valuation by means of primes p form a complete system of in-equivalent valuations of the rational number ﬁeld. 3. Examples 183

We shall denote the ordinary absolute valuation by . ||∞ Let us consider the case of a function field K over a ground field k and let L be the algebraic closure of k in K. L is called the field of 208 constants of the function field K. The valuations on K that we shall consider shall always be such that

a = o | | for a k. (We consider exponential valuation). Hence valuations of ∈ function fields are always non-archimedian, since the prime field is contained in k. Let, now, α be in L. Then α satisfies the equation

n n 1 α + a α − + + a = o 1 ··· n where a ,..., a k. If is a valuation of K then 1 n ∈ || n n 1 α min( α − ,..., 1 ). | | ≥ | | | | From this, it follows that α 0. Also, since 1/α is algebraic, | | ≥ α 0. Hence for all α L | |≤ ∈ α = 0. | | Thus the valuation is trivial on the ﬁeld of constants. We shall consider the simple case where K = k(x), x transcendental over k. Here L = k. Let be a valuation and let x < 0. Let f (x) = || | | a xn + + a be in k[x]. Then 0 ··· n n n 1 f (x) min( x , x − ,..., 1 ). ≥ | | | | | | Therefore f (x) = n x . | | g(x) This means that, if R(x) = is an element of K, then h(x) R(x) = (deg h(x) deg g(x))( x ). | | − −| | We denote this valuation by . ||∞ 184 8. Valuated ﬁelds

Suppose now that x 0. As in the case of rational integers, con- | | ≥ sider the subset Y of k[x] consisting of polynomials f (x) with f (x) > | | 0. Then, since for every φ(x) in k[x], ϕ(x) 0, it follows that Y is a | | ≥ maximal ideal generated by an irreducible polynomial p(x). As in the 209 case of the rational number ﬁeld

R(x) = λ p(x) | | | | A(x) where R(x) = p(x) λ where A(x) and B(x) are prime to p(x) and { } B(x) λ is a rational integer. If we denote, by , this valuation and put ||p(x) λ = ordp(x)R(x), then

R(x) = c ord R(x) | | p(x) where c = p(x) > 0. Every irreducible polynomial also gives rise to | |p(x) a valuation of this type. Hence

Theorem 3. A complete system of inequivalent valuations of k(x) is given by the valuations induced by irreducible polynomials in k[x] and the valuation given by the diﬀerence of degrees of numerator and denominator of f (x) in k(x).

4 Complete ﬁelds

Let k be a ﬁeld and a valuation of it. The valuation function deﬁnes || on k a metric and one can complete k under this metric. The method is the same as in the previous chapter and we give here the results without proofs. A sequence (a1, a2,...) of elements of k is said to be a Cauchy sequence if for every ε> 0, there exists an integer n = n(ε) such that

a a <ε , n , n > n. | n1 − n2 | 1 2 It is a null sequence if for every ε> 0, there is an integer n such that

a < ε, m > n. | m| 4. Complete ﬁelds 185

210 The Cauchy sequences form a commutative ring R and the null sequences a maximal ideal Y , therein. The quotient k¯ = R/Y is called the completion of k under . The mapping a (a, a, a,...) is an iso- || → morphism of k in k¯ and we identify this isomorphic image with k itself. We extend to k¯ the valuation in k, in the following manner. The real || number ﬁeld Γ¯ is complete under the valuation induced by the unique order on it. So, if a k¯, then a = (a , a ,...), a Cauchy sequence of ∈ 1 2 elements of k. Put a = lim an n | | →∞ | | That this is a valuation follows from the properties of limits in Γ¯. Also, the extend valuation is archimedian or non-archimedian, according as the valuation in k is archimedian or non-archimedian. For instance, if on k is non-archimedian and || a = (a1, a2,...)

b = (b1, b2,...) are two elements of k¯, then

a + b = (a1 + b1, a2, +b2,...) and

a + b Max( a , b ) = lim ( an + bn Max( an , bn )) n | |− | | | | →∞ | |− | | | | which is certainly 0, ≤ k¯ is thus a complete valuated ﬁeld. It may also be seen that the elements of k are dense in k¯ in the topol- 211 ogy induced by the metric. Let c + c + c + be a series in k¯. We denote by S the partial 1 2 3 ··· n sum S = c + c + + c . n 1 2 ··· n We say that c + c + + c + is convergent if and only if the 1 2 ··· n ··· sequence of partial sums S 1, S 2,..., S n,... converges. This means that, for every ε> 0, there exists an integer n = n(ε) such that

S S = c + + + c < ε, m, m′ > n. | m′ − m| | m 1 ··· m′ | 186 8. Valuated ﬁelds

Obviously c1, c2,... is a null sequence in k¯. In case the valuation is non-archimedian, we have the following property:- 1) The series c + c + + c + is convergent if and only if 1 2 ··· n ··· c1, c2,... is a null sequence. Proof. We have only to prove the suﬃciency of the condition. Suppose that c 0; then, for large n and m, n → S S = c + + c + + + c | n − m| | m 1 m 2 ··· n| max( c + ,..., c ) ≤ | m 1| | n| and so tends to zero. This proves the contention.

Note that this theorem is false, in case the valuation is archimedian. Let k be a field and a valuation on it. a a is a homomorphism || → | | of k∗ into multiplicative group of positive real numbers. Let G(k) denote 212 this homomorphic image. This is a group which we call the value group of k for the valuation. If k¯ is the completion of k by the valuation in k, then G(k¯) is value group of k¯. Suppose is an archimedian valuation. Then k has characteristic || zero and contains Γ, the rational number field as a subfield. On Γ, is || the ordinary absolute value. Since k¯ contains Γ¯, the field of real numbers, it follows that 1) G(k¯) is the multiplicative group of all positive real numbers. Also, because of the definition of the extended valuation, it follows that G(k) is dense in the group G(k¯). We shall now assume that is a non-archimedian valuation. We con- || sider the exponential valuation. Then G(k) is a subgroup of the additive group of all real numbers. We shall now prove 2) G(k¯) = G(k).

Proof. For, let 0 , a k¯. Then a = (a ,...) is a Cauchy sequence, ∈ 1 not a null sequence in k. By deﬁnition,

a = lim an | | n | | 4. Complete ﬁelds 187

Now a = a a + a so that n n − a min( a a , a ). | n|≥ | n − | | | But, for n large, a a > a so that a = a and our contention is | n − | | | | n| | | established.

G(k) being an additive subgroup of the real number ﬁeld is either dense or discrete. The valuation is then called dense or discrete ac- cordingly. In the second case, there exists π in k with smallest pos- 213 itive value π π is then the generator of the inﬁnite cyclic group | | · | | G(k) π is called a uniformising parameter. It is clear that π is not · unique. For, if u k with u = 0, then uπ = π and uπ is also a ∈ | | | | | | uniformising parameter. Consider in k the set O of elements a with a o.O is then an | | ≥ integrity domain. For,

a 0, b 0 a + b Min( a , b ) 0. | |≥ | |≥ ⇒ | |≥ | | | | ≥

Also ab = a + b 0. We call O the ring of integers of the | | | | | | ≥ valuation. Consider the set Y of elements a k with a > 0. Then ∈ | | Y is a subset of O and is a maximal ideal in O. For, if a r and ∈ b Y , then ab = a + b > 0. Also, if a r but not in Y , then ∈ | | | | | | ∈ a = 0 and a 1 = 0 so that if U is an ideal in O containing Y , then | | | − | U = Y or U = r. We call Y , the prime divisor of the valuation. Since Y is maximal, r/Y is a field. We call π/Y the residue class field. Exactly the same notions can be defined for k¯. We denote by O¯ the ring of integers of the valuation so that O¯ is the set of a k¯ with ∈ a 0.Y¯ is the maximal ideal in O¯, hence the set of a k¯ with | | ≥ ∈ a > 0. Also O¯/Y¯ is the residue class field. Clearly | | Y = Y¯ r ∩

We now have 188 8. Valuated ﬁelds

2) Every a k(k¯) has the property; a r(¯r) or a 1 Y (Y¯ ). ∈ ∈ − ∈ This is evident since if a < O(O¯), a < 0 so that a 1 > 0 and hence 214 | | | − | a 1 Y (Y¯ ). − ∈ 3) O(O¯) is integrally closed in k(k¯).

Proof. We should prove that every α in k(k¯) which is a root of a polynomial of the type xn + a xn 1 + + a , a ,..., a O(O¯), is 1 − ··· n 1 n ∈ already in O(O¯). For, suppose

n n 1 α + a α − + + a = o 1 ··· n and α < O(O¯). Then α 1 Y (Y¯ ). Therefore − ∈ 1 2 n 1 = (a α− + a α− + + a α− ) − 1 2 ··· n and so 1 n o = 1 min( a α− ,... a α− ) > o | |≥ | 1 | | n which is absurd

4) Every element in O¯ is the limit of a sequence of elements in O and conversely.

Proof. Let a = (a1,..., an,...) be in O¯, a1,..., an,... in k. Then, as we saw earlier, for suﬃciently large n

a = a . | n| | | But a 0 so that a 0. Thus, for suﬃciently large n, all a s are | | ≥ | n|≥ n′ in O. Converse is trivial. We now prove

5) There is a natural isomorphism of O/Y on O¯/Y¯ . 4. Complete ﬁelds 189

Proof. The elements of O/Y are residue classes a + Y , a r. We ∈ now make correspond to a + Y the residue class a + Y¯ . Then a + Y¯ 215 is an element of O¯/Y¯ . The mapping a + Y a + Y¯ is a homomor- → phism (non-trivial) of O/Y into ℧¯ /Y¯ . Since both are ﬁelds, this is an isomorphism into. In order to see that it is an isomorphism of O/Y onto O¯/Y¯ , leta ¯ + Y¯ be any residue class in O¯/Y¯ . By (4) therefore, given any N > 0, there exists a ℧ with ∈ a¯ a > N > 0 | − | ora ¯ a Y¯ . If we then take a + Y , thena ¯ + Y¯ is the image of − ∈ a + Y . This proves that the mapping is an isomorphism onto.

(5) enables us to choose, in k itself, a set of representatives of ℧¯ /Y¯ , the residue class ﬁeld. We shall denote this set by R and assume that it contains the zero element of O. Also it has to be observed that, in general, R is not a ﬁeld. It is not even an additive group. We now assume that is a discrete valuation. Let π in O be a uni- | | formising parameter. Then clearly

Y¯ = (π)

where u is a unit. Let R be the set deﬁned above and a ℧¯ . Then ∈ a a ( mod Y¯ ) ≡ 0 with a R. This means that (a a )π 1 is an integer (in Ω¯ ). 216 0 ∈ − 0 − 1 (a a )π− a (modY¯ ), − 0 ≡ 1 190 8. Valuated ﬁelds

where a R. Then 1 ∈ a a + a π(modY¯ 2) ≡ 0 1 In this way, one proves by induction that

a a + a π + + a πm(mod)Y¯ m+1) ≡ 0 1 ··· m where a , a ,..., a R. Put b = a + a π + + a πm. 0 1 m ∈ m 0 1 ··· m Then a b (modY¯ m+1) which means that ≡ m a b m + 1 | − m|≥ Consider the series

b + (b b ) + (b b ) + 0 1 − 0 2 − 1 ··· + Then, since b + b = a + πm 1, we see that b + b increases m 1 − m m 1 | m 1 − m| indeﬁnitely. Hence the above series converges. Also, since its elements are integers,

b = b + (b b ) + (b b ) + 0 1 − 0 2 − 1 ··· is an element of O¯.

Since b0 + (b1 b0) + + (bm bm 1) = bm, it follows that − ··· − −

b = lim bm. m

Thus we have

2 a = lim bm = a0 + a1π + a2π + m →∞ ··· By the very method of construction this expression for a is unique, once we have chosen R and π. 217 If a k¯, aπt O¯ for some rational integer t. Hence we have ∈ ∈ 4. Complete ﬁelds 191

6) Every element a in k¯ has the unique expression

∞ n a = anπ n= t X− a R and π in O¯ is a generator of Y . t is a rational integer. n ∈ If t > 0, we shall call 1 − n h(a) = anπ n= t X− the principal part of a. If t 0 we put h(a) = 0. Clearly, h(a) is an ≤ element in k. Also a h(a) O¯. − ∈ We now study the two important examples of the rational number field and the rational function field of one variable. Let Γ be the field of rational numbers. We shall denote by the ||∞ ordinary absolute value and by p the p-adic value for p, a prime. If λ || a is an integer, a = p n1, (p, n1) = 1. We put a = λ log p | |p and a = absolute value of a. | |∞ It is then clear that each of the non-archimedian valuations of Γ is discrete. Let us denote by Γ the completion of Γ by the archimedian ∞ valuation and by Γp the completion of Γ by the p-adic valuation. Γ is clearly the real number field. ∞

If Op denotes the set of integers of Γp and Y the prime ideal of the valuation then Y = (p)

A set of representatives of Op mod Y is given by the integers 0, 1, 2, 218 ..., p 1 as can be easily seen. Hence, by (6), − 192 8. Valuated ﬁelds

7) Every a Γ is expressed uniquely in the form ∈ p

∞ n a = an p n= t X− where a = 0, 1, 2,..., p 1. i − The elements of Γp are called the p-adic numbers of Hensel.

As before, we denote, by hp(a), the principal part of a at p. Clearly a h (a) is a p-adic integer. − p b Let a be a rational number, a ℧ a 0, that is a = , ∈ p ⇔ | |p ≥ c (b, c) = 1 and c is prime to p. Since only finitely many primes divide b and c, it follows that a ℧ for almost all p, that is except for a ∈ p finite number of p. Hence hp(a) = 0 for all except a finite number of primes. Hence for any a

hp(a) p X has a meaning. Also, a h (a) is a rational number whose denom- − p inator is prime to p. Hence a hp(a) is a rational number whose − p denominator is prime to every rationalP integer. Hence 8) For every rational number a

a hp(a) 0(mod1). − p ≡ X This is the so-called partial fraction decomposition of a rational number. Let now k be an algebraically closed field and K = k(x) the field 219 of rational functions of one variable x. All the valuations are non- archimedian. Every irreducible polynomial of k[x] is linear and of the form x a. With every a k there is the valuation associated, − ∈ ||a which is defined by f (x) = λ, | |a 4. Complete fields 193

where f (x) k[x], f (x) = (x a)λϕ(x), ϕ(a) , 0. If we denote by ∈ − the valuation by degree of f (x), then, for f (x) k[x], ||∞ ∈ f (x) = deg f (x) | |∞ −

Let Ka and K denote the completions, respectively, of K at a and ∞ || . If Oa and O are the set of integers of Ka and K , Ya and Y ||∞ ∞ ∞ ∞ the respective prime divisors, then 1 Ya = x a , Y = . { − } ∞ { x}

It is clear, then, that ℧a/Ya and ℧ /Y are both isomorphic to k and since K contains k, we may take k∞itself∞ as a set of representatives of the residue class ﬁeld. Any element f in Ka is uniquely of the form

∞ n f = an(x a) , n= t − X− an k. Similarly, if ϕ K , ∈ ∈ ∞

∞ n ϕ = bn x− . n= t X−

As before, if we denote by ha( f ) and h ( f ) the principal parts of f K for the two valuations, then ∞ ∈

ha( f ) + h ( f ) ∞ a X has a meaning since ha( f ) = 0 for all but a ﬁnite number of a

If we define ϕ K to be regular ar a( ) if ϕ ℧a(℧ ), then for 220 ∈ ∞ ∈ ∞ f K, ∈ f ha( f ) − a X where a may be infinity also, is regular at all, a k and also for the ∈ valuation . Such an element, clearly, is a constant. Hence ||∞ 194 8. Valuated fields

9) If f K then ∈ f ha( f ) = constant − a X Conversely, it is easy to see that there exists, up to an additive constant, only one f K which is regular for all a k except a ,..., a ∈ ∈ 1 n (one of which may be also) and with prescribed principal parts at ∞ these ai. 9) gives the partial fraction decomposition of the rational function f .

5 Extension of the valuation of a complete non-archimedian valuated ﬁeld

We shall study the following problem. Suppose k is compute under a valuation . Can this valuation be extended, and if so in how many || ways, to a finite algebraic extension K of k ? We prove, first Lemma 1. Let k be a field complete under a valuation , and K a finite || algebraic extension of k. Let ω ,...,ω be a basis of K/k. Let have 1 n || an extension to K and n αν = aivωi, aiv k, = ∈ Xi 1 221 v = 1, 2, 3,... be a Cauchy sequence in K. Then aiv i = 1, 2,..., n are Cauchy sequence in K. Proof. We consider the Cauchy sequence α , { v} m αv = aivωi, aiv , k = ∈ Xi 1 1 m n and we shall prove that the a are Cauchy sequences in k. ≤ ≤ { iv} We use induction on m. Clearly, if m = 1,

αv = aivω1 and α is a Cauchy sequence in K if and only if a is a sequence in { v} { iv} k. 5. Extension of the valuation of a complete. . . 195

Suppose we have proved our statement for m 1 1, instead of m. − ≥ Write m 1 − αv = aivωi + amvωm. = Xi 1 If a is a Cauchy in k, then α a ω is a Cauchy sequences { mv} { v − mv m} in K and induction hypothesis works. Let us assume that amv is not a Cauchy sequence in k. This means that there exists a λ> 0 and for every v, an integer µv such that µν > v and a µ a > λ. | m v − mv| Consider now the sequence β in K with { v}

αµv αv βv = − . a µ a m v − mv Because of the above property, we see that β is a null sequence in { v} K. Now m 1 − aiµv aiv βv ωm = − ωi − = amµv amv Xi 1 − ! and β ω is a Cauchy sequence in K. Induction hypothesis now 222 { v − m} works and so, if

aiµv aiv lim − = bi, v a µ a →∞ m v − mv ! then ωm = biωi, bi k. − = ∈ Xi 1 This is impossible because ω1,...,ωm are linearly independent over k. Therefore a is a Cauchy sequence and our Lemma is thereby { mv} proved. We shall now prove the following theorem concerning extension of 223 valuation. 196 8. Valuated ﬁelds

Theorem 4. If the valuation of a complete ﬁeld k can be extended to a || ﬁnite extension K, then this extension is unique and K is complete under the extended valuation.

Proof. That K is complete under the extended valuation is easy to see. For, if α is a Cauchy sequence in K and { v} α = a ω , a k., v iv i iv ∈ Xi by the lemma, the a ’s are Cauchy sequences. So, if { iv}

lim aiv = bi k, v →∞ ∈ then lim αγ = ω1 lim aiv = biωi v v →∞ →∞ Xi Xi which is again in K.

We shall now prove that the extended valuation is unique. From the lemma, it follows that if α is a null sequence in K and { v} αv = aivωi, aiv k, then the aiv’s are null sequences in k. i ∈ InP particular, if α K and α < 1 in some extension the valuation 2 3 ∈ | | m = (m) in k, then α,α ,α ,... is a null sequence in K. If α a1 ωi, || 1 a(m) k, then a(m), i = 1,..., n are null sequence in k. P i ∈ i If α = xiωi is a general element of x1,..., xn. Put

P t = (t) α xi ωi; = Xi 1 t (t) = t then NK/kα is the same polynomial in xt , i 1,..., n as NK/kα is in x ,..., x . If now α < 1, is an extended valuation, then the x(t) are 1 n | | { i } null sequences. Hence Nα, Nα2 is a null sequences in k, But Nαt = (Nα)t so that (Nα), (Nα)2,..., is s a null sequence in k. This means that Nα < 1. We have, thus, proved that if α in K is such that α < 1 is an | | | | extended valuation, then Nα < 1 in k. | | 5. Extension of the valuation of a complete. . . 197

In a similar manner, if α > 1, then Nα > 1. Thus we get Nα = | | | | | | 1 α = 1. ⇒ | | αn Let now β be in K and write β = where n = (K : k) Nα (Nα)n Then Nβ = . Thus (Nα)n Nβ = 1. | | By the above, it means that β = 1 in the valuation. Hence | | α = n Nα | | | | showing that the value of α in the extendedp valuation is unique ﬁxed. 224 Our theorem is thus completely proved. In order to prove that an extension of the valuation is possible, we shall consider the case where k is complete under a discrete non-archimedian valuation. Let O be the ring of integers Y , the prime divisor of the valuation and O/Y , the residue class ﬁeld. We shall now prove the celebrated lemma due Hensel

Lemma 2. Let f (x) be a polynomial of degree m in O[x], g0(x), a monic polynomial of degree r 1 and h (x), a polynomial of degree m r ≥ 0 ≤ − both with coeﬃcients in O such that 1) f (x) g (x)h (x) (modY ) ≡ o o 2) go(x) and ho(x) are coprime mod Y Then there exists polynomials g(x) and h(x) in O[x] such that

g(x) go(x) ≡ (mod Y ), h(x) h (x) ≡ o   = g(x) has the same degree as go(x) and f (x) g(x) h(x).  · Proof. We shall now construct two sequences of polynomials go(x), g1, (x),... and ho(x), h1, (x),... satisfying

n gn(x) gn 1(x)( mod Y ) ≡ − 198 8. Valuated ﬁelds

n hn(x) hn 1(x)( mod Y ) − ≡ + f (x) g (x)h (x)( mod Y n 1) n ≡ n n g (x) is a monic polynomial of degree r and h (x) of degree m r. All n n ≤ − the polynomials have coeﬃcients in O.

225 The polynomials are constructed inductively. For n = 0, go(x) and ho(x) are already given and satisfy the conditions. Assume now that go(x),..., gn 1(x) and ho(x),..., hn 1(x) have been constructed so as satisfy the requisite− conditions. − Since gn 1(x) go(x)(mod Y ) and hn 1(x) ho(x)(mod Y ) and − ≡ − ≡ g0(x) and ho(x) are coprime mod Y , there exists, for any polynomial fn(x) in O[x], two polynomials L(x) and M(x) with

fn(x) L(x)gn 1(x) + M(x)hn 1(x) (mod Y ). ≡ − − L(x) and M(x) are clearly not uniquely determined. We can replace L(x) by L(x) + λ(x)hn 1(x) and M(x) by M(x) + λ(x)gn 1(x). Let π be a generator− of the principal ideal Y .− By induction hypothesis, n fn(x) = π− ( f (x) gn 1(x)hn 1(x)) − − − is an integral polynomial, so in O[x]. Since gn 1(x) has degree r and is − monic and hn 1(x) degree m r, it is possible to choose M(x) and L(x) − ≤ − so that M(x) has degree < r and L(x) degree m r. Put now ≤ − n gn(x) = gn 1(x) + π M(x), − n hn(x) = hn 1(x) + π L(x). −

Then gn(x) is monic and of degree r, since M(x) has degree < r.hn(x) has degree m r. Now f (x) gn(x)hn(x) = f (x) gn 1(x)hn 1(x) n ≤ − − n+1 − − − − π (gn 1(x)L(x) + hn 1(x)M(x))( mod Y ) By choice of L(x) and M(x),− it follows that −

+ f (x) g (x)h (x)(mod Y n 1). ≡ n n 226 We have thus constructed the two sequences of functions. Put now 5. Extension of the valuation of a complete. . . 199

g(x) = go(x) + (g1(x) go(x)) + + (gn(x) gn 1(x)) + − ··· − − ··· h(x) = ho(x) + (h1(x) ho(x)) + + (hn(x) hn 1(x)) + − ··· − − ··· n Since gn(x) gn 1(x) = 0(modπ ), it follows that the correspond- − − ing coeﬃcients of the sequence of polynomials g0(x), g1(x),... form Cauchy sequences in O. Since k is complete and

g(x) = lim gn(x), n →∞ it follows that g(x) ℧[x]. It is monic and is of degree r. In a similar ∈ way, b(x) ℧[x] and has degree m r. ∈ ≤ − + Also since f (x) g (x)h (x) 0(modyn 1), it follows that the coef- − n n ≡ ﬁcients of ( f (x) g (x)h (x)) form null sequences. Hence − n n

f (x) = lim gn(x)hn(x) = hn(x) = g(x)h(x) n and our lemma is proved. We now deduce the following important .

Lemma 3. Let f (x) = xn +a xn 1 + +a be an irreducible polynomial 1 − ··· n k[x] , k satisfying hypothesis of lemma 2. Then f (x) O[x] if and only ∈ if a O. n ∈ Proof. It is clearly enough to prove the suﬃciency of the condition. Let an ℧, and if a1,..., an 1 (some or all of them) are not in ℧, then there ∈ − is a smallest power πa, a > 0, of π such that

a n n 1 π f (x) = b x + b x − + + b o 1 ··· n is a primitive polynomial in O[x]. Also, now b 0( mod Y ). and 227 n ≡ at least one of b0,..., bn 1 is not divisible by Y . Let br be the ﬁrst coeﬃcient from the right− not divisible by Y . Then

a r n r π (b x + b )x − (mod Y ) ≡ 0 ··· r

Since br . 0( mod Y ), Hensel’s lemma can be applied and we see that πa f (x) is reducible in O[x]. Thus f (x) is reducible in k[x] which contradicts the hypothesis. The lemma is therefore established. 200 8. Valuated ﬁelds

We are now ready to prove the important theorem concerning extension of discrete non-archimedian valuations, namely

Theorem 5. Let k be complete under a discrete non-archimedian valuation and K a ﬁnite algebraic extension over k. Then can be extended || || uniquely to K and then for any α in K,

1 α = N / α . | | (K : k)| K k |

Proof. Because of Theorem 4, it is enough to prove that the function deﬁned on K by 1 α = N / α | | (K : k)| K k | is a valuation function. Clearly , 0 = ; α is a real number for α , 0. | | ∞ | | Also, αβ = α + β . | | | | | | We shall now prove that

α + β min( α , β ). | |≥ | | | |

228 If α or β is zero, then the above is trivial. So let α , 0, β , 0. Since α β or is 0, it is enough to prove that if λ 0, 1 + λ 0. β α ≥ | |≥ | |≥

Let f (x) = xm + a xm 1 + + a be the minimum polynomial of λ 1 − ··· m in K over k, Then Nλ = (( 1)ma )(K:k(λ)). − m Also, N(1 + λ) = ( 1)n(1 a + a )(K:k(λ)). If Nλ 0 , − ± 1 ± ··· m | ≥ | then a 0 which , by lemma 3, means that a ,..., a . Hence | m| ≥ | 1| | m| N(1 + λ) 0. Our theorem is proved. | |≥ Incidentally it shows that the extended valuation is discrete also. 6. Fields complete under archimedian valuations 201

6 Fields complete under archimedian valuations

Suppose k is complete under an archimedian valuation. Then k has characteristic zero and contains, as a subfield, the completion Γ¯ of the rational number field. barΓ(i) is, then, the complex number field. Every complex number is to the form a + ib, a, b Γ¯. On Γ¯, we have the ∈ ordinary absolute value. Define in Γ¯(i) the function

2 2 1 z = (a + b ) 2 | | where z = a + ib. It is, then, easy to verify that is a valuation on Γ¯(i) || which extends the valuation in Γ¯. Also, by theorem 4, this is the only extension of the ordinary absolute value. We consider the case k Γ¯(i) ⊃ and prove the theorem of A. Ostrowski. Theorem 6. Let k Γ¯(i) be the complex number ﬁeld and let k be a ⊃ ﬁeld with archimedian valuation and containing Γ¯(i). If the valuation in k is an extension of the valuation in Γ¯(i), then k = Γ¯(i).

Proof. If k , Γ¯(i), let a k but not in Γ¯(i). Denote by the valuation in 229 ∈ || k. Consider a z for all k = Γ¯(i). Since a z 1 0, we have | − | | − | ≥ 1 ρ = g b a z 0. · 2 · | − |≤

There exists, therefore, a sequences z1,..., zn,.. of complex numbers such that lim a zn = ρ. n | − | →∞

But z = z a + a and so z z a + a which shows that the n n − | n| ≤ | n − | | | z , for large n, are bounded. We may therefore, choose a subsequence | n| zi1 , zi2 ,... converging to a limit point zo such that

ρ = lim a zi = a zo n n →∞ | − | | − | We have thus proved the existence of a z in Γ¯(i) such that b = a z 0 − 0 has b = ρ. Since, by assumption, a < Γ¯(i) we have | | b = ρ> 0. | | 202 8. Valuated ﬁelds

ρ, by deﬁnition, being g.l.b., it follows that

b z ρ | − |≥ for z Γ¯(i). ∈ Consider the set of complex numbers z with z < ρ. Let n > 0 be an | | arbitrary rational integer and ε, a primitive nth root of unity. Then

n n n 1 b z = (b z)(b εz) ... (b ε − z). − − − − 230 Therefore

n 1 n 1 n n b z ρ − b z b εz ... b ε − z = b z | − | ≤ | − || − | | − | | − | z b n + z n = ρn(1 + (| |)n). ≤ | | | | ρ Hence z n b z ρ 1 + | | . | − |≤ ρn ! But z < ρ and as n is arbitrary, it follows that b z ρ. | | | − |≤ We therefore have

z < ρ b z = ρ. | | ⇒ | − | We now prove that for every integer m > 0, b mz = ρ if z < ρ. | − | | | For, suppose we have proved this for m 1 instead of m, then we can − carry through the above analysis with b (m 1)z instead of b, then we − − can carry through the above analysis with b (m 1)z instead of b and − − then we obtain b mz = ρ. | − | Suppose now that z′ is any complex number. Then there is an integer z m > 0 such that ′ < ρ. Therefore |m|

z′ b m = b z′ = ρ. − m | − |

Now z′ = z′ b + b and so −

z′ b z′ + b 2ρ | | ≤ | − | | |≤ 6. Fields complete under archimedian valuations 203 which shows that all complex numbers are bounded in absolute value. This is a contradiction. Hence our assumption that a < Γ¯(i) is false . The theorem is thereby proved, Before proving theorem 7 which gives a complete characterization of all complete ﬁelds with archimedian valuation, we shall prove a cou- ple of lemmas.

Lemma 4. Let k be complete under an archimedian valuation and λ 231 || in k such that x2 + λ is irreducible in k[x]. Then 1 + λ 1. | |≥ Proof. If possible, let 1 + λ < 1. We construct, by recurrence, the | | sequence c0, c1, c2,..., in k, deﬁned as follows: -

c0 = 1 1 + λ cn+1 = 2 n = 0, 1, 2,... − − cn

It, then, follows that cn 1. For, if we have proved it upto cn 1, | | ≥ − then 1 + λ cn 2 | | 1. | |≥ − cn 1 ≥ | − | Thus cn does not vanish for any n. Also,

1 + λ cn cn 1 cn+1 cn = | || − − | ρ cn cn 1 | − | cn cn 1 ≤ | − − | | || − | where ρ = 1 + λ < 1. This means that the series | | c + (c c ) + (c c ) + 0 1 − 0 2 − 1 ··· converges in k. Let it converge to c in k. Then

c = lim co + + (cn on 1) = lim cn. n − n →∞ ··· − →∞ 1 + λ Therefore, by deﬁnition of c , we get c = 2 . But this means n − − c that λ = c2 + 2c + 1 = (c + 1)2 which contradicts the fact x2 + λ is − irreducible in k[x]. We now prove the 204 8. Valuated ﬁelds

Lemma 5. If k is complete under an archimedian valuation, , then this || valuation can be extended to k(i). Proof. If i k, there is nothing to prove. Let i < k. Then every element ∈ of k(i) is of the form a + ib, a, b k. ∈ 232 The norm from k(i) to k of α = a + ib is

Nα = a2 + b2.

By theorem 4, therefore, it is enough to prove that

2 2 1 α = (a + b ) 2 | | | | b2 is a valuation on k(i). By putting λ = in the lemma 4, we see that a2 a2 + b2 a2. Therefore | |≥ (1 + a)2 + b2 1 + a2 + b2 + 2 a | |≤ | | | |

We now obtain a complete characterization of complete archimedian fields, namely, Theorem 7. The only fields complete under an archimedian valuation are the real and complex numbers fields. Proof. k has characteristic zero and since it is complete, it contains the field Γ¯ of real numbers. If k contains Γ¯ properly, then we assert that k contains i. For, k(i), by lemma 5, is complete and k(i) contains Γ¯(i) and, by theorem 6, k(i) = Γ¯(i). 7. Extension of valuation of an incomplete field 205

Therefore Γ¯(i) = k(i) k Γ¯. ⊃ ⊃ But (Γ¯(i) : Γ¯) = 2 so that k = k(i) = Γ¯(i). We have thus found all complete ﬁelds with archimedian valuation.

7 Extension of valuation of an incomplete ﬁeld

Suppose k is a complete ﬁeld under a valuation and let Ω be its alge- 233 || braic closure. Then can be extended to Ω by the prescription ||

1 α = Nα n , | | | | where Norm is takes form k(α) over k and n = (k(α) : k). It is clear that it defines a valuation function. For, of K is a subfields of Ω and K/k is finite and K contains α then, by properties of norms,

1 α = N / α m , | | | K k | where m = (K : k). So, if α and β are in Ω, we may take for K a field containing α and β and with (K : k) finite. Furthermore, defined as such, the valuation on Ω is dense because, if α > 1, then α 1/n has value as near 1 as one wishes, by increasing n | | | | sufficiently. Also, for every n, α1/n is in Ω. Also, let σ be an automorphism of Ω/k, and α in Ω. Then, by definition of norm, Nα = N(σα) so that α = σα . Thus all conjugates of an element have the same | | | | value. We shall now study how one can extend a valuation of an incomplete field to an algebraic extension. Let k be a field and K a finite algebraic extension of it. Let be a || valuation of k and k¯ the completion of k under this valuation. Let k , k¯. 234 206 8. Valuated fields

Kk¯ uu uu uu K

k¯ uu uu uu uu k

Suppose it is possible to extend to K. Let K¯ be the completion of || K under this extended valuation. Since K¯ K k, it follows that K¯ ⊃ ⊃ contains K and k¯ and therefore the composite Kk¯. Thus

K¯ Kk¯. ⊃ On the other hand, Kk¯/k¯ is a ﬁnite extension, since K/k is ﬁnite. Since k¯ is complete, Kk¯ is complete also. Kk¯ contains K and hence its completion K¯ under this extended valuation. Thus

K¯ = Kk¯.

Thus if the valuation can be extended, then the completion of K by this extended valuation is a composite extension of K and k¯. Suppose now that Ω is an algebraic closure of k¯. Ω, then, contains an algebraic closure of k. We have seen above that the given valuation of k can be extended to Ω. Let σ be an isomorphism of K/k into Ω. The valuation in k¯ can be extended to σK k¯ which is a subﬁeld of Ω. · Therefore, there is a valuation on σK. Deﬁne now, for α in K,

α = σα | |o | | where is the extension of on k to σK k¯, which extension is unique. || || · It is now trivial to see that is a valuation on K and extends on k. ||o || Hence every isomorphism of K into Ω which is trivial on k, gives rise to a valuation of K. 235 We now inverstigate when two isomorphisms give rise to the same valuation on K. Let σ and τ be two isomorphisms of K/k into Ω giving the same valuation on K. σK and τK are subﬁelds of Ω and they have 7. Extension of valuation of an incomplete ﬁeld 207

1 the same valuation. Thus µ = στ− is an isomorphism of τK onto σK which preserves the valuation on τK. Now µ is identity on k and so on k¯. Since σK k¯ is the completion of σK, it follows that µ is an isomorphism · of the composite extensions σKk¯ and τKk¯. Hence, if σ and τ give rise to the same valuation on K, the corresponding composite extensions are equivalent. Suppose now that σ and τ are isomorphisms of K into Ω such that the composite extensions σKk¯ and τKk¯ are equivalent. There exists then a mapping µ of τKk¯ on σKk¯ which is identity on k¯ and such that

µτ = σ

σ induces a valuation on K such that α = σα and τ induces a ||1 | |1 | | valuation on K such that α = τα . But µ is such that µτα = σα or ||2 | |2 | | σα and τα are conjugates over k¯ in Ω. Hence

σα = τα | | | | or = which shows that σ, τ give the same valuation on K. ||1 ||2 We have hence the Theorem 8. A valuation of k can be extended to a finite extension K || of k only in a finite number of ways. The number of these extensions of 236 to K stand in a (1, 1) correspondence with the classes of composite || extensions of K and k¯ From what we have already seen, the number of distinct composite extensions of K and k¯ is at most (K : k). We apply these to the case where k = Γ, rational number field and K/Γ finite so that K is an algebraic number filed. From theorem 2, it therefore follows that K has at most (K : Γ) distinct archimedian valuations and that all the non-archimedian valuations of K, which are countable in number, are discrete. In a similar manner, if K is an algebraic function field of one variable over a constant filed k, then all the valuations of K are non-archimedian and discrete. This can be seen from the fact that if x K is transcenden- ∈ tal over k, then K/k(x) is algebraic and one has only to apply theorems 3 and 8.

Appendix

Abelian groups 1 Decomposition theorem 237 All the groups that we deal with here are abelian. Before proving the main decomposition theorem for ﬁnite abelian groups we shall prove some lemmas. Lemma 1. If a, b are elements in G and have orders m and n respectively and (m, n) = 1, then ab has order mn. Proof. Clearly if t is the order of ab, t mn, since | (ab)mn = (am)n(bn)m = e,

t t e being unit element of G. Also a = b− . Thus tm tm e = a = b− so that n/t. Similarly m t. Hence t = mn. | Lemma 2. Let p be a prime number dividing the order n of the group G. Then there is, in G, an element of order p. Proof. We use induction on n. Let a be an element in G of order m. If p m, then am/p has order p and we are through. Suppose p ∤ m. Let H | be the cyclic group generated by a G/H has then order n/m which is n · divisible by p. Since n, induction hypothesis applies, so that there is m| a coset Hb of order p. If b has order t then bt = e and so (Hb)t = H which means that p t and so bt/p has order p. | 209 210 8. Appendix

Lemma 3. Let G be a ﬁnite group and λ the maximum of the orders of elements of G. Then aλ = e

238 for all a G. ∈ Proof. Let b be an element of order λ. Let a be any element in G and let µ be its order. To prove the lemma, it is enough to prove that µ λ. If | not, there is a prime p which divides µ to a higher power than it does λ. Let pr be the highest power of p dividing µ and ps the highest power dividing λ. Then r > s. We will see that this leads to a contradiction.

µ µ s Since and pr are coprime, a pr has the order pr. Similarly bp has pr λ order . By lemma 1, ps µ s c = a pr bp · λ has order pr. > λ, which contradicts the definition of λ. Hence µ λ. ps | Lemmas 2 and 3 show that λ n where n is the order of the group and | that λ and n have the same prime factors. λ is called the exponent of the finite group G. A set a1,..., an of elements of a finite group G are said to be independent if ax1 axn = e 1 ··· n xi = = implies ai e, i 1,..., n. If G is a finite group and is a direct product of cyclic groups G1, ..., Gn and if ai, i = 1,..., n is a generator of Gi, then a1,..., an are independent elements of G. They are said to form a base of G. 239 We shall now prove

Theorem 1. Let G be a ﬁnite group of order n. Then G is the direct product of cyclic groups G1,..., Gl of orders λ1,...,λl such that λi λi 1, i = 2,..., l, λℓ > 1. | − Proof. We prove the theorem by induction on the order n of the group G. Let us therefore assume theorem proved for groups of order < n. Let 1. Decomposition theorem 211

G have order n. Suppose λ1 is the exponent of G. If λ1 = n, then G is cyclic and there and there is nothing to prove. Let therefore λ1 < n. There is an element a1 of order λ1. Let G1 be the cyclic group generated n by a1. G/G1 has order < n. Hence induction hypothesis works on λ1 G/G1.

G/G1 is thus the direct product of cyclic groups W2,..., Wl of order λ2,...,λl respectively and λl λl 1 ... λ2. Let Hi be a generator of Wi. | − | | Then Hi = Glbi for some bi in the coset Hi. Let the element bi in G have order t . Then t λ by lemma 3. But i i| 1 ti = ti ti = Hi bi G1 G1 which proves that λi λ1. Thus λl λl 1 ... λ1. | | − | | Let now bλi = axi . Put x = y z where (y , λ ) = 1 and all the prime i 1 i i · i i 1 factors of zi divide λ1. Choose ui prime to λ1 such that

u y 1(mod λ ). i i ≡ 1 uiλi = zi Then bi a1 . Since λ1 is the exponent of G, 240

= uiλ1 = z1 λ1/λi e bi (a1 ) .

Since a has order λ this means that λ z , i = 2, 3,..., l. 1 1 i| i Put now = ui zi/λi ai bi a−1 .

Since ui is prime to λ1 and so to λi, the coset Fi = Glai is also a generator of Wi. Thus G1a2,..., G1al is a base of G/G1. Let ai have order fi. Then

= fi = ui fi fizi/λi e ai bi a−i .

This means that ui fi = fizi/λi bi ai . Therefore by deﬁnition of b , λ u f . But (u λ ) = 1. Hence λ f . i i| i i i i i| i 212 8. Appendix

On the other hand

λi = ui λi zi = ai bi ai− e.

Hence fi = λi. We have thus elements a1,..., al in G which have orders λ1,...,λl satisfying λl λl 1 ... λ1. | − | | We maintain that a1,..., a1 are independent elements of G. For, if v1 vl = a1 ... al e, v2 vl v1 v2 vl 241 then a a = a− which means that F ... F = G . But since 2 ····· l 1 2 l 1 F2,..., Fl are independent, λi vi, i = 2,..., l. But this will mean that v | a 1 = e or λ v . 1 1| 1 Since λ1 ...λl = n, if follows that a1,..., al form a base of G and the theorem is proved. Let G be a group of group of order n and let G be direct product of cyclic groups G1, G2,..., Gl of orders λ1,...,λl. We now prove Lemma 4. Let µ be a divisor of n. The number N(µ) of elements a G ∈ with aµ = e is given by ℓ N(µ) = (µ, λi). = Yi 1 Proof. Let a ,..., a be a base of G so that a is of order λ . Any a G 1 l i i ∈ has the form a = ax1 axl . 1 ··· l If aµ = e, then e = ax1µ axlµ. Since a ,..., a is a base, this means that 1 ··· l 1 l x µ 0(mod λ ), i = 1, λ, l. Hence x has precisely (µ, λ ) possibilities i ≡ i i i and our lemma is proved. We can now prove the

Theorem 2. If G is the direct product of cyclic groups G1,... Gl, of orders λ1,...,λl respectively with λl λl 1 λ1 and G is also the direct | − |···| product of cyclic groups H1,..., Hm of orders µ1,...,µm respectively 242 with µm µm 1 µ1, then m = 1 and | − |···| λi = µi, i = 1,..., l. 2. Characters and duality 213

Proof. Without loss in generality let l m. Let a ,..., a be a base of G ≥ 1 l in the decomposition G G G . Since the number of elements 1 × 2 ×···× 1 a with aµ = e is independent of the decomposition

ℓ m N(µ) = (µ, λi) = (µ,µ j) = = Y1 1 Yj 1 Put now µ = λ . Then N(µ) = λ1. But since (µ,µ ) µ, it follows 1 1 j ≤ that λℓ λm, so that l m. This proves ℓ ≤ ℓ ≤ l = m.

Also it follows that each factor (λ ,µ ) = λℓ or λℓ µℓ. Inverting the 1 j | roles of λ and µ we get λℓ = µℓ.

Suppose now it is proved that λq+1 = µq+1,...,λ1 = µ1. Then by putting µ = λq, we have

q q N(µ) = λq.λq+1 λ1 = (λq,µ j)λq+1 λℓ. ··· = ··· Yj 1

By the same reasoning as before, it follows that λq = µq and we are, therefore, through. For this reason, the integers λ1,...,λl are called the canonical invariants of G. From theorems 1 and 2 we have the Corollary Two ﬁnite groups G and G′ are isomorphic if and only if they have the same canonical invariants.

2 Characters and duality 243 Let G be a group, not necessarily abelian and Z a cyclic group. A homomorphism χ of G into Z is called a character of G. Thus

χ(a)χ(b) = χ(ab). 214 8. Appendix

If we denote the unit element of G by e and that of Z by 1, then

χ(e) = 1.

The character χ deﬁned by χ (a) = 1 for all a G is called the o o ∈ Principal character. If χ1 and χ2 are two characters, we deﬁne their product χ = χ1χ2 by

χ(a) = χ1(a)χ2(a)

and the inverse of χ1 by 1 = 1 χ1− (a) (χ1(a))− . Under this deﬁnition, the characters form a multiplicative abelian group G∗ called the character group of G. Since χ is a homomorphism, denote by Gχ the kernel of the homomorphism χ of G into Z. Then G/Gχ is abelian. Denote by H the subgroup of G given by

H = Gχ, χ G∗. χ ∈ \ Then clearly G/H is abelian. We call Z an admissible group for G if H consists only of the identity 244 element. This means ﬁrst that G is abelian and furthermore that given any two elements a, b in G there exists a character χ of G such that, if a , b, χ(a) , χ(b). If χ is a character of G, then χ can be considered as a character of G/Go where χ is trivial on Go, by deﬁning χ(Goa) = χ(a), Goa being a coset of G modulo Go. In particular, the elements of G∗ can be considered as characters of G/H. Moreover Z is now an admissible group for G/H. We now prove the

Theorem 3. If G is a ﬁnite abelian group and Z is an admissible group for G, then G∗ is ﬁnite and G is isomorphic to G∗. 2. Characters and duality 215

Proof. In the first place Z is a finite group. For, if a G, a , e, there is ∈ a character χ such that χ(a) , 1. If G is of order n, then χ(an) = (χ(a))n = 1 so that χ(a) in an element of Z of finite order. Since Z is cyclic it follows that Z is finite.

From this it follows that G∗ is finite. Since (χ(a))n = 1 for every χ and every a, there is no loss in generality if we assume that Z is a cyclic group of order n. In order to prove the theorem let us first assume that G is a cyclic group of order n. Let a be a generator of G and b a generator of the cyclic group Z of order n. Define the character χ1 of G by 245

χ1(a) = b.

Since a generates G any character is determined uniquely by its effect on a. χ1 is an element of order n in G∗. Let χ be any character of G. Let χ(a) = bµ µ for some integer µ. Consider the characterχ ˜ = χ χ− . · 1 µ µ µ χ˜(a) = (χ1(a))− χ(a) = b− b = 1 which shows thatχ ˜ = χo is the principal character. Hence G∗ is a cyclic group of order n and the mapping

a χ → 1 establishes an isomorphism of G on G∗. Let now G be ﬁnite non-cyclic abelian of order n. G is then a direct product of cyclic groups G1,..., Gl of orders λ1,...,λl respectively. Let ai be a generator of Gi so that a1,..., al is a base of G. Since λ1,...,λl divide n, we deﬁne l characters χ1, χ2,...,χl of G by

χi(a j) = 1 j , i

χi(ai) = bi i = 1,..., l, 216 8. Appendix

where bi is an element in Z of order λi. These characters are then independent elements of the abelian group G . For, if χt1 χtl = χ , then, ∗ 1 ··· l 0 for any a G, ∈ t1 tl = χ1 (a) ...χℓ (a) 1. 246 Taking for a successively a ,..., a we see that λ t and so χ ,...,χ 1 l i| i 1 l are independent. Let χ be any character of G. Then χ is determined by its eﬀect on = λi = λi = λi = a1,..., al. Let χ(ai) si. Since ai e, (χ(ai)) 1. But (χ(ai)) λi λi = si . Thus si 1. Z being cyclic, there exists only one subgroup of order λi. Thus = = µi χ(ai) si bi , µ1 for some integer µ (mod λ ). Consider the characterχ ˜ = χχ− χ. It i i 1 ··· us clear thatχ ˜(ai) = 1 for all i so thatχ ˜ = χo or

µ µ χ = χ1 χℓ . 1 ··· ℓ Thus G∗ is the product of cyclic group generated by χ1,...,χ1. By Corollary to theorem 2, it follows that G and G∗ are isomorphic.

Corollary. If G is a finite group and G∗ its character group, then whatever may be Z, OrderG∗ OrderG. ≤ Proof. For, if H is the subgroup of G defined earlier, then G/H is abelian and finite, since G is finite. Also Z is admissible for G/H. Furthermore every character of G can be considered as a character of G/H, by definition of H. Hence by theorem 3

Order G Order G/H Order G. ∗ ≤ ≤ Let us now go back to the situation where G is finite abelian and Z 247 ia admissible for G. Then G G . Let us define on G G the function ≃ ∗ × ∗ (a, χ) = χ(a). For a fixed χ, the mapping a (a, χ) is a character of G and so an → element of G∗. By definition of product of character, it follows that, for fixed a, the mapping a¯ : χ (a, χ) → 3. Pairing of two groups 217

is a homomorphism of G∗ into Z and hence a character of G∗, Let G∗∗ denote the character group of G∗. By Corollary above,

OrderG∗∗ OrderG∗ = OrderG. ≤ Consider now the mapping σ : a a¯ → of G into G∗∗. This is clearly a homomorphism. Ifa ¯ is identity, then (a, χ) = 1 for all χ. But since Z is admissible for G, it follows that a = e. Hence σ is an isomorphism of G into G∗∗. Therefore we have Theorem 4. The mapping a a¯ is a natural isomorphism of G on G → ∗ ∗∗ Note that the isomorphism of G on G∗ is not natural. Under the conditions of theorem 3, we call G∗ the dual of G. Then G∗∗ is the dual of G∗ and theorem 4 shows that the dual of G∗ is naturally isomorphic to G. Theorem 4 is called the duality theorem for ﬁnite abelian groups.

3 Pairing of two groups

Let G and G′ be two groups, σ, σ′,..., elements of G′ and τ, τ′,..., elements of G′. Let Z be a cyclic group. Suppose there is a function 248 (σ, τ) on G G into Z such that for every σ, the mapping × ′ λσ : τ (σ, τ) → is a homomorphism of G′ into Z and for every τ, the mapping

µτ : σ (σ, τ) → is a homomorphism of G into Z, Thus λσ and µτ are characters of G′ and G respectively. We then say that G and G′ are paired to Z and that (σ, τ) is a pairing. For every σ, let Gσ′ denote the kernel in G′ of the homomorphism λσ. Put = H′ Gσ′ σ \ 218 8. Appendix

for all σ G. Then G /H is abelian. Also, H is the set of τ in G such ∈ ′ ′ ′ ′ that (σ, τ) = 1 for all σ G. Deﬁne in a similar way the subgroup H of G. ∈ We are going to prove

Theorem 5. If G/H is ﬁnite, then so is G′/H′ and both are then isomorphic to each other. Proof. From ﬁxed σ, consider the function

χσ(τ) = (σ, τ).

This is clearly a character of G′. By deﬁnition of H′, it follows that for each σ, χσ is a character of G′/H′. Consider now the mapping

σ χσ → of G into (G′/H′)∗. This is again a homomorphism of G into (G′/H′)∗. The kernel of the homomorphism is the set of σ for which χσ is the principal character of G′/H′. By deﬁnition of H, it follows that H is the kernel. Hence G/H a subgroup of (G′/H′)∗. (1) ≃ 249 In a similar way

(G′/H′) a subgroup of (G/H)∗. (2) ≃ Let now G/H be ﬁnite. Then by Corollary to theorem 3,

ord(G/H)∗ ordG/H. ≤ By (2), this means that

order(G′/H′) order G/H. ≤ Reversing the roles of G and G′ we see that G/H and G′/H′ have the same order. (1) and (2) together with theorem 3 prove the theorem. If, in particular, we take G∗ for G′, then H′ = (χo) and so we have 3. Pairing of two groups 219

Corollary. If G is any group, G∗ its character group and if G/H is ﬁnite then G/H G∗. ≃

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