These Notes Were Taken Down by Me (Boris Lerner) During the Galois Theory Course Taught by Dr Daniel Chan in the Second Session of 2007 at UNSW

These notes were taken down by me (Boris Lerner) during the Galois Theory course taught by Dr Daniel Chan in the second session of 2007 at UNSW. I made a few brief remarks of my own here and there but in general this is my best attempt to reproduce exactly what was written on the board. The last four sections were not examinable so contain far less detail then the other sections. I have also not reproduced all the diagrams which accompanied the text from those sections. If you ﬁnd any mistakes, please email them to me at [email protected]. Enjoy!

1 .

Contents

1 Genesis of Galois Theory 6 1.1 Symmetry and Roots of Polynomials ...... 7

2 Splitting Fields 7 2.1 Review of Field Homomorphisms ...... 8 2.2 Review of Simple Algebraic Extensions ...... 8 2.3 The Converse: Adjoining Roots of Polynomials ...... 8 2.4 SplittingFields ...... 9 2.5 Existence ...... 10 2.6 Uniqueness ...... 10

3 Algebraic Closure 10 3.1 Review of Algebraic Extensions ...... 10 3.2 SetTheoreticalFact ...... 11 3.3 Splitting Fields for Families of Polynomials ...... 11 3.4 ExistenceandUniqueness ...... 12

4 The Galois Group 13 4.1 Field Automorphism over F ...... 13 4.2 Galois Group as Group of Permutations ...... 13 4.3 Constructing Field Automorphisms Over F ...... 14

5 Weak Galois Correspondence 15 5.1 FixedFields...... 15

6 Technical Results 18

7 Galois Correspondence 20 7.1 GaloisExtension ...... 20

8 Normality 22 8.1 StabilityofSplittingFields...... 23 8.2 Galois Action on Galois Correspondence ...... 23 8.3 Normal Subgroups in Galois Correspondence ...... 24

2 9 Separability 25 9.1 Inseparability ...... 25 9.2 Separable Polynomials, Elements and Extensions ...... 25

9.3 Multiplicity of [K : F ]S ...... 26 10 Criterion for Galois 27 10.1 Galois SeparableSplittingﬁeld ...... 28 10.2 SplittingFields⇐⇒ ...... 28 10.3NormalClosure ...... 29

11 Radical Extension 29 11.1 RadicalExtensions ...... 30 11.2 Galois Group of Simple Radical Extension ...... 30 11.3Solvability ...... 31

12 Solvable Groups 32 12.1BasicFact...... 32 12.2 DerivedSeries...... 33 12.3 AnInsolvableGroup ...... 34

13 Solvability by Radicals 34 13.1 Solvability of polynomials ...... 35 13.2 General degree n monicpolynomial ...... 36

14 Some Applications 36 14.1 A polynomials not solvable by radicals ...... 36 14.2 Primitive element theorem ...... 37 14.3 Fundamental theorem of algebra ...... 37

15 Trace and Norm 38 15.1TraceandNorm...... 38 15.2 Linear independence of characters ...... 40

16 Cyclic Extensions 41 16.1 Hilbert’stheorem90 ...... 41 16.2 Artin-SchreierTheory ...... 42

17 Solvable Extension 43 17.1Generalcubic ...... 44

3 18 Finite Fields 45 18.1 FrobeniusHomomorphism ...... 45 18.2 Classification of finite fields ...... 46 18.3 Latticeoffinitefields ...... 47 18.4Someexamples ...... 47

19 Pro-Finite Groups 47 19.1 Topological groups ...... 47 19.2Inverselimits ...... 48 19.3 Pro-ﬁnitegroups ...... 50

20 Infinite Galois Groups 50 20.1 Inverse system of finite Galois groups ...... 51 20.2 InfiniteGaloisgroups...... 52 20.3 AbsoluteGaloisgroup ...... 53

21 Infinite Galois Theory 54 21.1 Basicfactsaboutpro-finitegroups...... 54 21.2 Infinite Galois correspondence ...... 55

22 Inseparability 56 22.1 Purely inseparable extensions ...... 56 22.2 Maximal separable and purely inseparable extensions . . . . . 57 22.3 Normalextensions ...... 58

23 Duality 59 23.1Thedualgroup ...... 59 23.2 Perfectpairings ...... 60 23.3 Some abelian extensions ...... 60

24 Kummer Theory 61 24.1 Aperfectpairing...... 61 24.2 Classiﬁcation of abelian extensions ...... 62

25 Galois Correspondence in Topology 63 25.1 (Unramiﬁed)covers ...... 64 25.2 Galoiscorrespondence ...... 64 25.3 Galoiscorrespondence ...... 65 25.4 Simply connected cover ...... 66

4 26 Riemann Surfaces 67 26.1 Riemannsurfaces...... 67 26.2 Field of meromorphic functions ...... 67 26.3 Functoriality ...... 68 26.4 Riemann’sTheorem ...... 69

27 Geometric examples of field automorphisms 69 27.1Galoisgroups...... 69 27.2 Unramified covers of elliptic curves ...... 70 1 27.3 Ramified Cyclic Covers of PC ...... 70

28 Ramiﬁcation Theory 71 28.1 Discrete valuation rings ...... 71 28.2 Motivation for Ramiﬁcation ...... 71 28.3 Ring-Theoretic Reformulation ...... 72

5 1 Genesis of Galois Theory

Q: What is Galois Theory? A: Study of ﬁeld extensions K/F via symmetry.

Original motivating example: We can ﬁnd roots of some polynomials /F (ﬁeld).

2 b √b2 4ac Example 1.1. f(X)= X + bX + c has roots − ± − if char F = 2. 2 6 Example 1.2. f(X)= X3 pX q has roots − − 1 1 q2 p3 X = 3 q + β + 3 q β where β = if char F = 2. r2 r2 − r 4 − 27 6 Quadratic formula also exists. Note 1.3. Formulae for roots are in terms of coefficients of polynomials f(X) and uses elementary operations +, , , and radical √n . − × ÷ A major question in the 18th century algebra: can we find a similar formula for a quintic? A: (Abel, Galois, early 1800’s) No! Modern approach: use field extensions. Let K/F be a field extension. Let α ,...,α K. Recall the field generated by F and α ,...,α is the 1 n ∈ 1 n unique smallest subfield F (α1,...,αn) K containing F, α1,...,αn. Example 1.1 (again). The roots of f(⊆X)= X2+bX+c lie in F √b2 4ac . Example 1.2 (again). The roots of f(X)= X3 pX q lie in F (β,γ,δ− ) − − where γ = 3 1 q + β, δ = 3 1 q β. 2 2 − We haveq a tower of fieldq extensions: F F (β) F (β,γ) F (β,γ,δ). ⊂ ⊂ ⊂ Definition 1.4. A field extension K/F is radical if there exists a tower of field extensions

F = F F F F = K where F = F (α ) 0 ⊂ 1 ⊂ 2 ⊂···⊂ n i+1 i i where αri F for some r 1. i ∈ i i ≥ Upshot: Suppose you can solve for roots of f(X) F [X] by radicals i.e. as in note above. Let α be such a root. Then α lies in∈ a radical extension of F . Galois theory gives a good criterion for checking this (in characteristic 0) in terms of symmetry.

6 1.1 Symmetry and Roots of Polynomials Q: How does symmetry help you understand roots of polynomials? Example 1.5. The non-real roots of a real polynomial occur in symmetric complex conjugate pairs.

Example 1.1 (again). Quadratic formula by symmetry: Let x1, x2 be roots of X2 + bX + c = 0. Key point: Any symmetric polynomial in roots x1, x2 is a polynomial in the coeﬃcients. E.g: x + x = b 1 2 − x1x2 = c

2 Note x1 x2 is antisymmetric, which implies (x1 x2) is symmetric. But (x x )−2 =(x + x )2 4x x = b2 4c - the discriminant.− Solve 1 − 2 1 2 − 1 2 − x + x = b 1 2 − x x = √b2 4c 1 − 2 − 3 Exercise: repeat for cubic f(X)= X pX q. Hint: Let x1, x2, x3 be roots. Let ω be a primitive cube root of 1. Suﬃces− − to ﬁnd

Σ1 : x1 + x2 + x3 = 0 2 Σ2 : x1 + ωx2 + ω x3 2 Σ3 : x1 + ω x2 + ωx3

Now

3 3 3 3 2 2 2 2 2 2 2 Σ2 = x1+x2+x3+3!x1x2x3+3ω x1x2 + x1x3 + x2x3 +3ω x1x3 + x2x3 + x2x1 .

Note 1.6. It has A3-symmetry, i.e. symmetric with respect to cyclic permutations of variables. Show it is the sum of a symmetric polynomials and an anti-symmetric polynomial. Recall: have an anti-symmetric polynomial

δ =(x x )(x x )(x x ). 1 − 2 2 − 3 1 − 3 2 Write Σ3 in terms of coeﬃcients and δ.

2 Splitting Fields

Aim: Show that given any polynomial, can factorise it into linear polynomials is some ﬁeld extension.

7 2.1 Review of Field Homomorphisms

Proposition/Definition 2.1. A map of fields σ : F F ′ is a field homomorphism if it is a ring homomorphism. → 1. σ is injective 2. There is a ring homomorphism

σ[X]: F [X] F ′[X] n −→ n f Xi σ(f )Xi i 7−→ i i=0 i=0 X X Proof. 2. Easy. 1. ker σ ⊳ F not containing 1, so ker σ = 0.

2.2 Review of Simple Algebraic Extensions Let K/F be a field extension and α K be algebraic over F , so it is a root of its minimal or irreducible polynomial∈ say p(X). Fact: There is a field isomorphism F [X] ∼ F (α) p(X) −→ h i X + p(X) α h i 7−→ a + p(X) a. h i 7−→ √3 Q[X] Example 2.2. Q 2 X3 2 . ≃ h − i 2.3 The Converse: Adjoining Roots of Polynomials Let F = field, and p(X) F [X] be irreducible. ∈ Proposition 2.3. Then K := F [X]/ p(X) is a field extension of F via the composite ring homomorphism h i F [X] F F [X] . −→ −→ p(X) h i Also, K = F (α) where α = x + p(X) is a root of p(X). h i Proof. p(X) is irreducible in PID which implies p(X) ⊳ F [X] is maximal and thus F [X]/ p(X) is a field. It is clear thath K =i F (α) and p(α) = p(X)+ p(X) =h 0 so iα is a root of p(X). h i 8 Have the following uniqueness result:

Proposition 2.4. Let σ : F ∼ F ′ be a field isomorphism. Let p(X) F [X] −→ ∈ be irreducible. Let α and α′ be roots of p(X) and (σp)(X) (in appropriate field extensions). Then there is a field isomorphism σ˜ : F (α) ∼ F ′(α′) such that: −→

1.σ ˜ extends σ˜ = σ |F

2.σ ˜ = α′

Proof. σ˜ is the composite of the following:

F [X] F ′[x] F (α) ∼ ∼ F ′(α′). −→ p(X) −→ (σp)(X) −→ h i h i

α X + p(X) X + (σp)(X) α′ = (2) a 7−→ a + ph(X) i 7−→σ(a)+h (σp)(Xi) 7−→ σ(a)⇒ 7−→ h i 7−→ h i 7−→ 2.4 Splitting Fields Definition 2.5. Let F be a field, f(X) F [X]. A field extension K/F is a splitting field for f(x) over F if ∈

(a) f(X) factors into linear polynomials over K.

(b) K = F (α1,...,αn) where α1,...,αn are the roots of f(X) in K.

i2π 3 3 Example 2.6. Let ω = e 3 . Q(√2) is not a splitting ﬁeld for X 2 over Q but Q(√3 2, √3 2ω, √3 2ω2)= Q(√3 2,ω) is. − Note 2.7. Consider the tower of ﬁeld extensions F K L and f(X) F [X]. ⊂ ⊂ ∈

1. If L is a splitting ﬁeld for f(X) over F then L is a splitting ﬁeld for f(X) over K.

2. If K is generated over F by roots of f(X), then the converse holds.

The proof of this is left as an exercise.

9 2.5 Existence Theorem 2.8. Let F be a ﬁeld, f(X) F [X]. Then there exists a splitting ﬁeld K of f(X) over F . ∈

Proof. By induction on the degree of f(X). By Proposition 2.3, there exists a simple algebraic extension F (α)/F where α is a root is a root of f(X). Factorise in F (α) to get f(X)=(X α)g(X). By induction, there exists a splitting field K of g(X) over F (α).− Now K is a splitting field for f(X) over F (α) which by the note above implies, it is also a splitting field of f(X) over F .

2.6 Uniqueness

Theorem 2.9. Let σ : F ∼ F ′ be a field isomorphism and f(X) F [X]. −→ ∈ Suppose K,K′ are splitting fields for f(X) and (σf)(X) respectively over F and F ′ (respectively). Then there is an isomorphism of fields σ˜ : K ∼ K′ which extends σ. −→

Proof. By induction on [K : F ] = dimF K using Proposition 2.4. Let f (X) F [X] be an irreducible factor of f(X). Let α K, α′ K′ be roots 0 ∈ ∈ ∈ of f0(X) and (σf0)(X) respectively. Then get diagram:

σ / F _ F′_

by Prop 2.4 ∃ / F (α ) F ′(α ′) _ ∼ _ splitting ﬁeld for f(X) by Note 2.7 isom by induction ∃ / K K′ ∼

3 Algebraic Closure

Aim: Prove the existence and uniqueness of algebraic closure of ﬁelds.

3.1 Review of Algebraic Extensions Let F be a ﬁeld.

10 Definition 3.1. A field extension K/F is algebraic over F is every α K is algebraic over F . A field K is algebraically closed is the only algebraic∈ extensions of K is K itself. We say K is an algebraic closure of F if it is an algebraic field extension of F such that K is algebraically closed.

Example 3.2. C is the algebraic closure of R.

Proposition 3.3. Let F K L be a tower of ﬁeld extensions. Then: ⊂ ⊂ 1. [L : F ] = [L : K] [K : F ];

2. L/F is algebraic if and only if both L/K and K/F are algebraic;

3. Finite extensions are algebraic;

4. If K = F (α1,...,αn) where α1,...,αn are algebraic over F then K is ﬁnite over F . (i.e. ﬁnite dimensional over F ).

3.2 Set Theoretical Fact Lemma 3.4. Let F be a field. There exists a set S such that for any algebraic field extension K/F , K < S . | | | | Remember: Either K is countable or K = F . | | | | Proof. There is a finite-to-one map

K F [X] −→ α monic mimimal 7−→ polynomial of α

= K F [X] N Let⇒S |be| the ≤ | power× set| of F [X] N. Then we are done since F [X] N < × | × | S . | | 3.3 Splitting Fields for Families of Polynomials

Let F be a ﬁeld, fi(X) i I F [X] 0. { } ∈ ⊂ −

Definition 3.5. A field extension K/F is a splitting field for fi(X) over F if { }

(a) Every fi(X) factors into linear factors over K;

(b) K is generated as a ﬁeld over F by all the roots of the fi(X).

11 Note 3.6. Consider a ﬁeld extension K/F . Let αi j J K. Then the { } ∈ ⊆ subﬁeld of K generated by F and the αj’s is

of all subﬁelds of K contating F and all the αj’s = F (αj1 ,...,αjn )

j1,...,jn J \ { [ }⊆ Why? ( ) is clear. ⊇ ( ) Suffices to check RHS is a subfield. But closure axioms are easy to check ⊆ (exercise). For example: let α F (αj1 ,...,αjn ) and β F (αl1 ,...,αlm ) then α + β, αβ F (α ,...,α α∈ ,...,α ), etc. ∈ ∈ j1 jn l1 lm Example 3.7. If K is a splitting field for all non-constant polynomials over F then K is an algebraic closure of F . Why? Check first K/F is algebraic. Any α K lies in some F (α1,...,αn) where α1,...,αn are roots of some non-constant∈ over F and therefore algebraic over F . Proposition 3.3 (part 4) is algebraic over F and thus α is algebraic over F . Therefore K/F is algebraic. Check K is algebraically closed. Suppose K/K˜ is and algebraic extension. Then K/F is algebraic, which implies by Proposition 3.3 (part 2) that K/F˜ is algebraic. Let α K˜ be arbitrary. Now α is algebraic over F so satisfies minimal polynomial∈ f(X). But K is a splitting field for all non-constant polynomials over F and thus α K. Therefore K˜ = K and hence K is algebraically closed. ∈ Converse is also true (exercise).

3.4 Existence and Uniqueness

Theorem 3.8. Let F be a field, fi(X) i I F [X] 0. { } ∈ ⊆ − 1. There exists a splitting field K for f (X) over F . { i } 2. Suppose there is a field isomorphism σ : F F ′ and K′ is a splitting → field for (σfi)(X) i I over F ′. Then there exists a field isomorphism { } ∈ σ˜ : K ∼ K′ which extends σ. −→ Proof. Same as Theorems 2.8 and 2.9, except we replace induction with Zorn’s Lemma. We will prove (1) here, (2) similar. Let S be the set as in Lemma 3.4. Let L be the set of subsets of S equipped with a field structure such that L L is a splitting field for some subset of fi(X) i I over F . ∈ Let’s partially∈ order L by L L if L is a subfield of L{ . Given} a totally 1 ≤ 2 1 2 ordered chain Lj j J of elements of L , we note it has an upper bound { } ∈ L L . Hence by Zorn’s Lemma, there exists a maximal K L . ∪ j ∈ ∈ Claim: K is a splitting field for fi(X) i I over F . Why? Suppose not. { } ∈ Some fi(X) is not split in K. Let K˜ be a splitting field for fi(X) over K. So

12 K˜ is algebraic over F . Lemma 3.4 implies there is a map K˜ K S K. This gives a map φ : K˜ S extending id on K. The image− of φ→is bigger− than K which contradicts→ maximality of K. This proves the ﬁrst part of the theorem. Corollary 3.9. For any ﬁeld F , there exists a unique (up to isomorphism) algebraic closure, denoted F¯.

4 The Galois Group

Aim: Encode the symmetry of ﬁeld extension K/F in the Galois group.

4.1 Field Automorphism over F

Let K, K′ be ﬁeld extensions of F .

Proposition/Definition 4.1. We say a field homomorphism σ : K K′ fixes F , or σ is a field homomorphism over F , if σ(α)= α for all α→ F . ∈ Such homomorphisms are linear over F . If, furthermore, K = K′, and σ is an isomorphism, then we say σ is a field automorphism over F . Proof. σ is additive and for α F, β K, σ(αβ) = σ(α)σ(β) = ασ(β). So it preserves scalar multiplication∈ by elements∈ of F . Example 4.2. K = C,F = R. c := conjugation: C C is a field automorphism over R because if α R it implies c(α) =α ¯ =→α. ∈ Proposition/Definition 4.3. Let K/F be a field extension and G a set of field automorphisms of K over F . Then G is a group when endowed with composition for group multiplication. It is called the Galois group of K/F and is denoted by Gal(K/F ). Proof. Suffices to check G Perm K. Let σ, τ G. στ is a field automorphism. It fixes any α F since≤ (στ)(α) = σ(τ(α∈)) = σ(α) = α. Therefore 1∈ στ G. Similarly σ− G and 1 G. ∈ ∈ ∈ 4.2 Galois Group as Group of Permutations Let F be a field. Lemma 4.4. Let f(X) F [X], K/F a field extension and α K a root ∈ ∈ of f(X). Let σ : K K′ be a field homomorphism over F . The σ(α) is also a root of f(X). →

13 n i i Proof. If f(X) = i=0 aiX then 0 = f(α) = σ(f(α)) = σ ( aiα ) = i i σ(ai)σ(α) = aiσ(α) = f(σ(α)). P P PRemark 4.5. PAny field homomorphism σ : F (α ,...,α ) K over F is 1 n → determined by the values of σ(α1),...,σ(αn). Example 4.6. Let K = Q √3 2 and F = Q. Then Gal (K/F ) = 1 since any σ Gal(K/F ) sends √3 2 to a cube root of 2 in Q √3 2 i.e. σ : √3 2 ∈ 7→ √3 2 = σ = id . ⇒ K Corollary 4.7. Let K be a splitting field for f(X) F [X] over F . Let α ,...,α be the roots of f(X) so K = F (α ,...,α ).∈ Then any σ G := 1 n 1 n ∈ Gal(K/F ) permutes the roots α1,...,αn so gives an injective group homo- . morphism G ֒ Perm(α ,...,α ) S → 1 n ≃ n Example 4.2 (again). Gal(C/R) = 1,c Z/2Z. Why? C is the splitting field for X2 + 1 over R (C = R(i{)). Let} ≃σ Gal(C/R). There are 2 possibilities. Either: ∈

(a) σ(i)= i so σ(a + bi)= a bi ∴ σ is conjugation; − − (b) σ(i)= i so σ(a + bi)= a + bi ∴ σ is the identity.

4.3 Constructing Field Automorphisms Over F Use Proposition 2.4 and Theorem 2.9.

Lemma 4.8. Let F be a field, f(X) F [X] F . Let K be the splitting field ∈ − of f(X) over F . Suppose α, α′ are two roots if an irreducible factor f0(X) of f(X) (over F ). Then there is a σ G := Gal (K/F ) such that σ(α) = α′. ∈ In particular G acts transitively on roots of f0(X). Proof. By Proposition 2.4, there exists a field isomorphism

σ˜ : F (α) ∼ / F (α ) _ _ ′

unique?? K ∃ / K over F . But K is a splitting field for f(X) over F (α) and F (α′). Therefore, by the uniqueness of splitting fields Theorem 2.9 we can extendσ ˜ to a field automorphism over F , σ : K K so that α α′. → 7→

14 3 3 3 2 i2π Example 4.9. K = Q √2, √2ω, √2ω , ω = e 3 ,F = Q,G := Gal(K/Q). K is a splitting field for X3 2 over Q. Conjugation, τ : K K 3 3 3 is a field automorphism over Q. It swaps− √2ω √2ω2 and fixes √2.→ Note ↔ τ G ֒ S3-order 6. Lagrange’s Theorem implies G = τ or S3. Lemma h4.8i implies≤ → there exists σ G such that σ(√3 2) = √3 2ω hτ i. Thus, G S . ∈ 6∈ h i ≃ 3 Example 4.10 (Biquadratic Extension). K = Q √2, √3 =ex Q(√2+ √3) is a splitting field for (X2 2) (X2 3) over Q. Ant σ G := Gal (K/Q) must map − − ∈

√2 √2 7−→ ± √3 √3 7−→ ± φ ∴ G / Z/2Z Z/2Z × Claim: φ is an isomorphism. Why? K is a splitting field for X2 3 over Q(√2). So Proposition 2.4 = there exists a field automorphism−σ of K fixing Q(√2) and sending √3 ⇒ √3, σ G. Similarly, there exists τ G 7→ − ∈ ∈ such that τ(√2) = √2 and τ(√3) = √3. Now, σ, τ generate Z/2Z Z/2Z. This shows Gal Q(−√2, √3)/Q Z/2Z Z/2Z. × ≃ × 4 Exercise: Compute Gal Q(√2,i)/Q . (Answer: D4) 5 Weak Galois Correspondence

Aim of next three sections: Show Gal(K/F ) gives info about intermediate fields L i.e. where F L K (subfields). ⊆ ⊆ 5.1 Fixed Fields Let K be a field, and G a group of field automorphism of K.

Proposition/Definition 5.1. The fixed field of G (in K) is

KG := α K σ(α)= α, σ G . { ∈ | ∀ ∈ } KG is a subﬁeld of K.

Proof. Simply check closure axioms (under +, 0, , , 1 and inverses). Let us just check closure under +. Let α, β KG and− ×σ G. σ(α + β) = σ(α)+ σ(β)= α + β. ∴ α + β KG. etc. ∈ ∈ ∈ c Example 5.2. c: C C is conjugation. Ch i = R. → 15 3 3 3 2 3 2πi Example 5.3. K = Q √2, √2ω, √2ω = Q(√2,ω), ω = e 3 . We saw in the previous section G := Gal (K/Q)= S (on identifying 1,2,3 with 1st, 3 2nd, 3rd root) e.g. (2 3) is conjugation. We will show

H G / KH K

≤ ⊆

Q ⊃ G ⊃ <

< < <

⊃ 3 ⊃ 3 3 2 ⊂

(1 2 3) (1 2) (2 3) (1 3) Q(ω) Q(√2ω ) ⊂ Q(√2) Q(√2ω)

> h i h i h i> h i > > ⊂ 1 ⊂ Q(√3 2,ω)

(2 3) 3 3 Exercise: Show Kh i = R Q(√2,ω)= Q(√2). Surprise: We will show in this∩ case, all intermediate fields are obtained from KH as above. Let K/F be a field extension and G := Gal (K/F ). We define two maps both called prime ’.

/ H H H′ := K

subgroups / intermediate o ﬁelds K/F of G

K′ := Gal (K/K ) o K K F 0 0 ⊇ 0 ⊇ Check well-defined. (i.e. codomains are what we say they are). 1) Know H H H′ := K is a subfield of K. Need K F . For α F, σ H G, σ Gal(K/F ), so fixes F ∴ σ(α) = α⊇ = α ∈KH ∴∈KH ≤is an intermediate∈ field. ⇒ ∈

2) Know K0′ := Gal (K/K0) this is a group of field automorphisms of K (which fix K0). They all fix K0 and therefore fix F (as F K0. So it is a group of automorphism fixing F . Therefore it is a subgroup⊆ o G. Example 5.4.

/ 1 H = 1 1′ := K = K

K′ = Gal(K/K) o K 1

F ′ = Gal(K/F ) o F G

16 But suppose K/F = Gal Q(√3 2)/Q so Gal (K/F ) = 1 (from Section 4) and 3 so G G′ = 1′ = Q(√2). Therefore prime is not always a bijection. 7−→ Let K/F be a ﬁeld extension, G = Gal(K/F ). Below we let H1,H2,... denote subgroups of G and K1,K2,... denote intermediate ﬁelds of K/F . Proposition 5.5. Priming reverses inclusion. i.e.

1. H H = H′ H′ . 1 ⊆ 2 ⇒ 1 ⊇ 2

2. K K = K′ K′ . 1 ⊆ 2 ⇒ 1 ⊇ 2 H2 Proof. 1. Suppose H1 H2 and let α H2′ := K For any σ H1 ⊆ H1 ∈ ∈ ⊆ H we have σ(α)= α. ∴ α K =: H′ . 2 ∈ 1

2. Suppose K K and let σ K′ := Gal (K/K ). For any α K 1 ⊆ 2 ∈ 2 2 ∈ 1 σ(α)= α (as σ ﬁxes K and ∴ K ) = σ Gal(K/K )= K′ . 2 2 ⇒ ∈ 1 1

Proposition 5.6. 1. H H′′. 1 ⊆ 1

2. K K′′. 1 ⊆ 1 H1 H1 Proof. 1. Let σ H1, α H1′ := K . H1′′ := Gal K/K . σ(α)= α H1 ∈ ∈ H1 H1 (as α K ) = σ ﬁxes all of K . So σ Gal K/K = H′′. ∈ ⇒ ∈ 1 Gal(K/K 1) 2. Let α K, σ K1′ =: Gal(K/K1) .K1′′ := K . Now σ(α)= α ∈ ∈ K′ (as σ Gal(K/K )) ∴ α is ﬁxed by any σ K′ = α K 1 = K′′. ∈ 1 ∈ 1 ⇒ ∈ 1

Deﬁnition 5.7. We say H1 is a closed subgroup of G if H1 = H1′′. We say K1 is a closed intermediate ﬁeld of K/F if K1 = K1′′. The closure of H1, respectively K1, is H1′′, respectively K1′′.

Proposition 5.8. 1. H1′ = H1′′′.

2. K1′ = K1′′′.

Proof. 1. Previous proposition applied to K = H′ = H′ H′′′. 1 1 ⇒ 1 ⊆ 1 Proposition 5.5 applied to H H′′ = H′ H′′′. Thus H′ = H′′′. 1 ⊆ 1 ⇒ 1 ⊇ 1 1 1 2. Proved identically.

We may restrict the Galois correspondence to closed objects to get:

17 Theorem 5.9. Priming induces a well deﬁned bijection:

/ H H H′ := K

closed subgroups / closed intermediate o ﬁelds K/F of G

K′ := Gal (K/K ) o K K F 0 0 ⊇ 0 ⊇ Proof. Check well deﬁned. By Proposition 5.8, prime maps anything to a closed object. Therefore codomains are legitimate. Thus by the deﬁnition of “closed” we get that priming induces a bijection on closed objects. Hence theorem.

6 Technical Results

Aim: Relate [H2 : H1] to [H1′ : H2′ ] and [K2 : K1] to [K1′ : K2′ ]. Setup: Fix, for this section, K/F - a ﬁeld extension with G := Gal (K/F ). Correspondence

/ H Hi Hi′ := Ki

subgroups / intermediate o ﬁelds K/F of G o Ki′ := Gal (K/Ki) Ki

Theorem 6.1. Let K K be intermediate ﬁelds with n := [K : K ] < , 1 ⊆ 2 2 1 ∞ then [K′ : K′ ] [K : K ]. 1 2 ≤ 2 1 Proof. We reduce to the case of a simple extension by: Claim: It suﬃce to prove the theorem when K2 = K1(α). Proof: Pick α K K . We argue by induction on n. We are assuming ∈ 2 − 1 [K′ : K (α)′] [K (α) : K ]. Note: K ( K (α) K and K′ K (α)′ 1 1 ≤ 1 1 1 1 ⊆ 2 1 ⊇ 1 ⊇ K2′ . [K1′ : K2′ ] = [K1′ : K1(α)′] [K1(α)′ : K2′ ] [K2 : K1]. ≤ ≤ (induction) ≤

[K1(α) : K1] [K2 : K1(α)] Back to proof of theorem when K = K (α) = K [X]/ p(X) . [K : K ] = 2 1 1 h i 2 1 n = deg. of min. ploy. p(X) K [X] of α over K . Let σ K′ = ∈ 1 1 ∈ 1 Gal(K/K1). It ﬁxes K1 so must send α to a root of p(X) in K be Lemma 4.4. ∴ There are n possibilities for σ(α). The theorem will be proved once ≤ 18 we show:

Lemma: Let σ, τ K1′ , then σK2′ = τK2′ = σ(α)= τ(α). ∈ 6 ⇒ 1 Proof: (of contrapositive). Suppose σ(α) = τ(α). Then τ − σ K1′ := 1 1 ∈ Gal(K/K1) sends α to τ − σ(α)= τ − τ(α)= α so fixes α and K1 ∴ it fixed 1 K2 = K1(α) ∴ τ − σ Gal(K/K2) = K2′ = σK2′ = τK2′ . This proves lemma. Theorem follows.∈ ⇒ Theorem 6.2. Let H H G. Suppose n = [H : H ] < . Then 1 ≤ 2 ≤ 2 1 ∞ [H′ : H′ ] [H : H ]. 1 2 ≤ 2 1 Proof. Uses: Lemma: Let σ, τ G be such that σH = τH . The for any α H′ we have ∈ 1 1 ∈ 1 σ(α)= τ(α). H1 Proof: Let ρ H be such that σ = τρ. Then σ(α)= τρ(α)= τα since α H′ = K . ∈ 1 ∈ 1 Remark: As a result, given any left coset C of H in G and α H′ we can 1 1 define (unambiguously) C(α) := σ(α) where σ is any element of∈C. Back to the proof of theorem, by contradiction. Let the distinct left cosets of H1 in H2 be C1 = H1,C2,C3,...,Cn. Suppose α1,...,αn+1 H1′ are linearly independent over H2′ . Consider n (n + 1) matrix over K∈: ×

C1(α1) C1(α2) ··· C1(αn+1)

 C (α )  A := 2 1 .. .  . .   .   .     C (α ) C (α )   n 1 ··· n n+1   n 1  There exists non-zero solutions x K + to Ax = 0. Pick a non-zero solu- ∈ n+1 tion x with the most number of 0 entries. We seek to construct x′ K 0 ∈ − with Ax′ = 0 and having more 0 entries. By scaling x and permuting α1,...,αn if necessary, we can assume x1 = 1. Also C1(αi) = H1(αi) = αi. So the ﬁrst row of A is (α1,...,αn+1) which are linearly independent over H2′ ∴ not all xi’s are in H2′ so permuting αi’s if necessary, we can as- H2 sume x H′ = K ∴ can pick σ H such that σ(x ) = x . Let 2 6∈ 2 ∈ 2 2 6 2 σ(x1) . σx =  .  which is a solution to σ(xn+1)     σ (C (α )) σ (C (α )) σ(x ) 1 1 ··· 1 n+1 1 . .. . .  . . .   .  = 0 σ (Cn (α1)) σ (Cn (αn+1)) σ(xn+1)  ···     σA   

| {z 19 } σA is A with rows permuted since σC1,...,σCn is just the left cosets permuted. Therefore Aσx = 0 = A (x σx) = 0. Now x =0 iﬀ σ(x ) = 0, ⇒ − i i x′ so xi =0 = xi′ = 0. Since x1 = 1, x1′ = x1 σ(x1) = 1 1 = 0. Also ⇒ | {z } − − x′ = x σ(x ) = 0 by choice of σ. Therefore x′ is a non-zero solution to 2 2 − 2 6 Ax′ = 0 with more 0 entries than x. This is a contradiction, and thus the theorem is proved.

7 Galois Correspondence

Same setup as in Section 6. Recall the bijection between closed objects (Theorem 5.9). Estimates from Section 6 gives: Corollary 7.1. (a) Let H H G. If H is closed and [H : H ] < 1 ≤ 2 ≤ 1 2 1 ∞ then H2 is closed and [H2 : H1] = [H1′ : H2′ ]. (b) Given intermediate ﬁelds F K K K, suppose K is closed ⊆ 1 ⊆ 2 ⊆ 1 and [K : K ] < . Then K is closed and [K : K ] = [K′ : K′ ]. 2 1 ∞ 2 2 1 1 2 Proof. (a) Theorem 6.1 and 6.2 give [H : H ] [H′ : H′ ] [H′′ : H′′]= 2 1 ≥ 1 2 ≥ 2 1 [H′′ : H ] (since H is closed). Proposition 5.6 shows H′′ H and so 2 1 1 2 ⊇ 2 equality must hold above and so H2′′ = H2. (b) Similar. Swap H’s with K’s.

7.1 Galois Extension Definition 7.2. An algebraic field extension K/F is Galois if F is closed. Gal(K/F ) i.e. F = F ′′ = K Example 7.3. We saw before that Q is not closed in K/Q = Q(√3 2)/Q therefore Q(√3 2)/Q is not Galois. Proposition 7.4. Let K/F be a field extension. 1. G := Gal (K/F ) is finite. 2. K/F Galois = [K : F ]= G . ⇒ | | 3. If G [K : F ] then K/F is Galois. | | ≥ Proof. 1. Write K = F (α1,...,αn). Each αi is algebraic over F so given σ G Lemma 4.4 implies there are only finitely many possibilities for σ(∈α ),...,σ(α ) ∴ G < . 1 n | | ∞ 20 2. K/F Galois implies F is closed. Corollary 7.1 (b) implies [K : F ] = [F ′ : K′] = [G : 1] (as F ′ = Gal(K/F ) and K′ = Gal(K/K)) = G | | 3. [K : F ] G = [G : 1] because 1 G is closed ≤ | | ≤ = [1′ : G′] by part 1 and Corrolary 7.1 (a) G = [K : F ′′] because G′ = K = F

Again F ′′ F so dimension considerations imply F = F ′′ so F is closed K/F is Galois.⊇

3 Example 7.5. We saw K/Q = Q(√2,ω)/Q has G := Gal (K/Q)= S3.

K = Q1 Q√3 2 Q√3 4 Qω Q√3 2ω Q√3 4ω. ⊕ ⊕ ⊕ ⊕ ⊕ So [K : Q]=6= S . Thus by Proposition 7.4 (3), K/Q is Galois. | 3| Theorem 7.6 (Fundamental Theorem of Galois Theory (Galois Cor- respondence)). Let K/F be a ﬁnite Galois extension and G := Gal (K/F ).

(a) There are inverse bijections

H / KH

subgroups / intermediate o ﬁelds of K/F of G o K0′ := Gal (K/K0) K0

(b) For intermediate field K0, K/K0 is Galois with Galois group Gal(K/K0)= K0′ . Proof. (a) By weak Galois correspondence suffice to show all objects are closed. Proposition 7.4 (1) implies G < so since 1 G is closed, Corollary 7.1 (a) implies any H G|is| also∞ closed. K/F Galois≤ = F ≤ ⇒ is closed. Therefore any intermediate field K0 is also closed by Corollary 7.1 (b).

Gal(K/K0) (b) We need to show K0 = K = K0′′ (w.r.t. K/F ). This holds because K0 is closed in K/F . Therefore K/K0 is Galois too.

21 Example 7.5 (again). K/Q = Q(√3 2, √3 2ω, √3 2ω2)/Q,G = Gal(K/Q)=

S3.

S Q ⊃ 3 ⊃ <

< < <

⊃ 3 ⊃ 3 3

2 ⊂

(1 2 3) (1 2) (2 3) (1 3) Q(ω) Q(√2ω ) Q⊂ (√2) Q(√2ω) > h i h i h i> h i > > ⊂ 1 ⊂ K = Q(√3 2,ω)

(1 2) 3 2 Let’s check Kh i = Q √2ω . We know ( ). Also ⊇

(1 2) (1 2) Kh i : Q = Q′ : Kh i ′ = [S : (1 2) ] = 3. 3 h i h i 3 2 (1 2) 3 2 Since Q √2ω : Q = 3 too, we must have Kh i = Q √2ω .

Theorem 7.7. LetK be a field and G be a finite group of field automorphisms of K. If K = F G, then K/F is Galois and its Galois group is G.

Proof. Let G = Gal(K/F ) and note G G. Now ≤ G Ge = [G : 1] e | | ≥ | | = [1′ : G′] by Corollary 7.1 (a), since 1 G is closed e ≤ = K : KG = [K : F ] Therefore [K : F ] < and so Proposition 7.4 (3) implies K/F is Galois. ∞ Also part 2 of the proposition, implies G = G = Gal(K/F ).

e 8 Normality

Aim: Determine when intermediate field L of finite Galois extension K/F is such that L/F is Galois. Galois = Splitting Field ⇒ Definition 8.1. K is a splitting field over F if it is the splitting field of some family of polynomials over F . We also say in this case that K/F is normal.

Proposition 8.2. Let K/F be a Galois extension (not necessarily ﬁnite) with Galois group G.

22 1. Let α K and p(X) F [X] its minimal polynomial. Then p(X) factors into∈ linear factors∈ over K.

2. K is the splitting field of f over F where f ranges over the minimal { i} i polynomials of all α K. ∈ Proof. (1) = (2) is obvious. Proof of (1): Let α K and p(X) its min. ⇒ ∈ poly. Let H := σ G σ(α)= α Stab α G. Consider { ∈ | } ≤ G ≤ q(X) := (X (σH) α) , − σH left cosetY which is well defined by the Remark from the proof of Theorem 6.2 and the product is finite because G-orbit of α is finite. But for any τ G, τq(X)= q(X) because it just permutes factors (X (σH)(α)) since q(X∈) KG[X]= F [X] (since K/F is Galois). ∴ p(X) q(X−) must factorise into linear∈ factors over K because q(X) does. |

8.1 Stability of Splitting Fields Lemma 8.3. Let K/F be an algebraic extension and σ : K K be a field homomorphism over F . Then σ is an isomorphism. → Proof. Suffice to show σ is surjective. Pick α K, let p(X) F [X] ∈ ∈ be its min. poly. and α = α1,...,αr the roots of p(X) in K. Let L = F (α ,...,α ). This is finite over F . But σ permutes α ’s = σ (L) = L. 1 r i ⇒ Dim. considerations = σ L : L ֒ L is surjective. ∴ im σ L α. ∴ σ is surjective. ⇒ | → ⊇ ∋

Proposition 8.4. Let K be the splitting field of fi over F . Given field extension L of K and field homomorphism σ : K{ } L over F , we have σ(K)= K. → Proof. By the previous lemma, it suffices to show σ (K) K. Let α ⊆ { j} be the roots of all the fi so K is generated over F by the αj’s. σ permutes these roots. So σ (K) K. ⊆ 8.2 Galois Action on Galois Correspondence Let K/F be a Galois extension, with G = Gal(K/F ). G acts on: 1. intermediate fields of K/F by K σ (K ) (where σ G) { } 1 7−→ 1 ∈ 1 2. subgroups of G by conjugation i.e. for H G, H σHσ− . { } ≤ 7−→ 23 Galois correspondence is compatible with G-action as follows:

Proposition 8.5. Notation as above.

1 1. If K1 is an intermediate ﬁeld σ (K1)′ = σK1′ σ− i.e. Gal(K/σ (K1)) = 1 σGal(K/K1) σ− .

1 2. For H G, σ (H′)=(σHσ− )′ ≤

Proof. 1. For τ G, τ Gal(K/σ (K1)) iﬀ for any α K1, τσ (α) = σ (α) ∈ ∈ ∈ 1 for any α K1, σ− τσ (α)= α ⇐⇒ 1 ∈ σ− τσ Gal(K/K1) ⇐⇒ ∈ 1 τ σGal(K/K ) σ− . ⇐⇒ ∈ 1 2. Similar to above and is left as an exercise.

8.3 Normal Subgroups in Galois Correspondence Theorem 8.6. Let K/F be a finite Galois extension. Let G = Gal(K/F ). If L is an intermediate field then L/F is Galois iff L′ = Gal(K/L) is a normal subgroup. In this case Gal(L/F )= G/L′.

Proof. Suppose L/F is Galois so normal by Proposition 8.2. Given any σ G, Proposition 8.4 implies σ (L) = L. But then Proposition 8.5 = ∈ ⇒ L′ E G. Also we have a group homomorphism

π : Gal(K/F ) Gal(L/F ) −→ σ σ 7−→ |L

Note ker π = Gal(K/L) = L′. Letσ ¯ Gal(L/F ). But K/L is a splitting field so uniqueness of splitting fields (Theorem∈ 3.8) = can extendσ ¯ to Gal(K/F ) ∴ π is surjective. 1st isomorphism theorem⇒ = Gal(L/F ) ⇒ ≃ Gal(K/F ) / ker π = G/L′. Conversely, suppose L′ = Gal(K/L) E G. Then Proposition 8.5 = for any σ G, σ (L) = L. To show L/F is Galois, it suffices to show⇒ F LGal(L/F∈ ). We know F KGal(K/F ) since K/F is Galois. Let α L F . ⊇ ⊇ ∈ − There exists σ Gal(K/F ) such that σ (α) = α. But σ L Gal(L/F ) since ∈ Gal(L/F6 ) | Gal(∈ L/F ) σ (L) = L. Also, σ L (α) = α so α L ∴ F L and L/F is Galois too. | 6 ∈ ⊇

24 3 Example 8.7. K/Q = Q √2,ω , Gal(K/Q)= S3.

S S Q ⊃

3 3 ⊃ < < h(1 2 3)i E <

⊃ 3 ⊃ 3 3

2 ⊂

(1 2 3) (1 2) (2 3) (1 3) Q(ω) Q(√2ω ) Q⊂ (√2) Q(√2ω) > h i h i h i> h i > > ⊂ 1 ⊂ K = Q(√3 2,ω)

9 Separability

Aim: Look at perversity in positive characteristic.

9.1 Inseparability Let F be a field of characteristic p = 0. 6 Proposition/Definition 9.1. Let K/F be a field extension. We say α K is purely inseparable over F if there is some n 1 with αpn F . In∈ this case, Gal(F (α) /F ) = 1. ≥ ∈

Proof. α satisﬁes Xpn αpn =(X α)n F [X]. Any σ Gal(F (α) /F ) sends α to α. Therefore,− Gal (F (α) /F− ) = 1.∈ ∈

p Example 9.2. Fp = Z/pZ, F = Fp(t), α = √t F , is purely inseparable over F and F (α) is a splitting ﬁeld for Xp t over F6∈with Gal (F (α) /F ) = 1, so is Galois. −

9.2 Separable Polynomials, Elements and Extensions

n j Let F be a ﬁeld, f(X) j=0 fjX F [X]. We can deﬁne its derivative df j ∈1 ∈ f ′(X)= = jf X − F [X]. dX j ∈P d pn Example 9.3.P If F has char. p = 0 then for a F dx X a = n pn 1 6 ∈ − p X − = 0.

Proposition 9.4. (a) (f +g)′ = f ′ +g′, (fg)′ = f ′g +fg′ for f,g F [X]. ∈

(b) α is a multiple root of f(X) F [X] iﬀ 0= f(α)= f ′(α) ∈ The proof is left as an exercise.

25 Definition 9.5. Let K/F be a finite field extension. Its separable degree ¯ is [K : F ]S := number of field homomorphisms σ : K K over F . (Note K¯ = F¯). → Proposition/Definition 9.6. Let f(X) F [X] be irreducible. The following are equivalent: ∈ (a) In the splitting field L of f(X) over F , f(X) factors into distinct linear factors;

(b) f ′(X) = 0; 6

(c) [F (α) : F ]S = [F (α) : F ] where α is a root of f(X). If these equivalent conditions hold, we say f(X) and α are separable over F . ¯ Proof. (a c) [F (α) : F ]S = is the number of roots of f(X) in F . This is deg f(X)⇐⇒ = [F (α) : F ] iﬀ there is no multiple roots. Prop 9.4 (b) (a b) f(X) has no multiple roots = f ′(X) = 0. (b ⇒ a) If (a) fails then there is a multiple⇒ root α of6 f(X). Proposition 9.4 ⇒ (b) implies f (α)= f ′ (α)=0. But f is the min. poly. of α and deg f ′

9.3 Multiplicity of [K : F ]S Theorem 9.8. Let K L F be a tower of finite field extensions. Then: ⊇ ⊇ (a) [K : F ]S = [K : L]S [L : F ]S (b) [K : F ] [K : F ] and equality occurs iff K/F is separable. S ≤ [K:L] S ¯ Proof. (a) Let S = σi i=1 be distinct field homomorphisms σ : K F over L. { } → [L:F ] S ¯ T = τj j=1 be distinct field homomorphisms τ : L F over F . Uniqueness{ } of algebraic closure (Theorem 3.8) = →we can extend τ : L F¯ to field hom.τ ˜ : F¯ F¯. (a) follows from:⇒ j → j → Claim: The distinct field hom. ρ: K F¯ over F are the τ˜jσi . Why? → { } 1 If ρ L = τj, then the τj is the unique element of T such thatτ ˜j− ρ fixes | 1 L. i.e. τ − ρ S. This proves claim. j ∈ 26 (b) By induction on [K : F ]. Pick any α K F . ∈ −

[K : F ]S = [K : F (α)]S [F (α) : F ]S ≤ ( ) ≤ ( ) † ∗ [K : F ] = [K : F (α)] [F (α) : F ] ( ) holds by induction. (†) holds because ∗ [F (α) : F ] = # of roots α in F¯ deg. of mi. poly. of α over F ≤ = [F (α) : F ]

∴ [K : F ]S [K : F ]. If K/F is separable, then by deﬁnition equality holds in ( )≤ and equality holds in ( ) by induction. ∗ † Conversely, if [K : F ]S = [K : F ] then equality holds in ( ) = α is separable over F . ∴ K/F is separable too. ∗ ⇒

Corollary 9.9. (a) Consider ﬁeld extensions K/L, L/F . The K/F is separable iﬀ K/L and L/F are separable.

(b) If ﬁeld K is generated over F by separable elements αi then K/F is separable. { }

Proof. (a)

[K : F ]S = [K : L]S [L : F ]S ≤ ≤ ≤ [K : F ] = [K : L] [L : F ] equality holds on LHS iﬀ equality holds on RHS.

(b) Any α K lies in some F (α ,...,α ). Now just apply (a) to F ∈ i1 in ⊆ F (α ) F (α , α ) . . . i1 ⊆ i1 i2 ⊆

10 Criterion for Galois

Aim: Show Galois extensions are separable splitting ﬁelds so separable extensions embed in Galois extensions.

27 10.1 Galois Separable Splitting field ⇐⇒ Lemma 10.1. Let K/F be a Galois extension. Let L F be a splitting field of some f(X) F [X] over F . The L/F is Galois too.⊆ ∈ Proof. Let α L F . Since K/F is Galois, F = KGal(K/F ) and can find σ Gal(K/F )∈ with− σ (α) = α. Proposition 8.4 on stability of splitting ∈ 6 Gal(L/F ) fields = σ L Gal(L/F ). σ L (α) = α ∴ L F L L so F LGal(⇒L/F ) |and∈L/F is Galois too.| 6 − ⊆ − ⊇ Theorem 10.2. K/F is a Galois extension iff K is a separable splitting field over F .

Proof. ( ) Proposition 8.2 = K is a splitting field over F and for any α K with⇒ min. ploy. f(X) ⇒F [X] over F , we know f(X) factors into linears∈ over K. Let L K be the∈ splitting field of f(X) over F . The above ⊆ lemma = L/F is Galois. So [L : F ] = Gal(L/F ) [L : F ]S. Theorem 9.8 = L/F⇒ is separable = α is separable| over F .| Since ≤ α was arbitrary K/F ⇒is separable. ⇒ ( ) Need to know F KGal(K/F ). Let α K F and f(X) F [X] its min.⇐ poly. over F . deg⊇ f(X) > 1 (else α∈ F−) and since α is∈ separable ∈ there is a root α′ of f(X) with α = α′. Proposition 2.4 = there is a field 6 ⇒ isomorphismσ ˜ : F (α) ∼ F (α′) over F such thatσ ˜ (α) = α′. Uniqueness −→ of splitting field K/F = we can extendσ ˜ to σ Gal(K/F ) ∴ σ (α) = Gal(⇒K/F ) ∈ Gal(K/F ) α′ = α and so α K . This shows K F K K so F 6 KGal(K/F ) which6∈ shows K/F is Galois too. This− proves⊆ theorem.− ⊇ 3 i2π Example 10.3. Q √2,ω = e 3 /Q is Galois and is separable (since char = 0) and is splitting field of X3 2 over Q. − 10.2 Splitting Fields Proposition 10.4. The following are equivalent:

(a) K/F is a normal extension (i.e. K is a splitting ﬁeld over F );

(b) for any α K, if fα(X) is its min. poly. over F , then fα(X) factors into linear∈ factors over K.

Proof. (b a) K is the splitting field of fα α K over F so K/F is normal. ⇒ { } ∈ (a b) Let α′ K¯ be any root of f (X). It suffice to show that α′ K. ⇒ ∈ α ∈ Proposition 2.4 = there is a field isomorphismσ ˜ : F (α) ∼ F (α′) such ⇒ −→ thatσ ˜ (α)= α′. Uniqueness of splitting fields Theorem 3.8 = can extend ⇒

28 σ˜ to a ﬁeld hom. σ : K K¯ . Stability of splitting ﬁelds (Proposition 8.4) → = σ (K)= K. ∴ α′ σ (K)= K. Therefore, all roots of fα (X) are in K and⇒ (b) holds. ∈

10.3 Normal Closure Theorem 10.5. Let K/F be an algebraic extension. For α K, let f (X) ∈ α ∈ F [X] be its minimal polynomial over F . Let L K¯ be the splitting ﬁeld of ⊆ fα (X) α K over F , so in particular L K. { } ∈ ⊇

(a) If L1 is a subﬁeld of K¯ such that L1/F is normal and L1 K then L L. ⊇ 1 ⊇ (b) If K/F is separable, so is L/F (so L/F is Galois).

(c) If K/F is ﬁnite, so is L/F and L is splitting ﬁeld of single polynomial f(X) F [X] over F . ∈ Proof. (a) Follows from previous proposition.

(b) α K sep. over F = f (X) sep. poly. = L is generated by ∈ ⇒ α ⇒ separable elements (the roots of fα (X)) over F . Corollary 9.9 = L/F is separable. ⇒

(c) Can write K = F (α1,...,αn). Let f(X)= fα1 (X)fα2 (X) ...fαn (X). Let F˜ be splitting field of f(X) over F . Note L/F˜ is a finite normal extension. Also, it contains K since it contains F, α1,...,αn. (a)= L˜ L. But definition shows L˜ L. ∴ L/F is finite too. ⇒ . ⊇ ⊆

Deﬁnition 10.6. The ﬁeld L from Theorem 10.5 is called the normal closure of K (over F ). If K/F is separable, we also call L the Galois closure of K (over F ).

3 3 i2π Example 10.7. The Galois closure of Q √2 /Q is Q √2,ω = e 3 /Q.

Remark 10.8. In characteristic zero, where separability is guaranteed, can always “embed” ﬁnite extensions in ﬁnite Galois extensions.

11 Radical Extension

Aim: See what Galois correspondence reveals about radical extensions.

29 11.1 Radical Extensions Recall K/F is a radical extension if there is tower of ﬁeld extensions F = p F0 F1 Fn = K, with Fi+1 = Fi √i αi for some pi 1, αi Fi. ⊆N.B.⊆···⊆ If such a tower exists, can replace it with tower where≥ all∈p are i prime.

Example 11.1. Q 2+ √5 /Q is radical. p 11.2 Galois Group of Simple Radical Extension Proposition 11.2. Let F be a ﬁeld of characteristic p and n 2 be such ≥ that p ∤ n. Then the group of nth roots of unity in F¯∗ is µ Z/nZ. n ≃ d n n 1 n Proof. First note that (X 1) = nX − so X 1 has no repeated dx − − roots. So the set of roots µn has n distinct elements. Use structure theorem of ﬁnitely generated abelian groups to see

µ Z/h Z Z/h Z Z/h Z where h h h and h h ...h = n. n ≃ 1 × 2 ×· · ·× r 1 | 2 | · · · | r 1 2 r

hr hr Now for any z µn we have z = 1. But no. of solns. to z = 1 is h . ∴ n = h .∈ Also must have then r = 1 so µ Z/nZ. ≤ r r n ≃ Lemma 11.3. Let p be a prime and F be a ﬁeld with all the pth roots of unity. For α F , let K = F (√p α). Then: ∈ (i) K/F is a splitting ﬁeld for Xp α over F ; − (ii) If char F = p and √p α F , then K/F is Galois with Galois group 6 11.2 6∈ Gal(K/F ) µ Z/pZ; ≃ p ≃ (iii) If char F = p or √p α F then Gal(K/F ) = 1. ∈

Proof. (i) Proposition 11.2 = µp is a cyclic group, say generated by ω. ⇒ p N.B. µp = 1 if char F = p (derivative is zero) ∴ the roots of X α are p p p 2 p n 1 p − √α, √αω, √αω ,..., √αω − F (√α) = K. ∴ K is splitting ﬁeld for Xp α over F . ∈ − p p n 1 (ii) Proposition 11.2 says if char F = p then the roots √α,..., √αω − 6 are distinct. ∴ K = F (√p α) /F is separable. Thus it is a separable splitting ﬁeld and hence is Galois. Let σ Gal(K/F ). Then σ (√p α) = √p αωi(σ) for some i (σ) Z. ∈ ∈ ∴ σ (√p αωj)= √p αωi(σ)ωj and so σ just multiplies roots by ωi(σ). Thus

30 we get an injective 1 group hom. ϕ: Gal(K/F ) / µ Z/pZ p ≃ σ ωi(σ) 7−→ Since √p α F, K = F and Gal(K/F ) = 1. Lagrange = ϕ is an isomorphism.6∈ 6 6 ⇒ (iii) Clear.

Lemma 11.4. Let p be a prime and F be a field. Let K be a splitting field of Xp 1 over F . Then Gal(K/F ) is abelian. − Proof. If char F = p then Xp 1=(X 1)p so K = F and Gal(K/F ) = 1. − − So assume char F = p and suppose ω µp generates µp. Note K = F (ω). Let σ Gal(K/F6 ) so σ (ω) = ωi(σ) ∈for some i (σ) Z. Thus σ (ωj) = σ (ω)j =∈ ωji(σ), i.e. σ raises root ωj to the power of ∈i (σ). It is left as an exercise to show that action of any σ, τ Gal(K/F ) commutes. ∈ 11.3 Solvability Galois correspondence and Lemma 11.4 suggests: Definition 11.5. Let G be a group. A normal chain of subgroups is a sequence of the form

1= G E G E G E E G = G. 0 1 2 ··· n We say G is solvable if there exists such a normal chain of subgroups with Gi+1/Gi cyclic of prime order (or trivial). In Section 13 we will show: Theorem 11.6. Let K/F be a field extension such that there exists a field extension L/F with L/F separable and radical. The Gal(K/F ) is solvable. The main key to the proof is: Theorem 11.7. Consider tower of field extensions

F = F F F F F = K, 0 ⊆ 1 ⊆···⊆ i ⊆ i+1 ⊆···⊆ n where F = F ( p√i α ) for some p and α F . Suppose K/F is Galois and i+1 i i i i ∈ i F contains all pith roots of unity. Then G := Gal (K/F ) is solvable. 1 Proving injectivity is left as an exercise

31 Proof. Galois correspondence gives:

G = F ′ = F ′ F ′ F ′ F ′ F ′ =1 ( ) 0 ≥ 1 ≥···≥ i ≥ i+1 ≥···≥ n ∗

Note K/Fi is Galois. Also Fi+1/Fi is a normal extension by Lemma 11.3 and separable because K/Fi is Galois. Thus Fi+1/Fi is Galois. There- fore Theorem 8.6 = Gal(K/Fi+1) = Fi′+1 E Fi′. And also Fi′/Fi′+1 11.3 ⇒ ≃ Gal(F /F ) Z/p Z. Therefore ( ) is a normal chain of subgroups show- i+1 i ≃ i ∗ ing G = F ′ is solvable.

12 Solvable Groups

Aim: Find ways to verify (not) solvable.

12.1 Basic Fact Proposition 12.1. Let G be a ﬁnite group and N E G, then G is solvable iﬀ N and G/N is solvable.

Proof. ( ) Suppose we have normal chain of subgroups ⇐ 1= N E N E N E E N = N 0 1 1 ··· m 1 = G /N E G /N E E G /N = G/N G/N 0 1 ··· n Gi+1/N Gi+1 and Ni+1/Ni and are cyclic of prime order. Then we get a Gi/N Gi normal chain of subgroups≃

1= N E N E E N E G E G E E G = G. 0 1 ··· 1 2 ··· n ( ) Suppose we have normal chain of subgroups ⇒ 1= G E G E G E G E E G = G 0 1 1 2 ··· r with Gi+1/Gi cyclic of prime order. If π : G G/N is the quotient map g gN, then consider normal chain of subgroups:→ 1)7→ 1= G N E G N E G N E E G N E N 0 ∩ 1 ∩ 2 ∩ ··· r ∩ 2) 1G/N = π (G0) E π (G1) E E π (Gr)= G/N. E ··· H HK Recall that H < Gi+1, K Gi+1 = H K K . Therefore ⇒ ∩ ≃ G N (G N) G G i+1 ∩ i+1 ∩ i i1 (G N) G ≃ G ≤ G i+1 ∩ ∩ i i 1

32 Gi+1 N Note that Gi+1/Gi is cyclic of prime order. Therefore Lagrange = ∩ ⇒ Gi ∩ N is cyclic prime order or trivial. Thus N is solvable. 2 Also

Gi+1 π (Gi+1)

Gi −→ π (Gi) gG π (g) π (G ) i 7−→ i is a surjective group hom. π (Gi+1) /π (Gi) is cyclic of prime order or trivial. Thus G/N is solvable too. This proves proposition. Proposition 12.2. A finite group G is solvable iff there is a normal chain of subgroups 1= G E G E G E E G = G with all G /G abelian. 0 1 2 ··· r i+1 i Proof. ( ) By definition. ( ) By induction⇒ and Proposition 12.1 suffice to show G is abelian then it is ⇐ solvable. Assume G is abelian and pick g G 1. Let p be a prime dividing the order n of g. Then gn/p is cyclic of prime∈ − order so solvable and G/ gn/p is solvable by induction,h so iG is solvable by Proposition 12.1. h i

12.2 Derived Series Deﬁnition 12.3. Let G be a group and g,h G. The commutator of g 1 1 ∈ and h is [g,h]= g− h− gh. The commutator subgroup of G is the group generated by all [g,h] as g,h range over G. It is denoted by [G, G]. Proposition 12.4. Let G be a group. (a) [G, G] E G.

(b) G/ [G, G] is abelian.

1 1 1 Proof. (a) Let g,h,k G. k [g,h] k− = [kgk− ,khk− ] [G, G]. Thus [G, G] E G. ∈ ∈

1 1 (b) & (c) Let g,h G. [g,h] N g− h− ghN = N ghN = hgN (gN)(hN∈)=(hN)(gN∈). [G,⇔ G] contains all [g,h] so⇔G/ [G, G] is abelian⇔ i.e. (b) holds. Also, if gNhN = hNgN g,h then N [G, G] giving (c). ∀ ⊇

2 Note that we never used the normality of N for this. This fact will be important later.

33 Deﬁnition 12.5. Let G be a ﬁnite group. Its derived series is the normal chain of subgroups

G = G(0) D G(1) D G(2) D D G(r) D . . . ( ) ··· ∗ where G(r+1) = G(r),G(r) .

Corollary 12.6. A ﬁnite group G is solvable iﬀ in the derived series ( ), G(r) = 1 for some r< . ∗ ∞ Proof. ( ) Clear by Proposition 12.4 (b) and Proposition 12.2. ( ) Consider⇐ normal chain of subgroups G = G0 D G1 D D Gr = 1 with⇒ Gi/Gi+1 abelian. Proof by induction on r, that G(r) ···Gr, G(i+1) = G(i),G(i) [Gi,Gi] (by induction) Gi+1 (by Proposition⊆ 12.4 (c)). ⊆ ⊆ 12.3 An Insolvable Group

Let A(n) be a subgroup of Sn generated by 3-cycles. Note A(n) E Sn, A(n) A (in fact A(n)= A ). ⊆ n n Theorem 12.7. Let n 5. ≥ (a) [A(n),A(n)] = A(n).

(b) A(n),An and Sn are not solvable. Proof. (a) (b) by 12.6. (a) Consider⇒e,f,g,h,k 1,...,n distinct.3 ∈ { } 1 1 (efg)− (ehk)− (efg)(ehk) =(egk). ∈ [A(n),A(n)] | {z } Therefore [A(n),A(n)] = A(n).

13 Solvability by Radicals

Aim: Determine solvability by radicals.

Lemma 13.1. Let L be a Galois closure of separable radical extension. Then L is radical. 3 Note that we need n 5 for this ≥

34 Proof. Consider tower of field extensions F = F0 F1 Fr = K pi ⊆ ⊆···⊆ where Fi = Fi 1(αi) and α Fi 1 with pi prime. Let L0 be subfield − − of L generated over F by σ (α∈ ) σ Gal(L/F ) . Note L K. Now { i | ∈ } 0 ⊇ L0/F is a radical extension so it suffices to show that L0 = L. For any τ Gal(L/F ), τ permutes generators of L /F so τ (L ) = L. Proposition ∈ 0 0 8.5 implies L0′ = Gal(L/L0) E Gal(L/F ). Theorem 8.6 implies L0/F is Galois. Minimality of Galois closure L L = L . ⇒ 0 Theorem 13.2. Let K/F be a field extension which “embeds” in a separable radical extension L/F in the sense that we have a tower of field extensions F K L. Then Gal(K/F ) is solvable. ⊆ ⊆ Proof. We can replace F with KGal(K/F ) since this leaves Galois group unchanged as well as the hypothesis. Thus we can assume K/F is Galois. Lets fix an algebraic closure L¯ of L in which to work. Lemma 13.1 implies we may replace L with Galois closure in L¯. Therefore can also assume L/F is Galois. Theorem 8.6 gives Gal (K/F ) Gal(L/F ) . So suffices to show ≃ Gal(L/K) Gal(L/F ) is solvable. Therefore can now also have L = K. Consider the tower of field extension F = F0 F1 Fr = L, pi ⊆ ⊆ ··· ⊆ where Fi = Fi 1 (αi) with αi Fi 1 for some pi prime. Note L/F separable − − ∈ pi r pi = char F . Let F1 be splitting field of X 1 i=1 over F . Let L1 ⇒ 6 { − }pi be subfield of L¯ generated over L by roots of these X 1. Note F1/F is a separable normal extension so is Galois. Proposition− 11.2 (or rather its proof) shows Gal(F1/F ) is abelian and hence solvable. Claim: L1/F is separable normal therefore Galois. Why? L1 is generated over F by the separable generators α ,...,α of L/F and roots of Xpi 1. Also if L and 1 r − F1 are splitting fields of f(X),g(X) F [X] over F then L1 is splitting field of f(X)g(X) over F , proves claim. ∈ F /F Galois F ′ = Gal(L /F ) E Gal(L /F ) = G. L /F Galois 1 ⇒ 1 1 1 1 1 L /F Galois. Also Theorem 11.7 and fact L = F (α ,...,α ) F ′ = ⇒ 1 1 1 1 1 r ⇒ 1 Gal(L1/F1) is solvable, so Proposition 12.1 implies G = Gal(L1/F ) is solvable. But Gal (L/F ) is a quotient of Gal (L1/F ) so is solvable too. Theorem is proved.

13.1 Solvability of polynomials Definition 13.3. Let F be a field and f(X) F [X]. The Galois group of f(X) is Gal(K/F ) where K is the splitting∈ field of f(X) over K. We say f(X) is solvable by radicals if K/F embeds in a separable radical extension L/F as in theorem above.

35 13.2 General degree n monic polynomial

Let α1,...,αn be indeterminants representing roots. Let K = Q (α1,...,αn). Let

a = (α + + α ) , a = α α , ..., a =( 1)n α α ...α . 1 − 1 ··· n 2 i j n − 1 2 n i

Aim: Give example of polynomial not solvable by radicals. Show ubiquity of simple extension.

14.1 A polynomials not solvable by radicals Fix p a prime.

Lemma 14.1. Let σ, τ S be a p-cycle and a 2-cycle respectively. Then ∈ p subgroup generated by σ, τ is Sp. Proof. Relabelling if need be, can assume τ = (1 2). Taking a power of σ if i i necessary and relabelling further, can assume σ =(12 ... p). σ (1 2)σ− = i i (σ (1) σ (2)) = (i + 1 i + 2) (mod p). But Sp is generated by such transpositions since any permutation of a row of elements can be gotten by switching neighbouring elements.

Proposition 14.2. Let f(X) Q[X] be irreducible of degree p. Suppose f ∈ has exactly 2 non-real roots. Then G = Galois group of f(X) is Sp. Thus f(X) is not solvable by radicals if p = 5. 6 Proof. Let K be splitting ﬁeld of f(X) over Q. Labelling roots as usual gives monomorphism φ : G ֒ S . Conjugation c is an element of Galois → p group and φ(c) S is a 2-cycle. Let α be a root of f(X). Then ∈ p [Q (α) : Q] [K : Q]= G | |

36 also p = deg f(X) = [Q (α) : Q]. Thus, Sylow’s theorem implies there is a Sylow p-subgroup of G which must have order p so is say generated by α G. Therefore φ (α) is a p-cycle. Thus φ(G)= S by Lemma 14.1. ∈ p Example 14.3. f(X) = X5 6X + 3 is irreducible over Q by Eisen- 4 − stein criterion. f ′(X) = 5X 6. Therefore 3 real and 2 non-real roots. Proposition 14.2 implies f(X) not− solvable by radicals.

14.2 Primitive element theorem Exercise: Use argument of Proposition 11.2 to show, if F is a ﬁnite ﬁeld, then F ∗ is a cyclic group.

Theorem 14.4. Let K/F be a finite separable extension. Then K = F (α) for some α K. ∈ Proof. If F is finite, so is K, so simply let α be a generator of the cyclic group K∗. Thus K = F (α). Suppose now F is infinite. Pick α K such that [F (α) : F ] is maximal. Suppose F (α) ( K so can pick β K∈ F (α). Note K/F has only finitely many intermediate fields since the∈ same− is of the Galois closure L/F by the Galois correspondence. Pigeon hole principle implies we can find distinct c ,c F with F (α + c β) = F (α + c β). Note F (α + c β) α + c β 1 2 ∈ 1 2 1 ∋ 2 and therefore β and also contains α. Therefore F (α + c1β) ) F (α). This contradicts maximality of [F (α) : F ]. Hence K = F (α).

14.3 Fundamental theorem of algebra Theorem 14.5. C := R √ 1 is algebraically closed. − Proof. We use the following facts about R:

(a) Any odd degree real polynomial has a real root.

(b) Any complex quadratic polynomial has a complex root.

Proof by contradiction. Let K/C be a non-trivial ﬁnite ﬁeld extension. Tak- ing Galois closure if necessary we can assume K/R is Galois with Galois group G P G | | G. Let P be a Sylow 2-subgroup of G so 2 ∤ P . Note K : R = K = | | [G : P ] is odd. Any α KP has [R (α) : R] KP : R so min. poly. of α ∈ | (a) over R is odd = α R so KP = R = KG soG = P . ⇒ ∈

37 Lemma 14.6. Let P be any 2-group. There is a chain of subgroups

P = P0 ⊲ P1 ⊲ P2 ⊲... 1

i with [P : Pi] = 2 . Proof. By induction on P , since P is a p-group (p = 2 here) Z(P ) = 1. It suffices by induction to| first| find P ⊳ P with [P : P ] = 2. If Z(P ) =6 P 1 1 6 we are done by applying inductive hypothesis, to P/Z(P ), to get P1/Z(P ). If Z(P ) = P then P1 is abelian so you can find P , by hand, e.g. using structure theorem for finitely generated abelian groups. This proves the lemma.

Back to proof of theorem. Pick P = P0 > P1 > P2 . . . as in lemma. P2 [P : P1] = 2 K is a degree 2 extension of C. This is impossible by (b). This proves theorem.⇒

15 Trace and Norm

Aim: Introduce two tools to study ﬁeld extension trace and norm.

15.1 Trace and Norm Let K/F be a ﬁnite ﬁeld extension of degree n.

Definition 15.1. Let α K and let multiplication by α be mα : K K an F -linear map. Represent∈ m with n n-matrix over F M . We define→ α × α the trace of α to be trK/F α = tr α = tr Mα, and the norm of α to be NK/F α = N α = det Mα. Note that tr, and N are well dell-define. α . Example 15.2. If α F then Mα = .. . Therefore tr α =nα ∈   α   and N α = αn.   Notation 15.3 (Only for this section). For a finite field extension L/F let G be the set of field homomorphisms L L¯ over F , so G = [L : F ] . L/F → | L/F | S Proposition 15.4. Let K/F be a finite separable extension. Then for α K ∈ tr α = σ (α) and N α = σ (α) α G α G ∈XK/F ∈YK/F

38 Proof. Easy case: Suppose ﬁrst K = F (α). Then (as in proof of Propo- sition 8.2) the min. poly. of α over F is p(X) = σ G (X σ (α)) ∈ K/F − because σ (α) are certainly roots of min. poly. and deg. p(X) = GK/F = separability Q | | [K : F ]S === [F (α) : F ]= deg. of min. poly. Thus p(X) is also the min. poly. of Mα (p (Mα) = 0). But the characteristic poly. of Mα has degree [F (α) : F ] p(X) is char. poly. of Mα. Therefore, tr α = tr Mα = σ (α) and⇒ N α = det M = σ (α). σ GK/F α σ GK/F ∈General case: want to reduce to easy∈ case. Recall from proof of Theorem P Q 9.8 the following fact. Suppose GF (α)/F = τ,...,τ[F (α):F ] and letτ ˜j : K¯ K¯ be arbitrary extension of τ . Then G = τ˜ ρ τ G , ρ G → . j K/F j j F (α)/F K/F (α) Therefore | ∈ ∈ σ (α) = [K : F (α)] σ (α) σ G σ G ∈XK/F ∈ XF (α)/F = [K : F (α)]trF (α)/F α [K:F (α)] σ (α)= σ (α)   σ G σ G ∈YK/F ∈ YF (α)/F   Let α be an F -basis for F (α) and suppose m : F (α) F (α) is repre- { i} α → sented by M¯ α with respect to this basis. Let βj be an F (α)-basis for K. With respect to F -basis for K α β , m : K { K} is represented by matrix { i j} α →

M¯ α M¯  α  ...  ¯   Mα      trK/F α = [K : F (α)]tr M¯ α =[K:F(α)]trF(α)/F α [K:F (α)] [K:F (α)] NK/F α = det M¯ α = NF (α)/F α so proposition follows from easy case. Example 15.5. Consider K = Q √d , d square free integer. K/Q is

39 Galois, with Galois group G = 1, σ : √d √d . For α, β Q. → − ∈ n o 15.4 trK/Q α + β √d = τ α + β√d τ G X∈ = α + β√d + α β√d − = 2α

N Q α + β√d = α + β√d α β√d K/ − = α2 dβ2 − Number theorists are often interested in solving equations like α2 dβ2 = γ for certain γ,α,β Q. − ∈ 15.2 Linear independence of characters Theorem 15.6 (E.Artin). Let F,K be ﬁelds. Consider the distinct linear homomorphism χ1,...,χn : F K. Then these are linearly independent over K. i.e. if there are a ,...,a→ K with 1 n ∈ a χ (α)+ a χ (α)+ + a χ (α)=0 ( ) 1 1 2 2 ··· n n ∗ for all α F then all the a = 0. ∈ i Proof. By induction. Suppose we have non-trivial relation ( ) and pick ∗ with as many ai = 0 as possible. (Can assume a1,a2 = 0). Pick 0 = β F with χ (β) = χ (β). ( ) 6 6 ∈ 1 6 2 ∗ ⇒ 0= a χ (βα)+ + a χ (βα) 1 1 ··· n n 0= a χ (β) χ (α)+ + a χ (β) χ (α) ( ) 1 1 1 ··· n n n † χ (β)( ) ( )) gives a shorter relation and thus the theorem is proved. 1 ∗ − † Corollary 15.7. Let K/F be a ﬁnite separable extension. There exists an element α K with tr α = 1. ∈ K/F Proof. (Obvious if char = 0) Linear independence of characters implies

σ G σ = 0. Pick α′ K with ∈ K/F 6 ∈ P σ (α′)= β = 0. 6 σ G ∈XK/F α′ α′ 1 If α = β , tr α = tr β = β tr α′ = 1.

40 16 Cyclic Extensions

Aim: Galois theory says study ﬁled extensions via the Galois group. Study implication of cyclic Galois group.

Deﬁnition 16.1. A Galois extension is cyclic (resp. abelian) if its Galois group is cyclic or abelian.

n 2πi 2πi Example 16.2. Q √2, e n /Q e n is cyclic. Q √2, √3 /Q is abelian. 16.1 Hilbert’s theorem 90 Theorem 16.3 (Hilbert’s theorem 90). Let K/F be a cyclic extension of degree n and Galois group G = σ . Then α K satisﬁes N α = 1 if and only if α = β for some β K.h i ∈ σ(β) ∈ Proof. ( ) ⇐ β β β σ (β) σ2 (β) σn 1 (β) N = τ = . . . − = 1 σ (β) σ (β) σ (β) σ2 (β) σ2 (β) σn (β) τ G Y∈ β

( ) Linear dependence of characters implies the function f :|K{z } K deﬁned by⇒ →

2 2 3 n 2 n 1 f = id+ασ+[ασ (α)] σ + ασ (α) σ (α) σ + + ασ (α) ...σ − (α) σ − = 0. ··· 6 We can ﬁnd γ K with f (γ) =: β = 0. ∈ 6 2 2 3 2 n 1 n σ (f (γ)) = σ (γ)+ σ (α) σ (γ)+ σ (α) σ (α) σ (γ)+ + σ (α) σ (α) ...σ − (α) σ (γ) ··· γ N(α)α−1=α−1 | {z } 1 = α− f (γ) | {z } β ∴ α = σ (β)

Theorem 16.4. Let K/F be a cyclic extension of degree n. Suppose char F ∤ n and F contains all n nth roots of 1. Then K = f (β) where β has min. poly. Xn b over F .4 − 4The β is not unique

41 Proof. Let σ generate the Galois group G and ω be a primitive nth root of n 1 n 1, so µn = id, ω,...,ω − . ω F so N ω = ω = 1 so Hilbert’s theorem { } ∈ β 1 90 implies there exists some β K with ω = σ(β) so σ (β)= ω− β. Let f(X) be the min. poly. of β over F .∈ Then

n 1 n 1 − − i i f(X)= X σ (β) = X ω− β − − i=0 i=0 Y Y since X σi (β) f(X) and deg f(X) [K : F ] = n. Note [F (β) : F ] = deg f(X)=− n |K = F (β). Direct expansion≤ shows f(X)= Xn b where b = βn. ⇒ −

16.2 Artin-Schreier Theory Theorem 16.5. Let K/F be a cyclic extension of degree n and with Galois group G = σ . Then α K satisﬁes tr α = 0 if and only if α = βσ (β) h i ∈ Proof. ( ) easy as in Hilbert’s theorem 90. ( ) Corollary 15.7 implies there is a γ⇐ K with tr γ = 1. ⇒ ∈ β = αγ + [α + σ (α)] σ (γ)+ α + σ (α)+ σ2 (α) σ2 (γ)+ . . . n 1 n 1 + α + σ (α)+ + σ − (α) σ − (γ) ··· ··· σ (β)= σ (α) σ (γ)+ σ (α)+ σ2 (α) σ2 (γ)+ + σ (α)+ σ2 (α)+ + α γ ···  ···  0=tr α 2 n 1 ∴ β σ (β)= α σ (γ)+ σ (γ)+ + σ − (γ)+ γ   − ··· | {z } = α tr γ = α

Theorem 16.6. Let K/F be a Galois extension of prime degree p =char F . Then K = F (β) where β has min. poly. Xp X b over F . − − Proof. Since [K : F ] is prime then the Galois group G = σ is cyclic of prime order. Note that tr 1 = p = 0 so Theorem 16.5 impliesh 1i = β σ (β) or σ (β)= β 1 for some β K. Since σ (β) = β, β KG = F . Therefore,− as [K : F ] is− prime, F (β) =∈ K. Let b = β6 p β so6∈ that β is a root of F (X) := Xp X b. Now σ (β) = σ (βp) σ −(β)=(β 1)p (β 1) = (βp 1 β +1)− =−βp β = b. By induction,− σi (β) −b K−G =−F . So f(X−) −F [X]. It has the− correct degree i.e. [K : F ] =⇒ [F (β∈) : F ] so is the min. poly.∈ of β.

42 17 Solvable Extension

Aim: Study implication of a solvable Galois group.

Deﬁnition 17.1. Let K/F be a ﬁnite separable extension. We say K/F is solvable if it is Galois closure has solvable Galois group.

Theorem 17.2. Let F be a ﬁeld of char. 0. Then any solvable extension K/F embeds in a radical extension.

Proof. We may pass to Galois closure of K/F and so assume K/F is Galois. Fix an algebraic closure K¯ in which to work. Let G = Gal(K/F ) and n = G . Let F and K be obtained from F and K respectively by | | 1 1 adjoining all nth roots of 1. Since F1/F is radical it satisﬁes to show K1/F1 embeds in a radical extension. As in proof of Theorem 13.2, K1/F is Galois so K1/F1 is Galois too.

⊂ N K1 K1

deg p ⊂ ⊃ ⊂ K F1 embeds ⊃ ⊂ in radical radical? F

Now we need:

Lemma 17.3. (i) G1 := Gal (K1/F1) is isomorphic to a subgroup of G, an so

(ii) is solvable. Proof. (i) (ii) by proof of Proposition 12.1. To prove (i), consider the restriction map⇒

ϕ: G Gal(K/F ) 1 −→ σ σ 7−→ |K which is well-defined since K/F is Galois. We check injectivity of ϕ as follows. Suppose σ ker ϕ so fixes K. But σ Gal(K/F ) so fixes also all nth roots ∈ ∈ of unity, and thus σ fixes K1 i.e. σ =1 in G1. This proves lemma.

Back to proof of theorem. Use induction on n = G (is a multiple of G1 ). Deﬁnition of solvable there is N ⊳ G with G /N| cyclic| of prime order| |p. ⇒ 1 1

43 Note p G . Theorem 8.6 KN /F is Galois with Galois group G /N. This | | ⇒ 1 1 cyclic extension is radical by Theorem 16.4 (and the fact that p G ). Also, | | K /KN is Galois and has solvable Galois group N. By induction, K /KN 1 1 1 1 embeds in a radical extension. We see from the picture that K/F embeds in a radical extension. Remark 17.4. (i) The above theorem is false in char. = p > 0 since there are cyclic extensions of deg p which can not be constructed by adjoining pth roots (since Xp a is not separable). But of we only assume K/F ﬁnite separable in− theorem, we do have another version of the theorem provided in our tower of ﬁeld extensions we allow not just nth roots but also roots of Xp X a. − −

(ii) In proof of theorem, it is easy to see that K1/F1 is in fact radical.

17.1 General cubic

2πi Let ω = e 3 . K + Q (ω, α1, α2, α3) where α1, α2, α3 are indeterminants. S3 S3 acts on K by permuting α1, α2, α3. Note if F = K then K/F is a solvable Galois extension with Galois group S3. Consider normal chain of subgroups 1 ⊳ A = (1 2 3) ⊳S . Galois correspondence gives tower of ﬁeld extensions 3 h i 3

Gal Gal K KA3 KS3 = F. A⊃3 S3⊃/A3

KA3 /F : is cyclic of degree 2 with Galois group S /A = (1 2)A 3 3 h 3i ≃ Z/2Z. Whats KA3 ? Let δ =(α α )(α α )(α α ) KA3 . Note that 1 − 2 1 − 3 2 − 3 ∈ the action of the Galois group of S /A is (1 2)A .δ = δ δ KS3 = F . 3 3 3 − ⇒ 6∈ Thus KA3 = F (δ) (it’s only deg. 2). Proof of Theorem 16.4 says δ2 F . ∈ Let’s check directly: δ2 KA3 . Also, (1 2), (2 3), (1 3) all ﬁxδ2 so δ2 ∈ ∈ KA3 = F . A3 K/K : is cyclic of degree 3 with Galois group A3 = (123). From proof of Theorem 16.4 we know K = KA3 (ǫ) with ǫ3 KA3 and ǫ can be any 1 ∈ 2 element of K with (1 2 3) .ǫ = ω± ǫ. We may as well take ǫ = α1 +ωα1 +ω α2 2 or α1 + ω α1 + ωα2. Now go back and look at ﬁrst section of these notes. What’s F : Let a1 = (α1 + α2 + α3) a2 = α1α2 + α2α2 + α3α1 a3 = S3 − α1α2α3 all in K = F . In fact, one can show F = Q (ω,a1,a2,a3). Also, − 3 2 K = F (δ, ǫ) so in principle you can solve the general cubic X + a1X + a2X + a3 = 0.

44 18 Finite Fields

Aim: Classify and study ﬁnite ﬁelds via Galois theory.

Proposition 18.1. Let K be a ﬁeld.

(i) char K = p> 0

(ii) K = p[K:Fp] | |

(iii) K is a subﬁeld of F¯p. Proof. (i) Pigeon hole principle.

(ii) K is a [K : Fp]-dim space over Fp. (iii) K K¯ F since K/F is algebraic. ⊆ ⊆ p p

18.1 Frobenius Homomorphism Fix a prime p.

Proposition/Deﬁnition 18.2. Let K be a ﬁeld of char. p. The Frobenius homomorphism is the map

ϕ: K K −→ x xp 7−→ (i) This is a ﬁeld homomorphism.

(ii) If K is ﬁnite or is Fp, then ϕ is an automorphism. Proof. (i) 1p =1, (xy)p = xpyp and (x + y)p = xp + yp since char. = p.

(ii) ϕ ﬁxes 1 and therefore ﬁxes all of Fp. But in both cases of (ii) K/Fp is algebraic so Lemma 8.3 implies ϕ is an automorphism.

45 18.2 Classification of finite fields Lemma 18.3. Let K be a field with pn elements, n a positive integer. Then any α K is a root of Xpn X = 0. Equivalently, if ϕ the Frobenius homomorphism∈ ϕn (α)= α. In− particular, ϕn is the identity on K.

pn Proof. α = 0 satisfies X X = 0. On the other hand, if α K∗, − n n ∈ since K∗ is a group of order (p 1) α has order dividing p 1. i.e. αpn 1 = 1 ∴ αpn = α. This proves− the lemma. − − Theorem 18.4. Let φ: F F be the Frobenius homomorphism. p → p n ϕ pn 1. The fixed field F n := F h i is the splitting field of X X over F . p p − p n 2. F n = p . | p | n 3. If K is a subfield of Fp with p elements, then K = Fpn .

4. F n /F is cyclic of degree n and Galois group ϕ F n Z/nZ. p p h | p i ≃

5. Fp/Fp is Galois. Proof. 1. The elements fixed by ϕn are precisely the roots of Xpn X. pn − Hence, Fpn is the smallest field containing roots of X X so is the splitting field of Xpn X over F . − − p pn pn 2. Fpn is the set of roots of X X and X X is separable over Fp so there are pn roots. − −

3. Lemma 18.3.

4. Note [Fpn : Fp] = n. Also, ϕ restricts to Frob. hom. on Fpn i.e. n ϕ Fpn Gal(Fp /Fp). So it suffices to show ϕ Fpn has order n. Note by | ∈ n m | m definition in (1) ϕ = 1. Suppose ϕ = 1. So ϕ fixes F n and (1) Fpn Fpn p ϕm | | F n F h i = F m . Therefore m n and n must be the order of ⇒ p ⊆ p p ≥ ϕ F n . Thus (4) holds. | p

5. Fp is closed in Fp since the only elements ﬁxed by ϕ Gal Fp/Fp is F by (1) and (2). ∈ p

Warning: Galois group of Fp/Fp is not ϕ.

46 18.3 Lattice of ﬁnite ﬁelds Let p be a prime.

Corollary 18.5. (i) The subﬁelds of F n are those of the form F d , d n. p p | n n (ii) Fp /Fpd is cyclic of degree d .

Proof. Let ϕ: F n F n be the Frobenius hom. which generates Gal (F n /F ) p → p p p by Theorem 18.4 (4). Now any subﬁeld K of Fpn contains Fp so Galois correspondence says it corresponds to a subgroup of ϕ Z/nZ. These have d d h i ≃ ϕ ex form ϕ where d n. Corresponding intermediate ﬁeld is Fhpn i Fpd . Also d h i | n ≃ ϕ E ϕ so F n /F d is a cyclic extension of degree . h i h i p p d

Remark 18.6. The lattice of finite subfields of of Fp is just the lattice of positive integers ordered by divisibility. E.g. if n, l are positive integers with l.c.m. m and g.c.d = d then smallest subfield containing Fpn and Fpl is Fpm and F n F l = F d . p ∩ p p 18.4 Some examples

4 4 3 F4/F2: F4 is splitting ﬁeld of X X over F2 and X X = X(X 1) = 2 − 2 − − X(X 1)(X +X +1) and so F4 F2 [X] / X X +1 = F2(α). This is an ““Artin-Schreier”− extension of degree≃ 2. Non-trivialh ± elementi of Galois group is ϕ is Frob. hom. ϕ (α) = α2 = α + 1 which agrees with Artin-Schreier theory. 3 F26 /F22 : Cyclic degree 3. Now F22 has all 3 roots of 1 ∴ F26 = F22 √2 . 19 Pro-Finite Groups

Aim: Inﬁnite Galois groups are best studied in the context of pro-ﬁnite groups which we introduce today.

19.1 Topological groups Deﬁnition 19.1. A topological group is a group G endowed with a topology such that:

(i) multiplication map µ: G G G is continuous; 5 × → 1 (ii) the inverse map η : G G : g g− is also continuous. → 7→ 5 G G is endowed with the product topology coming from G. ×

47 Example 19.2. R is a group with usual Euclidean topology is a topological group because (a, b) a + b and a a are both continuous. 7→ 7→ − Exercise 19.3. A subgroup of a topological group with the subspace topology is a topological group. Proposition 19.4. Let G be a topological group. 1. For g G, left and right multiplication by g are continuous. ∈ 2. If U G is open and g G then gU is open. ⊆ ∈ 3. Let U G be an open subgroup. Then U is closed. ≤ mult. Proof. 1. G / G G / G is continuous. × h / (g,h)

2. Follows from (1).

3. G U = gU is open. − g U [6∈

Proposition 19.5. Let Gα α A be a set of topological groups. Then G := { } ∈ Gα endowed with the product topology is a topological group. α A Y∈ mult. Proof. G G Gα Gα Gα is continuous, so G G G is continuous.× Similarly−→ the inverse× −→ map is continuous. × →

19.2 Inverse limits Consider data of: (i) a partially ordered set, A, ; ≤ (ii) for each α A a group G ; ∈ α

(iii) for each α, β A with α β a group homomorphism ϕαβ : Gα Gβ such that for all∈ α β γ≤ → ≤ ≤ ϕ ϕ αβ / βγ / Gα Gβ 6 Gγ

ϕαγ ϕ = ϕ ϕ αγ βγ ◦ αβ

48 Deﬁnition 19.6. These data are said to be: (i) an inverse system of groups if for any β,γ A, there is some ∈ α A with α β and α γ; ∈ ≤ ≤ (ii) a direct system if for any β,γ A there is some δ A with δ β and δ γ. ∈ ∈ ≥ ≥ Proposition/Deﬁnition 19.7. Let Gα be an inverse system of groups as above. Let { }

G = (gα) Gα ϕαβ (gα)= gβ . ( ∈ ) α A Y∈ Then G is a subgroup of Gα called the inverse limit of the inverse system Gα . It is denoted lim Gα. { } α A Q ←−∈ Example 19.8. Fix a prime number p. Let A = ..., 3, 2, 1 ordered by . { − − − } ≤

G 4 G 3 G 2 G 1 − − − − . . . / Z/p4Z / Z/p3Z / Z/p2Z / Z/pZ a + p3Z / a + p2Z / a + pZ

This inverse system has an inverse limit denoted Zˆ p called the ring of p- ∞ i adic integers. Consider a formal power series in p, q = aip where i=0 X a 0, 1,...,p 1 . Then q represents an element of Zˆ as follows: i ∈ { − } p . . . / Z/p3Z / Z/p2Z / Z/pZ 2 3 2 . . . / a0 + a1p + a2p + p Z / a0 + a1p + p Z / a0 + pZ q + p3Z q + p2Z q + pZ

Example 19.9. Let A be the set of positive integers. We order A by m n if n m. Define inverse system as follows: Gn := Z/nZ and if m n so n m we define| | ϕ : Z/mZ Z/nZ mn −→ a + mZ a + nZ. 7−→ Note that this is well-defined as mZ nZ. The corresponding inverse limit is denoted Zˆ. ⊆

49 Proposition 19.10 (Universal property of inverse limits). Let Gα, ϕαβ be an inverse system of groups and G = lim G G . Let π :{G G} α ≤ α α → α be the natural projection. Suppose H is a←− group and for each α A we have a group homomorphism ψ : H G Q ∈ α → α H C CC ψβ ψ CC α CC C! G / Gβ α ϕαβ satisfying ψβ = ϕαβ ψα whenever α β. Then there is a unique group homomorphism ψ : H◦ lim G such that≤ π ψ = ψ → α α ◦ α ←− The proof is left as an (easy) exercise.

19.3 Pro-finite groups Definition 19.11. A pro-finite group G is a group that is (isomorphic to) an inverse limit of an inverse system G of finite groups, i.e. all the G { α} α are finite. Put discrete topology on the Gα which makes Gα a topological group and hence G is also a topological group. Q Lemma 19.12. Let G = lim Gα be a profinite group with G+α finite. ←− (i) G G is closed. ≤ α (ii) G isQ compact. (iii) The open subgroups are the closed subgroups of finite index. Proof. (i) Let (g ) G G. So for some β γ, ϕ (g ) = g .Then α ∈ α − ≤ βγ β 6 γ U = (hα) Gα hβ = gβ, hγ = gγ is an open neighbourhood of (g ) disjoint∈ from GQ. α Q (ii) Follows from (i) and Tychoroff. (iii) Suppose U G is open. Then disjoint cosets form an open open cover which is finite≤ by compactness. Therefore [G : U] < . Conversely, if U G is closed and [G : U] < the complement is∞ a finite union of closed≤ cosets, so is closed. ∞

20 Inﬁnite Galois Groups

Aim: View Galois groups as pro-ﬁnite groups.

50 20.1 Inverse system of finite Galois groups Lemma 20.1. Let K/F be a Galois extension (not necessarily finite) with Galois group G. Let L be an intermediate field with L/F Galois too. Then the restriction map ρ: G Gal(L/F ) −→ σ σ 7−→ |L is a well defined group homomorphism with kernel L′ := Gal (K//) L. Also ρ is surjective. Proof. This is half of Theorem 8.6 which was stated with the additional hypothesis K/F finite. Proof doesn’t require this hypothesis. Consider the following set up: let K/F be a Galois field extension with Galois group G.

(i) A = set of subﬁelds Kα of K such that Kα/F is ﬁnite Galois. Order A by inclusion; i.e. A is “like”:

Kβ ... ⊆ ⊆ ... F ⊆ Kα Kδ ⊆ ⊆ Kγ ... This is a direct system. Why? Let L be the smallest field containing Kβ,Kγ A. Then L is finite over F being finitely generated by algebraic elements,∈ and Galois closure of L is also finite over F . (ii) Apply Galois correspondence.

... Nβ := Gal (K/Kβ) ⊇ ⊇ Nα := Gal (K/Kα) Nδ := Gal (K/Kδ) . . . ⊇ ⊇ ... Nγ := Gal (K/Kγ) If N N as above there’s a natural map β ⊆ α gNβ / gNα

ϕαβ : G/Nβ / G/Nα ≃ ≃

Gal(Kβ/F ) / Gal(Kα/F )

σ / σ |Kα 51 Note Gal (Kα/F ) is ﬁnite so get

Gal(K/Kβ) . . . l Ti T lll TTT lll TTTT lll TTT lu ll TTT Gal(K/Kα) Gal(K/Kδ) . . . Ri R j RRR jjj RRR jjjj RRR jjj RR ju jjj Gal(K/Kγ) . . .

This is an inverse system of ﬁnite groups.

20.2 Inﬁnite Galois groups Theorem 20.2. Let K/F be a Galois extension with Galois group G. Then

G lim Gal(K /F ) ≃ α ←− where the limit is taken over all the Kα subfields of K such that Kα/F is a finite Galois extension. In particular, G is pro-finite. Proof. We consider the restriction map in Lemma 20.1

ρ : G Gal(K /F ) . α −→ α Also, if K K are ﬁnite Galois “sub-extensions ” α ⊆ β Gal(K/F )= G RR RRR ρα ρβ RR RRR RRR ϕαβ ( Gal(Kβ/F ) / Gal(Kα/F )

∴ ρ = ϕ ρ . α αβ ◦ β So by the universal property of inverse limits we have a group homomorphism

ρ: G lim Gal(K /F ) −→ α σ σ←− 7−→ |Kα To show that ρ is injective, suppose ρ (σ) = 1. Thus for any finite Galois sub extension Kα/F , σKα = idKα . But K = Kα where the union is taken over all finite Galois sub-extensions Kα so σ = idK , so ρ is injective. To show that ρ is surjective considerS (σ ) lim Gal(K /F ). We wish α ∈ α to find σ G with ρ (σ)=(σ ). Pick ǫ K. There←− exists a finite Galois ∈ α ∈ 52 sub-extension Kβ/F with ǫ Kβ. We define σ (ǫ)= σβ (ǫ) Kβ K. This is well-defined, for if ǫ K ∈there is another finite Galois sub∈ extension⊆ K ∈ γ δ containing Kβ and Kγ.

σβ (ǫ) σδ K (ǫ) σδ (ǫ) β mm mmmm mmmm mmmm mmmm mmm σγ (ǫ)

Note that σδ = σβ. Therefore, σ is well-defined. A similar argument Kβ shows it is a field automorphism of K fixing F . Clearly ρ (σ)=(σα). Hence ρ is surjective and theorem is proved.

20.3 Absolute Galois group Proposition/Deﬁnition 20.3. Let F be a ﬁeld and F sep be the set of elements in F which are separable over F

(i) F sep is a ﬁeld.

(ii) F sep/F is Galois.

F sep is called the separable closure of F .

Proof. (i) Essentially because ﬁelds generated by separable elements are separable.

(ii) We need only show F sep is normal over F . But any separable polynomial (over F ) has all roots separable so if f(X) F [X] has a root in F sep then all roots are in F sep. ∈

Deﬁnition 20.4. Let F be a ﬁeld. The absolute Galois group of F is Gal(F sep/F ).

Example 20.5. Fix a prime p. What is the absolute Galois group of F ? Answer: Gal Fsep/F Zˆ. Why? The finite Galois sub-extensions p p p ≃ are F n /F which has Galois group Z/nZ. Lattice of finite fields is positive p p integers reverse ordered by divisibility. Hence the inverse system of finite Galois groups is F n F m with n m and Z/nZ Z/mZ. This is the p ⊆ p | ← inverse system we used to define Zˆ.

53 21 Inﬁnite Galois Theory

Aim: Exhibit the Galois correspondence in the inﬁnite case by determining closed objects.

21.1 Basic facts about pro-ﬁnite groups

Let G = lim Gα be a pro-ﬁnite group with all Gα ﬁnite. Let πα : G Gα α A → ←−∈ be natural projection map. Note 1Gα Gα is open. Hence Uα := ker πα = 1 { } ⊆ πα− (1) is an open subgroup of G. It is often called a fundamental open neighbourhood of 1. Proposition 21.1. With the above notation:

1. The Uα form a basis of open nbhds of 1 i.e. if U is an open nbhd of 1 then, U U for some α. ⊇ α 2. If σ G, then σU form a basis of open nbhds of σ. ∈ { α} 3. G is a Hausdorﬀ so in particular points are closed.

1 Proof. 1. Note that πβ− (gβ) G for some β A, gβ Gβ is a sub- ⊆ 1 ∈ ∈ base for the topology on G. Note π − (g ) 1 iff g = 1. ∴ it suffices β β G β n ∋ to show that any finite intersection U U for some α. But the αj ⊇ α j=1 \ Gα form an inverse system so you can find α A with α αj for j{ =} 1,...,n. But now ∈ ≤

G B BB BB π BB αj πα BB BB BB ϕαα B! j / Gα Gαj

1 1 π − (1) πα− (1) ⊆ αj

Uα Uαj n ∴ U U α ⊆ αj j=1 \ 2. (1) (2) since multiplication by σ is bicontinous. ⇒ 54 3. Suppose (gα) , (hα) are distinct elements of G so say gβ = hβ for some 1 1 6 β A. Then π − (g ) ,π − (h ) are disjoint open nbhds of (g ) and ∈ β β β β α (hα).

21.2 Inﬁnite Galois correspondence Let K/F be a Galois extension with Galois group G. We wish to show topologically closed subgroups of G are precisely the subgroups which are closed a la Section 5.

Lemma 21.2. With notation as above, let L be an intermediate ﬁeld of K/F .

1. The L′ := Gal (K/L) G is a topologically closed. ≤

2. The subspace topology on L′ is the same as that coming from its structure as a pro-ﬁnite group.

Proof. 1. We first prove the case where L/F is finite. Pick L˜ K a ⊆ Galois closure of L/F . Note L/F˜ is finite. F L L˜ K . Have ⊂ ⊂ ⊂ restriction map:

π : G = Gal(K/F ) Gal L/F˜ −→ 1 H := π− Gal L/L˜ Gal L/L˜ 7−→ Now Gal L/F˜ has discrete topology so Gal L/L˜ Gal L/F˜ is ⊆ 1 topologically closed. Therefore, H := π− Gal L/L˜ G is also ⊆ topologically closed. It suﬃces thus to showH = Gal(K/L). But π is

restriction so H Gal(K/L) as σ H σ ˜ Gal L/L˜ so fixes ≤ ∈ ⇒ |L ∈ L. Similarly, Gal (K/L) H. ⊆ Now if L/F is infinite, write L = L where L α/F are finite sub- α − extensions. If σ G, then to be in Gal(K/L) means σ fixes every Lα. ∈ S Gal(K/L)= Gal(K/Lα) so it topologically closed by the L/K finite case. T 2. Left as an exercise. Hint: Let f(X) F [X] and F˜ , L˜ K be the split- ∈ ⊆ ting fields of f over F and L respectively. Then U := ker G Gal F˜ /F 1 →

55 is a fundamental open neighbourhood of G and U := ker Gal(K/L) Gal K/L˜ 2 → is a fund. open nbhd. of 1Gal(K/F ) and observe U2 = Gal( K/L) U1. T Theorem 21.3. Let K/F be a Galois extension with Galois group G. There are inverse bijections

/ H H H′ := K

top. closed / intermediate ﬁeld o subgroups H G L ≤

L′ := Gal (K/L) o L

Proof. Use weak Galois correspondence. We first show any intermediate field L is closed. K/F is separable and normal K/L is also such. Thus Gal(K/L) ⇒ K/L is Galois i.e. L = K = L′′. So L is closed a la Section 5. Let H G be a top. closed subgroup. We wish to show it is closed a ≤ H la Section 5 i.e. H H′′ = Gal K/K . Of course H′′ H so by Lemma 21.2 we may assume⊇ F = KH . Let σ Gal K/KH and⊇ L/KH a finite Galois sub-extension of K/KH . Consider∈ fundamental open nbhd of 1. U := ker Gal K/KH π Gal L/KH −→ σU is a basic open nbhd of σ. Now by finite Galois theory we have π (σ) Gal L/LH = π (H). But Gal L/KH = L/LH = L/Lπ(H). Therefore∈ 1 σ− (H) ker π = U = . Thus H σU = . Since σU runs through basic open nbhds∩ of σ and 6 H ∅is top. closed∩ σ 6 H∅. Therefore H = Gal K/KH . This proves the theorem. ∈ 22 Inseparability

Aim: See what Galois group tells us about inseparable (i.e. not separable) extensions.

22.1 Purely inseparable extensions Proposition 22.1. Let K/F be a ﬁeld extension where char F = p 0. ≥ (i) The following are equivalent:

(a) α K is purely inseparable over F i.e. αpn F for some n 0. ∈ ∈ ≥ 56 (b) α is the only root of its minimal polynomial f(X) F [X]. ∈ (c) [F (α) : F ]S = 1. (ii) In particular, if α K is both separable and purely inseparable over F then α F . ∈ ∈ Proof. (i) (b) (c) easy. ⇔ (a) (b): α is a root of Xpn αpn F [X] so f(X) Xpn αpn = ⇒ n − ∈ − (X α)p . Therefore the only root of f(X) is α. − (b) (a): By induction on deg f. If deg f=1 then α F so (a) holds.⇐ Assume α F so no separable over F . f (X)=(∈ X α)n. 6∈ n− Have p n otherwise f ′(X) = 0 so f is separable. Write m = Z. 6 p ∈ f(X)=(X α)pm = (Xp αp)m. ∴ αp satisﬁes a polynomial in

F [X] of degree− m whose only− roots is αp. By induction, αp is purely inseparable over F . Thus α is purely inseparable over F too.

22.2 Maximal separable and purely inseparable extensions Let F be a field, which characteristic p> 0. Proposition/Definition 22.2. Let K/F be a field extensions. The intermediate field L := K F sep of all elts in K which are separable over F is such that K/L is purely∩ inseparable. L is called the maximal separable sub-extension. Proof. Let α K and f(X) be its min. poly. over L. Argue by induction on deg f(x). As∈ before deg f(X) = 1 α L. Suppose deg f(X) > 1, so ⇒ ∈ α L. Note α is not separable over F ∴ not separable over L ∴ f ′(x) = 0 6∈ f(X) = g (Xp). Thus αp has min. poly. over L of smaller degree so is purely⇒ insep. over L and hence α is purely insep. over L. Proposition/Definition 22.3. Let K/F be a field extension. The set of elements in K which are purely inseparable over F forms a field, say L, called the maximal purely inseparable sub-extension. Proof. Left as an easy exercise. Example 22.4. If φ : K K is Frobenius then L is just the union of →n increasing chain of subfields φ− (F ), n = 1, 2,...

57 22.3 Normal extensions Theorem 22.5. Let F be a ﬁeld of char. p> 0 and K/F a normal extension with Galois group G. Let Lsep be the maximal separable sub-extension and Lpi be the maximal purely inseparable sub-extension. Then: (i) K/KG is Galois with Galois group G.

G (ii) Lpi = K is Galois.

(iii) Lsep/F is Galois. (iv) L L = F . sep ∩ pi

(v) The smallest subﬁeld L˜ containing Lsep and Lpi is K.

(vi) The restriction map ρ : Gal(K/Lpi) Gal(Lsep/F ) is a well-deﬁned group isomorphism. →

G Proof. (i) G Gal K/KG Gal(K/F )= G. Therefore, KGal(K/K ) = KG and thus≤ K/KG is Galois≤ with Galois group G. (ii) Since K/F is normal, the G-orbits of any α K is precisely the set of roots of its min. poly. f (X) over F . But α∈ KG iﬀ the G-orbit of α is 1 element iﬀ (by Proposition 22.1) α is purely∈ inseparable over F .

(iii) Same as for F sep. If α L then its min. poly. f(X) is separable ∈ sep and K/F normal all roots of f(X) lie in Lsep. Therefore Lsep/F is normal too and hence⇒ Galois.

(iv) Proposition 22.1 (ii).

(v) K purely inseparable over Lsep so K is purely inseparable over L˜. (i) and (ii) K is separable over Lpi so K is separable over L˜. Therefore K/L˜ is separable⇒ and purely inseparable so by Proposition 22.1 K = L˜.

G (vi) Let σ Gal K/K . It must permute separable elements so σ (Lsep)= L so∈ρ is a well-deﬁned group homomorphism. sep To see that ρ is injective, note that if ρ (σ) = 1 then σ ﬁxes Lsep and Lpi and by (v) K. Thus σ = 1 and hence ρ is on-to-one.

Also, anyσ ¯ Gal(Lsep/F ) can by uniqueness of splitting ﬁelds be lifted to Gal (K/F∈ ) = Gal K/KG so ρ is surjective and the theorem is proved.

58 Corollary/Definition 22.6. A field F of char p > 0 is perfect is Frob F F is surjective. (Also fields of characteristic 0 are said to be prefect too).→ Any finite extension K/F of a prefect field is separable. Proof. Let φ : K K be the Frobenius map. We can assume K/F is → n normal. The maximal purely inseparable sub extension is φ− (F ) = F . n N So result follows from theorem. [∈

23 Duality

Aim: Prepare for classiﬁcation of ﬁnite abelian extension in the next section.

23.1 The dual group Deﬁnition 23.1. Let A be an abelian group and m be a positive integer. An element a A is m-torsion if ma = 0. We say A is m-torsion if every ∈ element in A is m-torsion. We say A is torsion if every element of A has ﬁnite order.

p Example 23.2. Q/Z is torsion since q + Z is q-torsion. Proposition/Notation 23.3. Let G be a group and A be an abelian group. Let Hom (G, A) be the set of group homomorphisms ϕ : G A. This is an → abelian group when endowed with addition (ϕ1 + ϕ2)(g) := ϕ1 (g)+ ϕ2 (g). Proof. Easy. Deﬁnition 23.4. Let G be a torsion abelian group. The dual of G is G∧ := Hom (G, Q/Z).

Example 23.5. (Z/nZ)∧ Z/nZ. Why? Any group homomorphism ϕ : Z/nZ Q/nZ is determined≃ by a := ϕ (1 + nZ) Q/Z. This can be → 1 2 ∈ n 1 any n-torsion element of Q/Z i.e. 0+Z, n +Z, n +Z,..., −n +Z. One easily veriﬁes that the hom. ψ : Z/nZ Q/nZ mapping i + nZ i + Z is the → 7→ n generator of (Z/nZ)∧. Exercise 23.6. Let A, B be torsion abelian groups. Show

A∧ B∧ ∼ (A B)∧ × −→ × (ϕ, ψ) ((a, b) ϕ(a)+ ψ(b)) 7−→ 7→ is a group isomorphism. Hence by Example 23.5 and structure theorem for ﬁnitely generated abelian groups, A A∧ for A ﬁnite abelian. ≃ 59 23.2 Perfect pairings Let A, B be torsion abelian groups.

Deﬁnition 23.7. A pairing between A and B is a function ψ : A B Q/Z such that: × →

(i) ψ (a, ), ψ ( , b) are group homomorphisms for all a A, b B. We say ψ −is perfect− if: ∈ ∈

(ii) ψ(a, ) = 0 a = 0 and ψ ( , b) = 0 b = 0. − ⇒ − ⇒ Exercise 23.8. For a ﬁnite abelian group A, show

A A∧ Q/Z × −→ (a, ϕ) ϕ(a) 7−→ is a perfect pairing. Conversely, we have:

Proposition 23.9. Let A, B be ﬁnite abelian groups and ψ : A B Q/Z × → is a perfect pairing between them, then A B∧. ≃

Proof. Let Ψ : A B∧ be the map a ψ(a, ). It is a group homomor- → 7→ − phism by Definition 23.7 and injective by part (ii). Therefore, A B∧ = B . Switching roles of A and B shows B A so Ψ is an isomorphism.| | ≤ | | | | | | ≤ | | 23.3 Some abelian extensions Fix positive integer m. Let F be a field of characteristic not dividing m and such that F µm (note µm = m). Kummer theory classifies abelian extensions of F with⊇ m-torsion| Galois| groups. Here we construct some examples. m m m m m F ∗/F ∗ : Let F ∗ = α α F ∗ . F ∗ abelian F ∗ F ∗ and F ∗/F ∗ { | ∈ } ⇒ ≤ is m-torsion abelian group. m m m m m √αF : Let a F ∗/F ∗ , say a = αF ∗ . We “define” √a to be any m-th ∈ root of α. Fact: √m a is well defined up to scalar multiple by some scalar m β F ∗. Why? Changing choice of m-th root of α, changes √a by an ele- ∈ m m ment of µm F ∗. Changing α to αβ where β F ∗ changes √α by β. m ⊆ m m ∈ F √J : Let J F ∗/F ∗ . Define F √J to be the splitting field over ≤ F of the family of polynomials

m m X α αF ∗ J . { − | ∈ } This is generated over F by the √m a, a J. ∈

60 m m Proposition 23.10. For J F ∗/F ∗ ,F √J /F is Galois. ≤ m m Proof. For αF ∗ J, X α has m distinct roots which must be separable ∈ − over F . Therefore, F √m J is generated by separable elements over F . Since it is also normal, it’s Galois.

m m Proposition 23.11. Let J F ∗/F ∗ . Let σ Gal F √J /F ,a J. ≤ ∈ ∈ σ( m√a) Then ψ (σ, a)= m µ and is independent of choice of m-th root of a. √a ∈ m m m m m Proof. Suppose √a = √α, where α F ∗ . Then σ maps √α to some m ∈ m m σ( √a) σ( √aβ) other root of X α. ψ (σ, a) = m µ . Also, for β F ∗ m = − √a ∈ m ∈ √aβ σ m√a σ(β) ( ) m m√aβ so independence of choice of √a follows from this fact.

m m Proposition 23.12. For J F ∗/F ∗ ,G := Gal F √J /F is m- ≤ torsion abelian group. Proof. The action of σ, τ G on F √m J is determined by its action on ∈ the generators √m a a J of F √mJ /F. Just as in the case of cyclic { | ∈ } extension Proposition 23.11 implies σ, τ act on √m a by multiplication by ψ (σ, a) ,ψ (τ,a) and the fact that F µm these actions commute. There- ⊇ ⇒ m fore, G is abelian. Finally, ψ (σ, a) µm implies σ acts as identity so G is also m-torsion. ∈

24 Kummer Theory

Aim: Classify abelian extensions a la Kummer. Fix for this section a positive integer m and ﬁeld F of characteristic not dividing m such that F contains all m, m-th roots of 1.

24.1 A perfect pairing

m m Let J F ∗/F ∗ and G = Gal F √J /F . Recall from propositions in ≤ section 23, we have a well-defined map ψ : G J µ × −→ m σ ( √m a) (σ, a) 7−→ √m a Note also that µ 1 Z/Z < Q/Z. m ≃ m 61 ψ Theorem 24.1. The map G J µm ֒ Q/Z is a prefect pairing. If J m × −→ → is finite then Gal F √J /F J ∧. ≃ Proof. We first check ψ is a paring. Let σ, τ G , a, b J. ∈ ∈ σ( √m ab) σ ( √m a) σ( √m b) ψ(σ, ab) := = = ψ (σ, a) ψ (σ, b) √m ab √m a √m b thus ψ (σ, ) is a homomorphism. Also, − (στ)( √m a) σ(ψ(τ,a)) √m a ψ (σ,τ,a) := = = ψ(τ,a)ψ(σ, a) √m a √m a so ψ( ,a) is multiplicative. We− now check that ψ is perfect. If ψ(σ, ) is 1, then σ acts as the − identity on the generator √m a,a J if F √m J /F ∴ σ = 1. Suppose ∈ m m m now that a J 1 . So α F ∗ ∴ √a F so as F √J /F is Galois ∈ − J 6∈ 6∈ (by Proposition 23.10) there exists σ G such that σ(√m a) = √m a and so σ( m√a) ∈ 6 ψ(σ, a)= m = 1. Therefore ψ( ,a) = 1 a = 1. Thus ψ is perfect too. √a 6 − ⇒ Now Proposition 23.9 shows if J is finite (so is Galois too) then G J ∧. This proves theorem. ≃

24.2 Classiﬁcation of abelian extensions Theorem 24.2. For a positive integer m, F a ﬁeld of characteristic not dividing m containing primitive m-th root of unity, then:

(i) There is a bijection

Abelian extensions subgroups Ψ m of F with m-torsion J F ∗/F ∗ −→   ≤  Galois group  J F ( √m J) 7−→   (ii) Finite subgroups correspond to ﬁnite extensions.

Proof. (ii) Do finite case first. m Ψ injective: Let J1,J2 F ∗/F ∗ be finite and J1J2 be the groups ≤ m generated by J1 and J2 (which is still finite). Suppose F ( √J1) = m m m F ( √J2). Then F ( √J1J2) = F ( √J1) since J1J2 consists of products

62 of elements of J1 and J2. Thus it suﬃces to show J1 = J1J2 for then by symmetry J2 = J1J2. By Theorem 24.1

m m J ∧ Gal F ( J )/F = Gal F ( J J )/F (J J )∧. 1 ≃ 1 1 2 ≃ 1 2 Therefore J = J ∧p= (J J )∧ = J J pso J = J J . | 1| | 1 | | 1 2 | | 1 2| 1 1 2 Ψ surjective: Suppose K/F is a finite abelian extension with mtorsion Galois group say G G G G where G is a cyclic group ≃ 1 × 2 ×···× r i of order ni m. Let πi : G = G1 Gr Gi be the natural | ×···× → Hi projection and Hi = ker π ⊳ G. Note, Hi ⊳ G K /F is a Galois extension with Galois group G/Hi G i.e.⇒ is a cyclic extension ≃ i of degree ni. The theory of cyclic extensions (Theorem 16.4) tells H m us K i = F ( √m a ) for some a F ∗/F ∗ . Let J = a ,...,a . i i ∈ h 1 ri m m m m Then F ( √J) = F ( √a1, [a2],..., [ar]) K. We wish to show m ⊆ m equality by computing GalpK/F ( √Jp) . Now F ( √J) is the smallest subfield containing KH1 ,...,K Hr therefore by Galois correspondence Gal K/F ( √m J) is H H H = 1. Thus K = F ( √m J) and Ψ 1 ∩ 2 ∩···∩ r is surjective. m (i) do now infinite case. As before, suffice to show if J1 J2 F ∗/F ∗ m m ≤ ≤ with F ( √J1)= F ( √J2) then J1 = J2. Let a J2. Then there exists m m m ∈ 0 a1,...,an J, such that √a F ( √a1,..., √an). If J1 = a1,...,an 0 ∈ ∈ m 0 m 0 h i and J2 = a,a1,...,an and F ( J1 ) = F ( J2 ). So by finite case J 0 = J 0 soha J ). Thusi J = J and so Ψ is injective. 1 2 ∈ 1 1 2p p To Ψ is surjective: let K/F be an abelian extension with m-torsion Galois group G. Write K = Ki where Ki/F are finite Galois extensions. These must have Galois∩ groups which are quotients of G so are m m m-torsion abelian. Thus K = F ( √Ji) for some J1 F ∗/F ∗ ¿ if J is m ≤ a subgroups generated by all the Ji then F ( √J)= K.

25 Galois Correspondence in Topology

In this section all topological spaces X will satisfy the following. X is path connected X is locally contractible. i.e for all x X there exists open nbhd U ( )  ∈ x ∗ and continous function κx : Ux [0, 1] Ux such that κx( , 0) =idUx   κ ( , 1) is constant. × → − x − Example 25.1. X is a manifold. 

63 25.1 (Unramiﬁed) covers Deﬁnition 25.2. Let X be a topological space satisfying ( ). A cover of X is a continuous map of topological spaces πY X such that∗ for any x X there is an open nbhd U of x with the property→ that any component V∈ of 1 π− (U) is open and π : V U is a homeomorphism. |V → Example 25.3.

π : R R/Z S1 −→ ≃ α α + Z 7−→

( ) α +• 2

( ) α +• 1

( ) α•

S1 R/Z ≃ ( ) • 25.2 Galois correspondence Proposition/Deﬁnition 25.4. Let π : Y X be a cover. A deck transformation of π is a homeomorphism σ : →Y Y such that the following →

64 diagram commutes σ Y / Y A AA ~~ A ~~ π AA ~~ π A ~~ X In other words, π σ = π. The set of deck transformations of π is called the Galois group of ◦π and is denoted Gal(Y/X). It’s a group under composition. Remark 25.5. (i) Deck transformation σ preserves fibres of π since 1 y π− (x) π(σ(y)) = π(y)= x. ∈ ⇒ (ii) X is path connected means cardinality of fibres is constant and is called the degree of the cover. Example 25.6 (Continued from previous example). For n Z, we can define σ : R R with σ (α)= α + n. In fact, Gal (R/S) Z ∈ n → n ≃ Definition 25.7. A cover π : Y X is Galois if Gal(Y/X) acts transitively on every fibre of π. → Example 25.8. From before R S is a Galois cover. → 25.3 Galois correspondence Definition 25.9. Let π : Y X be a cover. An intermediate cover is a factorisation of π i.e. commutative→ diagram Y @ @ ϕ @@ @@ @ π Z ~ ~~ ~~ψ ~~ X such that ψ is a cover (and therefore ϕ is a cover too). Two intermediate covers Y Z X and Y Z X are equiv- → 1 → → 2 → alent if there exists a homeomorphism h :! ∼ Z such that the following → 2 diagram commutes Y A }} AA }} AA }} AA }~ } A Z1 ∼ / Z2 A AA }} AA }} AA }} A }~ } X

65 Theorem 25.10. Let π : Y X be a Galois cover with Galois group G. There is a lattice anti-isomorphism→

H / (Y Y/H X) → →

subgroups / intermediate o of G covers of π Gal(Y/Z)o (Y Z X) → → The normal subgroups correspond to intermediate covers Y Z X where Z X is Galois. → → → Example 25.11 (Continuing from all previous examples). 0 nZ Z = Gal(R/S) corresponds to ⊇ ⊇

R JJ JJ JJ JJ JJ$ S R/nZ t≃ tt tt tt z tt S t

25.4 Simply connected cover Theorem 25.12. Let X be a topological space satisfying ( ). Then there is ∗ a unique cover π : Xsc X with Xsc simply connected6. This is called the simply connected cover→ of X. It satisﬁes the following universal property: Consider any cover ψ : Z X, then there is a commutative diagram of continuous maps: → Xsc / Z D DD D ψ π DD DD! X Corollary 25.13. Any cover π : Y X of a simply connected space X is → trivial. i.e. π = id.

Theorem 25.14. Let X be a topological space. Then Xsc X is Galois sc → and Gal(X /X) is isomorphic to the fundamental group of X, π1(X).

6i.e. all loops in Xsc are contractile to a point.

66 26 Riemann Surfaces

Aim: See how ﬁeld extensions arise in the theory of Riemann surfaces.

26.1 Riemann surfaces Deﬁnition 26.1. A Riemann surface is a 1 dimensional complex manifold i.e. a Hausdorﬀ topological space X with an atlas

φ U α V C α −→ α ⊆ n o 1 where X = Uα is an open cover such that the transition functions πα πβ− are holomorphic.∪ ◦

1 Example 26.2. Riemann sphere of complex projection line PC = X. As a topological space, X = spheres S2 = 1 pt compactiﬁcation of C. Atlas given by identifying −

(i) X with C (with complex variable z); − ∞ (ii) X 0 with C (with complex variable y) − 1 and transition function y = z− . Example 26.3. Elliptic curve. Pick τ C with im z > 0. Consider lattice Λ := Z + Zτ < C. Let X = C/Λ. This a∈ Riemann surface.

26.2 Field of meromorphic functions Definition 26.4. Let X be a Riemann surface. The field of meromorphic functions C(X) is the set of meromorphic functions on X i.e. functions which are holomorphic except for a finite number of poles.

Note that meromorphic functions on X are just holomorphic functions X P1 . → C 1 Example 26.5. The meromorphic functions on X = PC are the meromorphic functions on C which are neither holomorphic at or have a pole there. Thus meromorphic functions are just rational functio∞ ns. Therefore 1 C (PC)= C (z). Example 26.6. As before Λ = Z + Zτ is a lattice in C and X = C/Λ. A meromorphic functions on X is a meromorphic function f(x) on C which double periodic with periods 1,τ i.e. f(z +1) = f(z), f(z + τ)= f(z).

67 Proposition/Deﬁnition 26.7. The Weierstrass ℘ function

1 1 1 ℘(z) := + z2 (z w)2 − w2 w Λ 0 X∈ − − is meromorphic on C and doubly periodic. Therefore C(℘(z)) = C(X). ⊆ Facts:

2 3 (i) ℘′(z) = 4℘(z) g ℘(z) g for some g ,g C depending on Λ. − 2 − 3 2 3 ∈ 3 (ii) C(X)= C(℘(z), ℘′(z)) C(t, 4t g t g ). In fact, ≃ − 2 − 3 C/Λ pP2 −→ C z +Λ (℘(z) : ℘ :(z):1) 7−→ embeds C/Λ into P2 as the the curve y2 = 4x3 g x g C − 2 − 3 26.3 Functoriality Proposition 26.8. Let σ : X Y be a non-constant map of Riemann surfaces. →

1 (i) Let f : Y P be a meromorphic function Y . Then σ∗f := f σ is → C ◦ meromorphic on X. We thus obtain a ﬁeld homomorphism σ∗ : C(Y ) C(X). This is the pull back of functions. →

(ii) If we have another non-constant holomorphic map τ : W X of → Riemann surfaces then (σ τ)∗ = τ ∗ σ∗ : C(Y ) C(W ). ◦ ◦ →

(iii) (idY )∗ = idC(Y ). Proof. (i) Clear.

(ii) (σ τ)∗ (f)= τ ∗(σ∗f). ◦ (iii) Clear.

Example 26.9. Elliptic curve X = C/Λ

℘ X / P1 ? C ?? ?? f, hol. f(z) C(z) ∗ ?? ∈ ℘ f=f(℘(z))) ? 1 PC

68 Therefore, writing t for ℘(z)

℘ : C(P1 ) C (C/Λ) ∗ C −→ C(t) / C t, 4t3 g g − 2 − 3 t t p 7−→ 26.4 Riemann’s Theorem Theorem 26.10. Let K be a finite extension of C(z). Then there is a unique compact Riemann surface with C(X) K. Furthermore, if L/K is a finite extension and Y is the compact Riemann≃ surface with C(Y ) L, then there ≃ is a non-constant holomorphic map σ : Y X such that σ∗C(X) C(Y ) is the original field inclusion K L. → → ⊆ 27 Geometric examples of field automorphisms

Proposition 27.1. Let X,Y be a Riemann surfaces and π : Y XX be an unramiﬁed cover. Then Y has a unique structure as a Riemann→ surface in such a way that π is holomorphic.

27.1 Galois groups Proposition 27.2. Let π : Y X be a non-constant holomorphic map of → compact Riemann surfaces. We will use π∗ : C(X) C(Y ) to identify C(X) with a subﬁeld of C(Y ). → (i) Let σ : Y Y be a biholomorphic map such that → σ Y / Y A AA ~~ A ~~ π AA ~~ π A ~~ X

commutes π σ = π. Then σ∗ : C(Y ) C(Y ) Gal(C(Y )/C(X)). ◦ → ∈ (ii) The elements of Gal(C(Y )/C(X)) are the σ∗ where σ is as in (i).

(iii) Let G be the pro-ﬁnite completion π1(X) i.e.

\ G = lim π1(X)/N ⊳ π1(X) N E←−π1(X) [π1(X) : N] ≤ ∞

69 Then G is a quotient of the absolute Galois group Gal C(X)/C(X) corresponds to the “maximal unramiﬁed extension”.

Proof. (i) Functoriality implies σ∗ π∗ = π∗. Thus σ∗ ﬁxes C(X). ◦ (ii) This follows from Riemann’s theorem.

(iii) Any ﬁnite Galois unramiﬁed cover Xsc Y X corresponds to → → N⊳π1(X), Proposition 27.1 implies Y is a Riemann surface. Parts (i) and (ii) give Gal (C(Y )/C(X)) Gal(Y/X) π (X)/N. ≃ ≃ 1

27.2 Unramified covers of elliptic curves Consider lattice in C,Λ= Z + Zτ, im τ > 0. Elliptic curve X = C/Λ. C is simply connected so C C/Λ is the simply connected cover. Translation by → 2 λ C is a deck transformation. We see that π1(C/Λ) Λ Z . Therefore by∈ Proposition 27.2 (iii) the absolute Galois group Gal≃ (C(X≃)/C(X)) has a quotient corresponding to π \(G/Λ) Z2. What are finite unramified covers 1 ≃ of X = C/Λ? Pick Λ′ < Λ of finite index Y = C/Λ′ C/Λ = X. Note → Λ′ is still a Riemann surface, in fact anb elliptic curve. Also Gal (Y/X) = Gal(C(Y )/C(X))=Λ/Λ′.

1 27.3 Ramified Cyclic Covers of PC Definition 27.3. Let πY X be a non-constant holomorphic map of com- → pact Riemann surfaces. If is surjective, so we will call it a ramified cover if it not unramified. We will say it is Galois or cyclic if C(Y )/C(X) is.

top. sp. 1 2 Now PC S which is simply connected so any unramified cover is trivial. Let’s find≃ all ramified cyclic covers using Kummer theory, say with 1 Galois group G Z/mZ,m > 0. Recall, C(PC) C(Z). Need to know m ≃ ≃ C(z)∗/C(z)∗ . Fact:

∼ (i) ϕ: C∗ λ C Z C(z)∗. ⊕ ∈ −→ m (ii) C(z)∗/C(zL)∗ λ C Z/mZ. ≃ ∈ Why? L n ϕ: α + n + + n α (z λ) λ1 . . . (z λ )nλr λ1 ··· λr 7→ − − 1

70 Need cyclic, order m subgroups of λ C Z/mZ. Pick some (nλ1 ,...,nλr ) Zr which is order m, modulo m. Kummer∈ theory says if f(z) = (z ∈ L − nλ nλ m λ1) 1 . . . (z λr) r then C z, f(z) /C(z) is a cyclic extension of degree m. − p What’s happening geometrically? Let X P1 be corresponding map → C of compact Riemann surfaces. Then away from λ1,λ2,...λr π is unramified, λ ,...,λ (and maybe ) are ramified points. 1 r ∞ Example 27.4. If X P1 corresponds to C z, z(z 10(z λ) /C(z) → C − − then X is essentially zeroes of y2 = z(z 1)(z λ) in C2 compactified in P2 . − − p C 28 Ramification Theory

28.1 Discrete valuation rings Proposition/Definition 28.1. A discrete valuation ring (DVR) is a PID R which is not a field and has a unique maximal ideal (i.e. a local ring) tR = t . In this case, any non-zero ideal I has form tn , for some n N. h i h i ∈ This is not the standard definition, but is equivalent to it.

i Example 28.2. Let F be a field. Then F JtK = i 0 αit αi F which is the ring of formal power series in t with coefficients≥ in F .| This∈ is a DVR P i with maximal ideal t . Why? Given a(t) = i 0 αit where α0 = 0 then h i ≥ i 6 a(t) is invertible. This means that any element a(t)= i n αit with αn = 0 is an associate of tn and so any non-zero ideal hasP form t≥n for some n 6 N. Ph i ∈ Example 28.3. C t < CJtK is the subring of convergent power series. This is a DVR C t { ring} of germs of holomorphic functions at some point of a Riemann surface.{ } ≃ Example 28.4. Zˆ = lim Z/pnZ is a DVR, with maximal ideal p . p n h i ←− 28.2 Motivation for Ramification Let π : Y Z be a (un)ramified cover of a Riemann surface. Essentially, → ramification is where the sheets come together. Lets study π locally at y0 Y and z π(y ). Using charts we may identify open nbhds of y with open∈ an 0 ∈ 0 0 nbhd U of 0 in C with variable y and open nbhd of z0 with open nbhd of 0 in C with variable z. Then π is given by some holomorphic f : U C say with f(0) = 0. There are 2 cases: →

71 2 Unramified case: f ′(0) = 0 i.e. f(y) = α1y + α2y + . . . where α1 = 0. Inverse function theorem implies that f is a local isomorphism. 6 n n+1 Ramified case: f ′(0) = 0 so f(y)= αny + αn+1y + . . . with αn = 0, n 6 n 2. Locally at y = 0, f behaves like f(y) αny . ≥ Generically in nbhd of 0, f is n : 1. But≈ at y = 0, the n sheets come together. Can’t possibly be unramified there. The upshot is π : Y Z is → unramified everywhere except a finite number of points.

28.3 Ring-Theoretic Reformulation

Let π : Y Z be a non-constant map of Riemann surfaces and y0 Y . Let z = π(y →) Z. Write C y , C z for the ring of germs of holomorphic∈ 0 0 ∈ { } { } functions at y0 and z0. Consider

π∗ : C z C y { } −→ { } z α yn + α yn+1 + ...,n 1, α = 0. 7−→ n n+1 ≥ n 6 n π∗(z)C y = y . π is unramified at y0 iff π∗(z) generates the maximal ideal of{C} y . h i { } Definition 28.5. Let R,S be DVRs with maximal ideals tR and uS. Sup- 1 pose ϕ : R S is a ring homomorphism such that ϕ− (uS) = tR. We say ϕ is unramified→ if

(i) π∗(t) generates the maximal ideal uS of S and

(ii) S/uS is a separable ﬁeld extension of R/tR