Galois group
Chapter 7
Galois Theory
7.1 Extension Automorphism Group
03/25/19 Lemma 7.1.1 Let E/F be a field extension. The set of automorphisms of the extension E/F, i.e. automorphisms of E that fix the elements of F , ' Aut(E) '(a)=a for all a F { 2 | 2 } is a group, a subgroup of Aut(E). Definition 7.1.1. Let E/F be a field extension. The group of automor- phisms of this extension, is denoted by AutF (E)orbyGal(E : F ), and is sometimes called the Galois group of the extension E/F.1
Note that any automorphism of E fixes 1 and therefore it fixes the prime
subfield of E.Ifchar(E) = 0 then Aut(E) = AutQ(E). If char(E)=p,then
Aut(E) = AutZp (E). Examples 7.1.1. 1. Aut (F )= 1 . F { F } 2. Denote by ⌧ the complex conjugation map ⌧ : C C a + bi ! a bi 7! 1Some authors reserve the name “Galois group” for extensions which are Galois exten- sions. See Definition 7.4.3.
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It is straight forward to check that ⌧ is an autormorphism of C and it fixes the elements of R.Therefore,⌧ Aut (C). It follows from 2 R Proposition 7.1.6 below that 1C and ⌧ are the only automorphisms of the extension C/R. Thus Aut (C)= 1 ,⌧ . R { C }
We will often need to consider homomorphisms from a field of the form 03/26/19 F (↵1,...,↵n), where F is a field and ↵1,...,↵n come from an extension of F , to another field or ring. The next lemma tells us that such homomorphism is completely determined by where it maps the elements of F ,aswellas,where it maps ↵1,...,↵n.
Lemma 7.1.2 Let F be a field, ↵1,...,↵n elements from some extension E of F , and R a commutative ring with unity. If '1,'2 : F (↵1,...,↵n) R are homomorphisms such that ' (a)=' (a) for all a F and ' (↵ )=!' (↵ ) 1 2 2 1 i 1 i for i =1,...,n, then '1 = '2. Exercise 7.1.1. Prove Lemma 7.1.2.
Corollary 7.1.3 Let E = F (↵1,...,↵n) be an extension of F , and K an- other extension of F . Any F -homomorphism ' : E K is completely ! determined by the values '(↵1),...,'(↵n) in K.
In order to work out examples of automorphism groups of extensions, we will use a converse of Lemma 6.4.13, as well as a special cases of that lemma and its converse.
Lemma 7.1.4 Let F and Fˆ be fields, and ' : F Fˆ a homomor- '˜ K / Kˆ phism. Let p(x) F [x] be an irreducible polynomial,! and denote by pˆ(x) its image '(2p(x)) Fˆ[x].If↵ is a root of p(x) in some exten- sion K of F , Kˆ an extension2 of Fˆ, and '˜ : K Kˆ a homomorphism ! ' that extends ', then '˜(↵) is a root of pˆ(x). F / Fˆ
Proof. Let↵ ˆ =˜'(↵). Then
pˆ(ˆ↵)='(p)(' ˜(↵)) =' ˜(p)(' ˜(↵)) =' ˜(p(↵)) =' ˜(0) = 0
The following propositions are special cases of Lemma 6.4.13, and Lemma 7.1.4, by taking Fˆ = F , ' = IF ,andKˆ = K for the second one. 7.1. EXTENSION AUTOMORPHISM GROUP 83
Proposition 7.1.5 Let K and E be extensions of F . Let ↵ K be algebraic 2 over F , and E be a root of minF (↵). There is a F -homomorphism (i.e. it fixes every a2 F ) ' : F (↵) E, such that '(↵)= . 2 !
Proposition 7.1.6 Let K/F be an extension and ↵ K algebraic over F . For each ' Aut (K), '(↵) is a root of min (↵) in 2K. 2 F F
Now, we are ready to complete Example 7.1.1.2, and consider other ex- amples.
Example 7.1.2. 1. Since C = R(i), any automorphism of C/R is com- pletely determined by where it maps i.FromProposition7.1.6we know that i has to be mapped to a root of min (i)=x2 +1,i.e. to i, R ± confirming the statement that Aut (C)= 1 ,⌧ . R { C }
2 2. We know that minQ(p2)=x 2. A similar argument to that of the previous example shows that there are only two automorphisms of the p extension Q( 2)/Q,namelytheidentitymap1Q( p2),andthemap
: Q(p2) Q(p2) ! a + bp2 a bp2 7!
Thus Aut (Q(p2))= 1, . Q { }
3 3 3 3. Consider now the extension Q(p2)/Q.SinceminQ(p2)=x 2, has 3 3 3 3 roots p2 ,!p2 ,!2p2, but the last two are not in Q(p2)(theyarenot 3 even real numbers), the only automorphism of Q(p2) istheidentity 3 map. Thus, Aut (Q(p2))= 1 . Q { }
3 4. Let’s consider now the extension E := Q(p2 ,!)ofQ. Note first that since Q is the prime field, any automorphism of E will fix Q.The 3 3 3 3 2 3 minimal polynomial minQ(p2) = x 2hasrootsp2 ,!p2 ,! p2, 2 2 and the minimal polynomial minQ(!)=x + x +1hasroots!,! .By Proposition 7.1.6, there are only six potential automorphisms of the
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3 extension Q(p2 ,!)/Q,namely:
3 3 3 3 I : p2 p2 1 : p2 p2 ! 7! ! ! 7! !2 7! 7! 3 3 3 3 ⇢ : p2 !p2 2 : p2 !p2 ! 7! ! ! 7! !2 7! 7! 2 3 2 3 3 2 3 ⇢ : p2 ! p2 3 : p2 ! p2 ! 7! ! ! 7! !2 7! 7! Proposition 7.1.5, guarantees the existence of automorphisms doing what each of the above does. Therefore
3 2 Aut (Q(p2 ,!)) = I,⇢,⇢ , 1, 2, 3 Q { } Straightforward calculation of the table for this group, shows that it is isomorphic to D3.
The following Corollary to Lemmas 7.1.4 and 6.4.13, gives us both an upper bound, and an exact count, on the number of homomorphisms from asimplealgebraicextension,extendingagivenhomomorphismfromthe ground field. Corollary 7.1.7 Let F and Fˆ be fields, and ' : F Fˆ a homomorphism. Let K be an extension of F , and ↵ K algebraic over!F , and Kˆ an extension of Fˆ. The number of homomorphism2 '˜ : F (↵) Kˆ which extend ' is at ! most degF (↵). Moreover, let p(x)=minF (↵), and pˆ(x)='(p(x)). The number of homomorphism '˜ : F (↵) Kˆ which extend ' is the number of distinct roots of pˆ(x) in Kˆ . !
Proof. Any homomorphism' ˜ : F (↵) Kˆ that extends ' is completely ! determined by the value fo' ˜(↵). Take p(x) = minF (↵)inthelemma.Since degFˆ(ˆp(x)) = degF (p(x)) = degF (u), the polynomialp ˆ(x)hasatmostthis many distinct roots, so there are at most degF (↵)choicesfor˜'(↵). Scholium 7.1.8 Let F and Fˆ be fields, and ' : F Fˆ a homomorphism. Let E be an extension of F , and u E algebraic over!F , and Kˆ an extension 2 of Fˆ. Let f =minF (u), and fˆ = '(F ). The number of homomorphism '˜ : F (u) Kˆ which extend ' is the number of distinct roots of fˆ in Kˆ . ! 7.1. EXTENSION AUTOMORPHISM GROUP 85
As a particular case of Corollary ?? we get: enough automorphisms Proposition 7.1.9 Let E = F (↵) be a simple finite extension of F . Then
Aut (E) [E : F ]. (7.1) | F |
Proof. Let f(x)=minF (↵), and let ↵1,...,↵k be the distinct roots of f(x) in E.Letn =deg(f(x)), so k n.ByCorollary6.2.8,weknowthat [E : F ]=n,andbyCorollary7.1.7, Aut (E) = k. | F | 03/27/19 Note that in order to get equality in Proposition 7.4.1 it is necessary and su cient that f(x)hasallitsrootsinE,andthatithasnomultipleroots.
Corollary 7.1.10 Let E = F (↵) be a finite simple extension. Aut (E) = | F | [E : F ] i↵ f(x)=minF (↵) has no multiple roots, and E is the splitting field of f(x).
In Examples 7.1.1 we have
Aut (F ) =1 =[F : F ], | F | AutR(C) =2 =[C : R], | | 3 3 AutQ(Q(p2)) =1< 3=[Q(p2):Q], and | 3 | 3 Aut (Q(p2 ,!)) =6 =[Q(p2 ,!)) : Q]. | Q |
When equality holds in (7.1), we say that the extension E/F has enough automorphisms. When an extension has enough automorphisms, the auto- morphism group carries information about the extension. See, for example, Theorem 7.4.10, The Fundamental Theorem of Galois Theory.
The two conditions needed for a (simple) extension to have enough auto- morphims, are:
The minimal polynomial min (↵)hasnomultipleroots, • F the field E is the splitting field of that minimal polynomial. •
We are now going to study these two properties separately, and then jointly.
Version 2019.5.9 86 CHAPTER 7. GALOIS THEORY separable 7.2 Separable Extensions separable separable inseparable Definition 7.2.1. An irreducible polynomial p(x) F [x]issaidtobe separable over F ,ifithasnomultipleroots.Anon-constant,polynomial2 f(x) F [x]isseparable over F it each of its irreducible factors is separable over F2.ForanextensionE/F and ↵ E,wesaythat↵ is separable over 2 F if its minimal polynomial minF (↵) is separable over F ,andwesaythat E is separable over F if every element of E is separable over F . When a polynomial, an extension, or an element is not separable over F ,wesaythat it is inseparable over F .
Proposition 7.2.1 An irreducible polynomial p(x) F [x] is separable i↵ 2 p0(x) =0. 6
Proof. Assume p0(x) =0.Sincep(x)isirreducibleanddeg(p0(x)) < deg(p(x)), 6 no root of p(x)canbearootofp0(x). By Proposition 6.6.2, p(x)hasnomul- tiple roots, hence it is separable. Conversely, assume p(x)hasnomultipleroots,thenforanyroot↵ of p(x) we must have p0(u) =0,andthereforep0(x) =0. 6 6 Corollary 7.2.2 In characteristic 0, all irreducible polynomials, all non- constant polynomials, all algebraic elements, and all algebraic extensions are separable.
Corollary 7.2.3 Let char(F )=p, and p(x) F [x] be irreducible. Then p(x) is inseparable i↵ p(x) is of the form g(xp)2for some g(x) F [x]. 2
Proof.
Proposition 7.2.4 For any n N, the finite field Fpn is separable over Zp. 2
pn Proof. Note that the polynomial f(x)=x x is separable since f 0(x)= 1, and f(x) has no multiple roots. Any element of ↵ Fpn is a root of f(x), so 2 minZp (↵)isafactoroff(x), has no multiple roots, hence it is separable. Proposition 7.2.5 Let K/E/F be an extension tower. If K is separable over F then K is separable over E and E is separable over F . 7.2. SEPARABLE EXTENSIONS 87
Proof. rational functions field of rational functions The converse of Proposition 7.2.5, is true, but we will not prove it.
03/29/19 Proposition 7.2.6 Every algebraic extension of a finite field is separable.
Proof. Let F be a finite field of characteristic p,andE an algebraic extension of F .Let↵ E.Since↵ is algebraic over F ,thefieldF (↵)isafinite extension of F2, hence a finite field. From Proposition 7.2.4 we know that
minZp (↵) is separable. But minF (↵)isafactorofminZp (↵), and therefore it is also separable.
From Proposition 7.2.6, and Corollary 7.2.2, we see that the only way to find an inseparable extension, is by looking at infinite fields of prime characteristic. We now define one such field. Definition 7.2.2. Let F be a field, and F [x]theringofpolynomialswith coe cients in F .SinceF [x] is an integral domain, it has a field of quotients. We denote by F (x)thefieldofquotientsofF [x]. Each element in F (x)is f(x) of the from with f(x),g(x) F [x], and g(x) =0.Theelementsof g(x) 2 6 F (x)arecalledrational functions over F , and we call F (x)thefield of rational functions over F .
Every polynomial is a rational function, and every rational function is a quotient of two polynomials. Sometimes we will use a variable other than x for the rational functions, so that we can consider polynomials on x over a field of rational functions. Lemma 7.2.7 Let 0 = h(x) F (x) be a rational function over F . The 6 2 f(x) integer deg(f(x)) deg(g(x)) where h(x)= , with 0 = f(x),g(x) F [x] g(x) 6 2 is well-defined.
Proof. Use the fact that for non-zero polynomials, the degree of a product is the sum of the degrees. Definition 7.2.3. For 0 = h(x) F (x)arationalfunctionoverF ,we 6 2 denote by deg(h(x)) the integer deg(f(x)) deg(g(x)) where h(x)= f(x) , g(x) with 0 = f(x),g(x) F [x]. 6 2 Version 2019.5.9 88 CHAPTER 7. GALOIS THEORY
Example 7.2.1. Let F = Z2(t)bethefieldofrationalfunctionsoverZ2,on the variable t.Considerthepolynomialp(x)=x2 t F [x]. Note that for f(t) 2 any 0 = h(t)= F ,wehave 6 g(t) 2
f 2(t) deg(h2(t)) = deg g2(t) ✓ ◆ =deg(f 2(t)) deg(g2(t)) =2deg(f(t)) 2deg(g(t)) =2(deg(f(t)) deg(g(t))) =2deg(h(t)) is an even integer. In order to have h(t)asarootofx2 t we would need h2(t) t =0,i.e. h2(t)=t,butthisisimpossiblesincethelefthandside, as a rational function on t,hasevendegree,andtherighthandsidehas degree 1. Being quadratic, with no roots in F ,thepolynomialp(x)=x2 t is irreducible over F .Let↵ be a root of this polynomial in some extension E of F .Wehave↵2 = t,andtherefore(x ↵)2 = x2 ↵2 = x2 t.The irreducible polynomial p(x)=x2 t has ↵ as a root of multiplicity 2, hence it is inseparable. The element ↵ is inseparable over F ,andtheextensionE is inseparable over F .
The condition on the extension E/F to be simple, in Corollary 7.1.10, is somewhat restrictive, and it will be relaxed (See Proposition 7.4.1 below). However, the next theorem gives us weak conditions su cient for an extension to be simple. Theorem 7.2.8 [Primitive Element Theorem] If E/F is a finite, sep- arable extension, then it is a simple extension, that is, there is u E such that E = F (u). 2
As immediate consequence of this Theorem together with Chorollary 7.2.2 and Proposition 7.2.6, we get the following corollaries. Corollary 7.2.9 Every finite extension in characteristic 0 is simple. Corollary 7.2.10 Every finite extension of a finite field is simple. 7.2. SEPARABLE EXTENSIONS 89
The statement in Corollary 7.2.10 is something we have already proved in Corollary 6.6.10.
04/01/19 Before we prove the Primitive Element Theorem, we prove the following lemma.
Lemma 7.2.11 Let F be an infinite field, E and extension of F and , E, both algebraic over F , and separable over F . There is such that2 F ( , )=F ( ).
Proof. Let p(x)=minF ( ), and q(x)=minF ( ). Let K be a spliting field of p(x) q(x)thatcontainsF ( , ). Let = ,..., be the distinct roots · 1 m of p(x)inK,and = 1,..., n the distinct roots of q(x)inK.Since is separable over Fq(x)hasnomultipleroots,i.e.q(x)=(x 1) (x n). ··· There are finitely many elements i with i =1,...,m and j =2,...,n. j Since F is infinite, we may choose a F di↵erent from all those quotients. 2 Let = a .ClearlyF ( ) F ( , ). We want to show the F ( , ) other inclusion. Consider the tower F ( , )/F ( )/F .Letr(x)= min ( ). We have r(x)isafactorofq(x), since q( )=0,sothe F ( ) F ( ) roots of r(x) are among 1,..., n,andinclude 1.Considernow f(x)=p( + ax) F ( )[x]. We have f( )=p( + a )=p( )=0, F so r(x)isafactorof2 f(x), and every root of r(x)isarootoff(x). For any i =1,...,m,andj =2,...,n,wehavea( ) =( ), so j 6 i + a j =( a )+a j = a( j) = ( i)= i,andtherefore, for j =2,...,n , f( )=p( + a ) = 0.6 So, among all ,theonlyrootof j j 6 j f(x)is 1 = .andthereforer(x)=x ,whichyieldsdegF ( )( )=1,and F ( ). Since = + a ,wealsoget F ( ), completing the proof that F (2 , ) F ( ). 2
E Proof of Primitive Element Theorem. We consider two cases, depending on whether F is finite or not. F If F is finite of characteristic p,thenE is also finite of characteristic p. By Corollary 6.6.10, E is a simple extension of Zp,i.e. E = Zp(u)for Zp some u E.ItfollowsthatE = F (u). Now, consider2 the case when F is infinite. By Proposition 6.4.9, E = F (↵1,...,↵n)forsome↵1,...,↵n E,algebraicoverF . Use induction on n and Lemma 7.2.11. 2
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We now explore some additional examples of the automorphism group of an extension.
2 Example 7.3.1. For ! =cis(2⇡/3) we have minQ(!)=x + x +1,since !3 =1andx3 1=(x 1)(x2 + x +1). Therootsofx2 + x +1are! and !2,bothofwhichareinQ(!), so we have two automorphisms of Q(!), which depend on where they map ! to. There is the identity I : ! ! and 2 2 4 7! ⌧ : ! ! .Clearly,⌧ : ! ! = ! is the identity, and AutQ(Q(!)) = I,⌧ 7!is the cyclic group of order7! 2. { }
We generalize this example in the following proposition. 04/02/19 Problem Set 5
Proposition 7.3.1 Let p be a prime number, and let ⇠p =cis(2⇡/p). Then p 1 p 2 min (⇠p)=x + x + + x +1, and Aut (Q(⇠p)) Up. Q ··· Q ⇡
Proof. ⇠ satisfies ⇠p =1,soitisarootofxp 1. This polynomial factors as p p
p p 1 p 2 x 1=(x 1)(x + x + + x +1), (7.2) ···
p 1 p 2 and since ⇠p =1,itisarootof'p(x):=x +x + +x+1. Wewantto 6 ···p show that 'p(x)isirreducibleoverQ. Note that (x +1) 1=x '(x +1), and we have ·
x '(x +1) = (x +1)p 1 · p p = xi 1 i i=0 ✓ ◆ ! pX p = xi i i=1 ✓ ◆ X p p i 1 = x x , · i i=1 X ✓ ◆ 7.3. CYCLOTOMIC AND FINITE FIELD EXTENSIONS 91
so, cancelling the common factor x,weget cyclotimic polynomial p p i 1 ' (x +1) = x p i i=1 X ✓ ◆ p p p p 2 p p 1 = 1 + 2 x + + p 1 x + p x ··· p The leading coe cient is =1,andfori =1,...,p 1thebinomial p ✓ ◆ p p coe cient is divisible by p.Theconstantterm, = p is divisible i 1 ✓ ◆ ✓ ◆ by p,butnotbyp2.ByProposition4.6.2[EisensteinCriterion],weget that '(x +1)isirreducibleoverQ.Sinceanyfactorizationof'(x)yieldsa factorization of '(x+1), and viceversa, it follows that '(x)isalsoirreducible over Q. 04/03/19 Let’s consider now the automorphisms of the extension Q(⇠p)/Q.Theroots p k of x 1are⇠p for k =0,...,p 1. From Equation 7.2 we see that the k roots of 'p(x)are⇠p for k =1,...,p 1. Each automorphism of Q(⇠p)is completely determine from where it maps ⇠p,andithastomapittooneof k the roots of 'p(x), i.e. some ⇠p for k =1,...,p 1. For each such k,let k k(⇠p)=⇠p . That means AutQ((Q(⇠p))) = 1,..., p 1 . Note that { } j j k j kj k j(⇠p)= k(⇠p)= k(⇠p) =(⇠p ) = ⇠p = kj(⇠p),
so, we have k j = kj,andthemap
: Up AutQ(Q(⇠p)) k ! 7! k is an isomorphism.
Definition 7.3.1. For a prime p,thepolynomial
p 1 p 2 ' (x):=x + x + + x +1 p ··· is called the p-th cyclotimic polynomial.
There is a generalization of cyclotomic polynomials from 'p(x)whenp is aprime,to'n(x)foranynaturalnumbern.
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Proposition 7.3.2 Let p be a prime number, n 1, and E = Fpn . The au- n tomorphism group AutFp (Fp ) is cyclic of order n, generated by the Frobenius automorphism p.
Proof. Since E is finite and the Frobenius endomorphism : E E is p ! n injective and fixes the prime subfield, we have p AutFp (Fp ). The elements pn n 2n of E all satisfy x = x,i.e p (x)=x,so p = IE.SinceE/Fp is a simple extension, for a primitive element u E of this extension, n is the smallest integer such that upn = u,soitisalsothesmallestintegersuch2 n that p = IE,and p has order n.ButProposition7.1.9tellsusthat n n Aut p (Fp ) [E : Fp]=n. Therefore Aut p (Fp )iscyclicoforden, | F | F generated by the Frobenius automorphism p.
Example 7.3.2. Consider the extension Q(p2 , p3)/Q. Q(p2 , p3) The minimal polynomials are x2 2andx2 3, with roots p2and p3,respectively.For p2 Q(p2) Q(p3) there are two± choices,± and for p3therearealso two choices, for a total of 4 potential automor- Q phisms. So we get that Aut (Q(p2 , p3)) 4. | Q | (Note also, that [Q(p2 , p3):Q]=4)Thepotentialautomorphismsare given by I : p2 p2 : p2 p2 7! 1 7! p3 p3 p3 p3 7! 7! : p2 p2 : p2 p2 2 7! 3 7! p3 p3 p3 p3 7! 7!
Since Q(p2) is the splitting field of x2 2overQ,andQ(p2 , p3) isthe splitting field of x2 3overQ(p2). Corollary6.4.14,tellsusthatthethere are automorphisms doing the above. Moreover, notice that 1, 2, 3 have order 2. Therefore, AutQ(Q(p2 , p3))istheKlein4-group.
4 4 Example 7.3.3. The extension Q(p2)/Q has degree 4, since minQ(p2)= x4 2 (use Eisenstein’ criterion to check irreducibility). However, only two of the four roots, namely p4 2areintheextension,andtherefore 4 ± AutQ(Q(p2)) 2. Proposition 7.1.5 guarantees the existence of both auto- | 4 | 4 4 4 4 morphisms, (p2 p2),and(p2 p2 ). Thus, we have Aut (Q(p2)) 7! 7! Q 7.4. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 93
is cyclic of order 2. 4 The splitting field of x4 2 Q[x]isQ(p2 ,i), which has degree 8 over Q. In this case, there are four choices2 of where to map p4 2, i.e. p4 2 , ip4 2. There are two choices of where to map i,i.e. i,sincetheminimalpoly-± ± 4 ± nomial for i over Q(p2) is x2 +1. Thus,therearea totalof8 potential 4 automorphism of this extension. Therefore, AutQ(Q(p2 ,i)) 8. As in the previous example, Corollary 6.4.14, guarantees| the existence| of those 8 4 automorphisms, so AutQ(Q(p2 ,i)) =8.Wedescribethe8automorphisms by indicating where| they map p4 2and| i.
! 0 1 2 3 4 5 6 7 p4 2 p4 2 p4 2 ip4 2 ip4 2 p4 2 p4 2 ip4 2 ip4 2 i i ii ii ii i order 12 4 2 2 2 4 2
2 3 0 DD Note that 2 = 4,and 2 = 6.Moreover, 2 3 = 1, and 3 2 = 5,sothegroupisnotabelian.Therearetwo non-abelian groups of order 8: the dihedral group D4,andthe quaternion group Q8. Q8, however, has three elements of order 4 4, so by elimination we must have AutQ(Q(p2 ,i)) D4.We can do better. The four roots of x|4 2formasquareonthe⇡ 4 complex plane, and Aut (Q(p2 ,i)) act by symmetries on this | Q square. More precisely, 2 = ⇢ is a counterclockwise rotation by ⇡/2, 1 = H is a reflection with respect to the real axis, 5 = V is a reflection with respect to the imaginary axis, 3 = D is a reflection with respect to one diagonal, and 7 = D0 is a reflection with respect to the other diagonal.
04/05/19 Test 2
7.4 The Fundamental Theorem of Galois The- ory
04/08/19 The following proposition generalizes Proposition 7.1.9, by removing the hypothesis of simple extension. The first part is a weaker but useful version.
Version 2019.5.9 94 CHAPTER 7. GALOIS THEORY field of rational Proposition 7.4.1 1. Let E = F (↵ ,...,↵ ) be a finite extension of F . functions 1 n Let ri =degF (↵i).
Aut (E) r r r . | F | 1 · 2 ··· n 2. Let E/F be a finite extension. Then
Aut (E) [E : F ]. (7.3) | F |
Proof. (1) By Lemma 7.1.2, any Aut (E)iscompletelydeterminedby 2 F the choice of (↵1),..., (↵n). Let mi(x)=minF (↵i). Proposition 7.1.6 says that (↵i)isarootofmi(x), so, there are at most ri choices for (↵i). (2) Choose ↵1,...,↵n E,recursivelysuchthat↵i / F (↵1,...,↵i 1). Let 2 2 mi(x)=minF (↵1,...,↵i 1)(↵i), and ri =deg(mi). We have [F (↵1,...,↵i): F (↵1,...,↵i 1)] = degF (↵ ,...,↵ )(↵i)=ri,andthemultiplicativeproperty 1 i 1 of tower extensions yields
[E : F ]=r r r . 1 · 2 ··· n Let Aut (E), and let be the restriction of to F (↵ ,...,↵). Note 2 F i 1 i that 0 is the inclusion mape ◆F : F E,and n = Since all i : F (↵ ,...,↵) E agree with ,wegetthat! : F (↵ ,...,↵) E extends 1 i ! i 1 i ! i 1 : F (↵1,...,↵i 1) E.ByCorollary7.1.7thereareatmostri ways to ! do this extension. As is built from 0 = ◆F ,stepbystep,weendupwith at most r r r possible = Aut (F ). 1 · 2 ··· n n 2 E To see that Proposition 7.4.1.2 is indeed more general than Proposi- tion 7.1.9, we need an example of a finite extension that is not simple. The Primitive Element Theorem (7.2.8) tells us that any finite separable exten- sion is simple, so we need to look at inseparable extensions. Example 7.2.1 gives us an example of a finite inseparable extension. However, that exten- sion is simple. But we can extend the idea of that example, to create the example we need.
Example 7.4.1. Let R = F2[s, t]betheringofpolynomialsintwovariables s and t over F2. As R is an integral domain (it is, in fact, a UFD) it embeds in its field of quotients, that we denote F2(s, t), and call the field of rational functions in the two variables s and t.Letf(x)=x2 s, g(x)=x2 t F [x]. The same argument used in Example 7.2.1, shows that f(x)and g2(x)are 7.4. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 95
irreducible over F .Let↵ be a root of f(x)and arootofg(x). We can write ↵ = ps and = pt . F (↵)isspannedby 1,↵ over F ,soforany u F (↵), u = a + b↵ for some a, b F .Wehaveu{2 = a}2 + b2↵2 = a2 + b2s, 2 2 2 2 and therefore degt(u )iseven.Thus,u = t,and /F (↵). It follows that [F (↵, ):F (↵)] = 2 and [F (↵, ):F ]=4,6 with basis2 1,↵, ,↵ .Take v F (↵, ). We can write v = a + b↵ + c + d↵ ,with{ a, b, c, d } F .It follows2 that v2 = a2 + b2↵2 + c2 2 + d2↵2 2 = a2 + b2s + c2t + d22st F . 2 So, we have degF (v) 2, and therefore we cannot have F (↵, )=F (v). In other words, F (↵, )isnotasimpleextensionof F .
04/09/19 Theorem 7.4.2 [Dedekind-Artin Theorem] Let E be a field, G a finite sub- group of Aut(E). Let EG be the subfield of E fixed by G,
E = u E (u)=u for all G . G { 2 | 2 }
Then E/EG is a finite extension and [E : EG]= G . Moreover, AutEG (E)= G. | |
Proof. Let F = E .WehaveG Aut (E), so if E/F is finite, Propo- G F sition 7.4.1.2 yields G AutF (E) [E : F ]. It remains to show that E/F is indeed finite,| and| [E|: F ] |G.Letn = G .Bywayofcontradic- | | | | tion, assume [E : F ] >n.Thereareu0,u1,...,un E,distinct,suchthat I = u ,u ,...,u is linearly independent over F .Write2 G = ,..., { 0 1 n} { 1 n} with 1 =1E. Consider the n (n+1) matrix M =( i(uj)), with coe cients in E, ⇥ u0 u1 ... un 2(u0) 2(u1) ... 2(un) M = 2 . . . 3 . . . 6 7 6 n(u0) n(u1) ... n(un) 7 6 7 and the system of equations4 M X =0. 5 · u x + u x + + u x =0 0 0 1 1 ··· n n 2(u0)x0 + 2(u1)x1 + + 2(un)xn =0 . . ··· . . (7.4) . . . . (u )x + (u )x + + (u )x =0 n 0 0 n 1 1 ··· n n n Since there are more variables than equations, the system has a non-trivial solution↵ ¯ =(↵ ,↵ ,...,↵ ) En.Pick¯↵ non-trivial, with the smallest 0 1 n 2 Version 2019.5.9 96 CHAPTER 7. GALOIS THEORY normal extension number, (r +1) > 0, of non-zero entries. Reorder the elements of I (if needed) so that↵ ¯ =(↵0,↵1,...,↵r, 0,...,0), and ↵0,↵1,...,↵r =0.Since the set of solutions of (7.4) is a linear space over E,wemayalsochoose¯6 ↵, so that ↵0 =1.Thefirstoftheequationsin(7.4)yields
u ↵ + + u ↵ =0, 0 0 ··· r r and by the linear independence of the set I we must have that at least one of ↵0,...,↵r,say↵k,isnotinF .BydefinitionofF ,thismeansthatthere is ⌧ G such that ⌧(↵k) = ↵k.Thefactthat¯↵ is a solution of (7.4) means that2 for each i =1...n, 6
(u )↵ + (u )↵ + + (u )↵ =0. (7.5) i 0 0 i 1 1 ··· i r r Let = ⌧(↵ ), and = ↵ . Note that = ⌧(↵ )=⌧(1) = 1 = ↵ ,so j j j j j 0 0 0 0 = 0. Also, k = ↵k k = ↵k ⌧(↵k) =0.Let l = ⌧ i. Applying ⌧ to Equation (7.5) yields 6
(u ) + (u ) + + (u ) =0, (7.6) l 0 0 l 1 1 ··· l r r
As i ranges over G, l also ranges over all of G.So,foreachi =1,...n,
(u ) + (u ) + + (u ) =0, (7.7) i 0 0 i 1 1 ··· i r r and subtracting (7.7) from (7.5), we get
(u ) + (u ) + + (u ) =0. (7.8) i 0 0 i 1 1 ··· i r r
Since 0 =0and k =0,thisisanon-trivialsolutionto(7.4)withlessthan r + 1 non-zero entries,6 contradicting the choice of↵ ¯.
7.4.1 Galois Extensions
The property in Theorem 7.4.2 that [E : EG]= G is one several equivalent conditions thatr a finite extension may have. Proposition| | 7.4.3 below gives us a list of such equivalent properties. Before stating this proposition, we need to define several terms, and introduce some convenient notation. Definition 7.4.1. A finite extension E/F is said to be a normal extension, if E is the splitting field of a polynommial f(x) F [x]. 2 7.4. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 97
Definition 7.4.2. Let E be a field and G a subgroup of Aut(E). Let fixer subgroup Galois Connection F = E = a E (a)=a for all G , G { 2 | 2 } be the subfield of E fixed by G.WedefinetwomapsbetweenSub(G), the lattice of subgroups of G,andSubF (E), the lattice of intermediate fields of the extension E/F as follows. The maps go one in each direction, and both maps are denoted by ⇤.Thecontextwilltellwhichiswhich.
⇤ SubF (E) ) * Sub(G)(7.9) ⇤ For L Sub (E), let 2 F
L⇤ = G (a)=a, for all a L . { 2 | 2 } For H Sub(G), let 2
H⇤ = E = a E (a)=a, for all H . H { 2 | 2 }
Note that as defined F = G⇤.RecallthatweshowedinCorollary6.6.4,that EH ,thesubfieldfixedbyH,isindeedasubfieldofE.Itisobviousthatany H fixes the elements of F ,soE Sub (E). Similarly, it is easy to 2 H 2 F show that L⇤ is a subgroup of G.Itiscalledthefixer subgroup of L.(see Exercise 7.4.1 below) The pair of posets (SubF (E), Sub(G)), together with the maps just defined, is called a Galois Connection,asitsatisfiesthepropertiesinLemma7.4.4 below.
Exercise 7.4.1. Let E be a field, G a subgroup of Aut(E), F = EG,and L Sub (E). Show that L⇤ = Aut (E), and it is a subgroup of G. 2 F L
Proposition 7.4.3 Let E/F be a finite extension, and G = AutF (E). TFAE:
1. [E : F ]= G | | 2. E is the splitting field of a separable polynomial over F .
3. E is the splitting field of an irreducible separable polynomial over F .
4. The extension E/F is separable and normal.
Version 2019.5.9 98 CHAPTER 7. GALOIS THEORY
Galois extension 5. G⇤ = F , i.e. F ⇤⇤ = F Galois group
The proof appears below, after some examples, and lemmas.
Definition 7.4.3. Let E/F be a finite extension. We say that E/F is a (finite) Galois extension if it satisfies the properties in Proposition 7.4.3. The group AutF (E)iscalledtheGalois group of the extension, and is denoted GalF (E).
Examples 7.4.2. 1. In Examples 6.4.2 and 7.1.2.3, we have seen that 3 Q(p2 ,!)isthesplittingfieldofx3 2 Q[x]. Hence, it is a Galois 2 extension. It has degree 6, and Galois group isomorphic to D3.Onthe 3 3 other hand, the extension Q(p2)/Q has degree 3, but AutQ(Q(p2) is 3 trivial, so this extension is not Galois. It follows that Q(p2) cannot be the splitting field of any polynomial over Q.
2. Let E = Q(p2 , p3). Example 6.4.1 shows that the extension E/Q is a Galois extension, with Galois group isomorphic to K4,theKlein 4-group.
4 3. Let E = Q(p2). Example 7.3.3 shows that E/Q is not aGalois extension, as [E : Q]=4,but Aut (E) =2. | Q |
4 4. On the other hand, Q(p2 ,i)isthesplittingfieldofx4 2overQ, 4 and this polynomial is separable. Thus, Q(p2 ,i)/Q is a Galois exten- 4 sion, and therefore AutQ(Q(p2 ,i)) has order 8, as we have shown in Example 7.3.3.
5. Proposition 7.3.2 shows that Fpn /Fp is a Galois extension. Its Galois 04/10/19 group is cyclic of order n, generated by the Frobenius automorphism.
6. Proposition 7.3.1 shows that the extension Q(⇠p)/Q is a Galois exten- sion, with Galois group cyclic of order p 1. 7. The Dedkind-Artin theorem tells us that for G afinitesubgroupof Aut(E), the extension E/EG is a Galois extension, with Galois group G. 7.4. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 99
7.4.2 Galois Connection subgroup lattice partially ordered set Recall that given a group G,thesubgroup lattice of G as the set reflexive transivite Sub(G):= H H G anti-symmetric { | } lattice of all subgroups of G,orderedbyinclusion.Thisisanexampleofapartially meet join ordered set,posetforshort,sincethebinaryrelationofinclusionisreflexive, intermediate field transivite,andanti-symmetric.Itiscalledalattice as it has the following two lattice properties. Given H1,H2 Sub(G), there is a greatest lower bound of H1 Galois Connection and H in Sub(G), given by2 H H ,whichwecallthemeet of H and H . 2 1 \ 2 1 2 And a least upper bound of H1 and H2 in Sub(G), given by H1 H2,the join of H and H ,whichisthesubgroupgeneratedbytheunionH_ H . 1 2 1 [ 2 H H := H H H H := H H . 1 ^ 2 1 \ 2 1 _ 2 h 1 [ 2i
Given a field extension E/F,wedefineisasimilarway,theintermediate field lattice of the extension E/F,astheset
Sub (E):= L F L E F { | } of all subfields L of E that contain F .Itisorderedbyinclusion,andfor any L ,L Sub (E), there is a greatest lower bound in Sub (E)givenby 1 2 2 F F L1 L2, and a least upper bound given by L1 L2,thesubfieldofE generated by\ the union L L . _ 1 [ 2 L L := L L L L := L L . 1 ^ 2 1 \ 2 1 _ 2 h 1 [ 2i
The following lemma gives us properties of the maps in (7.9), that jus- tify calling them a Galois Connection.Inotherwords,aGaloisConnection consists of two posets, and two maps between them, satisfying the following properties.
Lemma 7.4.4 Let E be a field, G a subgroup of Aut(E), and F = EG. For any H, H ,H Sub(G), and any L, L ,L Sub (E) 1 2 2 1 2 2 F
1. If H H , then H⇤ H⇤. ( ⇤ is order reversing) 1 2 2 1 2. If L L , then L⇤ L⇤. ( ⇤ is order reversing) 1 2 2 1 Version 2019.5.9 100 CHAPTER 7. GALOIS THEORY
3. H H⇤⇤ (1 ⇤⇤)
4. L L⇤⇤ (1 ⇤⇤) Exercise 7.4.2. Prove Lemma 7.4.4.
Corollary 7.4.5 Let E be a field, G a subgroup of Aut(E), and F = EG. For any H Sub(G), and any L Sub (E) 2 2 F
1. H⇤⇤⇤ = H⇤
2. L⇤⇤⇤ = L⇤
Proof. Let H Sub(G). By part 3 of the lemma, we have H H⇤⇤,and 2 applying part 1, we get H⇤⇤⇤ H⇤.SinceH⇤ Sub (E), by part 4 we have 2 F H⇤ H⇤⇤⇤. Asimilarargumentworksfor L Sub (E). 2 F We often refer to the properties in Corollary 7.4.5 as the “3 = 1” property.
Lemma 7.4.6 Let E be the splitting field of a separable polynomial f(x) 04/12/19 F [x]. Given ' : F E, such that ' fixes the coe cients of f(x), there are2 [E : F ] di↵erent extensions! of the homomorphism ' to 'ˆ : E E. !
'ˆ Proof. By induction on n =[E : F ]. The statement is clear when E / E n =1.Letn>1. Let u E F be a root of f(x)andm(x)= = F 2 '˜ minF (u). Since m(x)isanirreduciblefactoroff(x), it is separa- ble, and it splits in E.ByScholium7.1.8,therearer =deg(m(x)) di↵erent extensions of ' to' ˜ : F (u) E that extend '.The F (u) ' extension E/F(u)hasdegree[E : F (u)]! = n/r < n,andE is the splitting field of the separable polynomial f(x) F (u)[x]. More- ' ' 2 f x over, ˜ being an extension of ,fixesthecoe cientsof ( ). By F induction, there are n/r di↵erent extensions of' ˜ to' ˆ : E E. Therefore, there are r (n/r)=n di↵erent extensions of ' to!' ˆ : E E. · !
Proof of Proposition 7.4.3. Recall that we have E/F afiniteextension,andG = AutF (E)=F ⇤. 7.4. THE FUNDAMENTAL THEOREM OF GALOIS THEORY 101
(1) (5) Assume [E : F ]= G . Note that F ⇤⇤ = G⇤ = E .SinceF F ⇤⇤, ) | | G we may consider the tower E/EG/F ,andhave
[E : F ]=[E : EG][EG : F ].
By assumption, [E : F ]= G , and, using the Dedekind–Artin Theorem, [E : E ]= G .Therefore[E| :|F ]=1,andE = F ,asdesired. G | | G G
(5) (4 and 3) Assume EG = F .WefirstprovethatE/F is separable. Write) G = 1, ,..., .Letu E,andconsidertheelements { 2 n} 2
u, 2(u),..., n(u)(7.10) in E.Thislistmaycontainrepetitions.Letu, u2,...,ur be the distinct elements in (7.10). Consider the polynomial
f(x)=(x u)(x u ) (x u ). 2 ··· r Each ⌧ G permutes the list (7.10), and therefore it also permutes the 2 elements u, u2,...,ur,andthefactorsoff(x). It follows that when we apply ⌧ to f(x), its coe cients are unchanged. In other words, the coe cients of f(x)areinE = F ,andf(x) F [x]. Since f(x)hasnomultipleroots, G 2 it follows that minF (u), which is a factor of f(x), is separable, hence u is separable over F .Moreover,minF (x)splitsinE.Sinceu E was arbitrary, we have E/F is separable. 2 Now, by the Primitive Element Theorem, there is u E such that E = F (u). 2 Since minF (u)splitsinE, E is the splitting field of minF (u), an irreducible separable polynomial in F [x], hence (3), and E/F is a normal extension, hence (4).
(4) (2) Assume E/F is normal and separable. By normality, let f(x) F [x] be) such that E is the splitting field of f(x). To see that f(x)isaseparable2 polynomial, let q(x) F [x]beanymonicirreduciblefactoroff(x), and let 2 u E be a root of q(x). Then q(x) = minF (u), and by separability of E/F, u 2is separable over F ,i.e.q(x)isseparableoverF .
(3) (2) is clear. ) (2) (1) Assume E is the splitting field of a separable polynomial f(x) F [x)], and let n =[E : F ]. Apply Lemma 7.4.6 to the inclusion map i2:
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