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Chapter 7 Galois Theory 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. 81 82 CHAPTER 7. GALOIS THEORY 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 β E be a root of minF (↵). There is a F -homomorphism (i.e. it fixes every a2 F ) ' : F (↵) E, such that '(↵)=β. 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 γ : Q(p2) Q(p2) ! a + bp2 a bp2 7! − Thus Aut (Q(p2))= 1,γ . 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 Version 2019.5.9 84 CHAPTER 7. GALOIS THEORY 3 extension Q(p2 ,!)/Q,namely: 3 3 3 3 I : p2 p2 σ1 : p2 p2 ! 7! ! ! 7! !2 7! 7! 3 3 3 3 ⇢ : p2 !p2 σ2 : p2 !p2 ! 7! ! ! 7! !2 7! 7! 2 3 2 3 3 2 3 ⇢ : p2 ! p2 σ3 : 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,⇢,⇢ ,σ1,σ2,σ3 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 sufficient 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.
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