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Notes on Galois Theory II

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Notes on Galois Theory II 2 The isomorphism extension theorem We begin by proving the converse to Lemma 1.7 in a special case. Suppose that E = F (α) is a simple extension of F and let f = irr(α; F; x). If : F ! K is a homomorphism, L is an extension field of K, and ': E ! L is an extension of , the '(α) is a root of (f). The following is the converse to this statement. Lemma 2.1. Let F be a field, let E = F (α) be a simple extension of F , where α is algebraic over F and f = irr(α; F; x), let : F ! K be a homo- morphism from F to a field K, and let L be an extension of K. If β 2 L is a root of (f), then there is a unique extension of to a homomorphism ': E ! L such that '(α) = β. Hence there is a bijection from the set of homomorphisms ': E ! L such that '(a) = (a) for all a 2 F to the set of roots of the polynomial (f) in L, where (f) 2 K[x] is the polynomial obtained by applying the homomorphism to coefficients of f. Proof. Let β 2 L be a root of (f). We know by basic field theory that there is an isomorphism σ : F (α) =∼ F [x]=(f) with the property that σ(a) = a + (f) for a 2 F and σ(α) = x + (f). Let evβ ◦ be the homomorphism F [x] ! L defined as follows: given a polynomial g 2 F [x], let (as above) (g) be the polynomial obtained by applying to the coefficients of g, and let evβ ◦ (g) = (g)(β) = evβ( (g) be the evaluation of (g) at β. Then evβ ◦ is a homomorphism from F [x] to K. For a 2 F , evβ ◦ (a) = (a), and evβ ◦ (x) = β. Moreover f 2 Ker evβ ◦ , since (f)(β) = 0 by hypothesis. Thus (f) ⊆ Ker evβ ◦ and hence (f) = Ker evβ ◦ since (f) is a maximal ideal and evβ ◦ is not the trivial homomorphism. Then there is an induced homomorphism e: F [x]=(f) ! L. Let ' be the induced homomorphism e ◦ σ : F (α) ! L. It is easily checked to satisfy: '(a) = (a) for all a 2 F and '(α) = β. 9 Next we claim that ' is uniquely specified by the conditions '(a) = (a) for all a 2 F and '(α) = β. In fact, every element of E = F (α) can be PN i written as i=0 aiα for some ai 2 N. Then N N N X i X i X i '( aiα ) = '(ai)'(α) = (ai)β : i=0 i=0 i=0 Thus ' is uniquely specified by the conditions above. In summary, then, every extension ' of satisfies: '(α) is a root of (f), ' is uniquely deter- mined by the value '(α) 2 L, and all possible roots of (f) in L arise as '(α) for some extension ' of . Thus the function ' 7! '(α) is a function from the set of extensions ' of to the set of roots of (f) in L. This function is injective (by the uniqueness statement) and surjective (by the existence statement), and thus defines the bijection in the second paragraph of the statement of the lemma. Corollary 2.2. Let E be a finite extension of a field F , and suppose that E = F (α) for some α 2 E, i.e. E is a simple extension of F . Let K be a field and let : F ! K be a homomorphism. Then: (i) For every extension L of K, there exist at most [E : F ] homomor- phisms ': E ! L extending , i.e. such that '(α) = (α) for all α 2 F . (ii) There exists an extension field L of K and a homomorphism ': E ! L extending . (iii) If F has characteristic zero (or F is finite or more generally perfect), then there exists an extension field L of K such that there are exactly [E : F ] homomorphisms ': E ! L extending . Proof. Let n = deg f = [E : F ]. Then deg (f) = n as well. Lemma 2.1 implies that the extensions of to a homomorphism ': F (α) ! L are in one-to-one correspondence with the β 2 K such that β is a root of (f), where f = irr(α; F; x). In this case, since (f) has at most n = [E : F ] roots in any extension field L, there are at most n extensions of , proving (i). To see (ii), choose an extension field L of K such that (f) has a root β in L. Thus there will be at least one homomorphism ': F (α) ! L extending . To see (iii), choose an extension field L of K such that (f) factors into a product of linear factors in L. Under the assumption that the characteristic of F is zero, or F is finite or perfect, the irreducible polynomial f 2 F [x] 10 has no multiple roots in any extension field, and the same will be true of the polynomial (f) 2 (F )[x], where (F ) is the image of F in K, since (f) is also irreducible. Thus there are n distinct roots of (f) in L, and hence n different extensions of to a homomorphism ': F (α) ! L. The situation of fields in the second and third statements of the corollary can be summarized by the following diagram: EL F / K Let us give some examples to show how one can use Lemma 2.1, espe- cially in case the homomorphism is not the identity: p Examplep p 2.3. (1) Consider the sequence of extensions Q ≤ Q( 2) ≤ Q(p2; 3). As we have seen, therep are two differentp automorphisms of Q( 2), Id and σ, where σ(a +p b 2) = a − b 2. We have seen that 2 f = x − 3 is irreducible in Q( 2)[x]. Since in fact f 2 Q[x], σ(pf) = f, and clearly Id(f) = f. In particular, thep roots of σp(f) = f arep ±p 3. Ap- plying Lemma 2.1 to the case F = Q( 2), E = F ( 3) = Q( 2; 3) = K, and = Id or = σ, we see that there are two extensions of Id to a homomorphismp (necessarilyp an automorphism) ': E ! E. Onep of thesep satisfies: '( 3) = 3, hence ' = Id, and the other satisfies '( 3) = − 3, hence ' = σ2 in the notation of (4) of Example 1.11. Likewise, there are two extensionsp p of σ to an automorphism ': E ! E. One ofp these satis-p fies: '( 3) = 3), hence ' = σ1, and the other satisfies '( 3) = − 3), hence ' = σ3 inp thep notation of (4) of Example 1.11. In particular, we see that Gal(Q( 2; 3)=Q) has order 4, giving another argument for (4) of Example ??. p p 3 3 (2) Taking F = Q, E = Q( 2), and K = Q( 2;!), we see that there are three injective homomorphisms fromp E to K since therep arep three rootsp in 3 3 3 3 2 3 K of the polynomial x − 2 = irr( 2; Q; x), namely 2, ! 2,p and ! 2. 3 On the other hand, consider also the sequence Q ≤ Q(!) ≤pQ( 2;!).p As 3 3 3 we have seen,p if the roots of x − 2 in C are labeled as α1 = 2, α2 = ! 2, 2 3 and α3 = ! 2 and σ is complex conjugation, then σ corresponds to the 3 permutation (23). We claim that f = x − 2 is irreducible in Q(!). In fact, since deg f = 3, f is reducible in Q(!) () there exists a root α of f 11 in Q(!). But then Q ≤ Q(α) ≤ Q(!) and we would have 3 = [Q(α): Q] 3 dividing 2 = [Q(!): Q], which is impossible.p Hence x − 2 is irreducible 3 in pQ(!)[x]. (Alternatively, note that !2 = Q( 2) since ! is not real but 3 Q( 2) ≤ R, hence p3 p3 p3 p3 [Q( 2;!): Q] = [Q( 2;!): Q( 2)][Q( 2) : Q] = 6 p3 = [Q( 2;!): Q(!)][Q(!): Q]; p 3 and so [Q( 2;!): Q(!)] = 3.) p 3 Considering the simple extension K = Q( 2;!) of Q(!), we see that the homomorphisms of K into K (necessarily automorphisms) which are the identity on Q(!), i.e. the elements of Gal(K=Q(!), correspond top the roots 3 3 of xp − 2 in K. Thus for example, therep is an automorphismp ρ: Q( 2;!) ! 3 3 3 Q( 2;!) such that ρ(!) = ! and ρ( 2) = ! 2. This completely specifies ρ. For example, the above says that ρ(α1) = α2. Also, p p p p 3 3 3 2 3 ρ(α2) = ρ(! 2) = ρ(!)ρ( 2) = ! · ! 2 = ! 2 = α3: Similarlyp ρ(α3) = α1. So ρ corresponds to the permutation (123). Then 3 Gal(Q( 2;!)=Q) is isomorphic to a subgroup of S3 containing a 2-cycle and a 3-cycle and hence is isomorphic to S3. p p p 4 4 4 (3) Considerp the case of Gal(p Q( 2; i)=Q), with β1 p= 2, β2 = i 2, 4 4 4 β3 = − 2, and β4 = −i 2. Then if ' 2 Gal(Q( 2; i)=Q), it follows that '(β1)p = βk for some k, 1 ≤ k ≤ 4 and '(i) = ±i. In particular 4 #(Gal(Qp( 2; i)=Q)) ≤ 8. As in (2), complex conjugation σ is an element of 4 4 Gal(Q( 2; i)=Q) corresponding to (24) 2 S4. Next we claim that x − 2 is 4 irreducible in Q(i). In fact, there is no root of x − 2 in Q(i) by inspection 4 (the βi are not elements of Q(i)) or because x − 2 is irreducible in Q[x] 4 4 and 4 = deg(x − 2 does not divide 2 = [Q(i): Q].
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