Kummer Theory April 13, 2010 1 HOMEWORK DUE APRIL 22: Exercise 1 Please hand in (to me in Tuesday's class) correct answers to all the problems you got wrong in the April 8 exam. Exercise 2 1. Let L=K be a Galois extension with Galois group G = Gal(L=K). Let H ⊂ G be a subgroup of G. Let M := LH be the subfield of L fixed by H; i.e. M = fm 2 L j h(m) = m for all h 2 Hg: Show that M=K is Galois if and only if H is normal; in G and if M=K is Galois then Gal(M=K) = G=H. 2. Let q be a power of prime number, k := Fq a field of cardinality q, n > 1 an integer and V a vector space over k of dimension n. Aside: Recall that Autk(V ), the group of automorphisms of the k vector space V , is iso- morphic to GLn(k) and for each k-basis of V we may write down an explicit isomorphism ∼ Autk(V ) = GLn(k): Choosing such a basis we will identify Autk(V ) = GLn(k): (You needn't prove what we have said in this little \Aside.") Now consider the vector space V as a finite (abelian) group|forgetting scalar multiplication by k|and consider the semi-direct product G := V ×~ GLn(k) where the action is the natural action described above. That is, for v; v0 2 V and α; α0 2 GLn(k) the multiplication law is (v; α) · (v0; α0) = (v + α(v0); α · α0): 1 Show that for every proper sub-vector space Vo ⊂ V (\proper" means that Vo is neither trivial nor V ) we have an example of a sequence of finite groups: Vo ⊂ V ⊂ G where Vo is normal in V and V is normal in G and yet Vo is not normal; in G. 3. Show that the following is an example of the above construction: the group G is S4, the vector space V is the subgroup consisting of the identity element and all products of two disjoint transpositions; i.e., (12)(34), (13)(24), and (14)(23). 4. Find a sequence of fields K ⊂ M ⊂ L such that M=K is Galois and L=M is Galois but L=K is not Galois. NOTE: If you remember the above type of example you will avoid falling into \Trap # 1" for students of Galois theory. Exercise 3 Artin Page 581 (Kummer Extensions) Exercise 6. Reading: Artin 14.7 2 Beginning Kummer Theory Let K be a field. Let n be a positive integer prime to the characteristic of K. Hypothesis: K ∗ contains µn := µn(K¯ ) all the n-th roots of unity. For a 2 K form the splitting field L=K of the polynomial f(t) := tn − a. Note that L=K really depends only on a modulo n-th powers, i.e., on the image of a in K∗=(K∗)n. Let G := Gal(L=K) and note that we have a canonical injection G,! µn: More generally, let A ⊂ K∗=(K∗)n be a finitely generated abelian subgroup (necessarily of exponent i=s n) and for a generating set of elements faigi=1 form p p n n L = KA := K( ai; i = 1; 2; : : : ; s) = K( a; a 2 A): Define the bilinear pairing: G × A −! µn by setting (for any g 2 G and a 2 A) p p n n hg; ai := g( a)= a 2 µn: Proposition 1 This is a nondegenerate bilinear pairing. 2 3 Artin's Theorem on Independence of characters 4 HT90 Let L=K be a cyclic Galois extension with g : L ! L a generator of the Galois group G = Gal(L=K). Let n := jGj = [L : K]. ∗ ∗ Theorem 1 If β 2 L has the property that that NL=K (β) = 1 then there exists an α 2 L such that β = g(α)/α: Proof: People who give this proof like to use exponential notation for the action of Gal(L=K) on L, so if γ 2 Gal(L=K) and x 2 L , xγ will denote γ(x). You'll see why that's convenient. Also if γ; γ0 2 Gal(L=K) we write 0 xγ+γ = xγ · xγ: OK, we're ready to go. Consider the following operator on L: 2 2 n−2 1 + βg + β1+gg2 + ··· + β1+g+g g3 + ··· + β1+g+g +···+g gn−1: By Artin's theorem (on independence of characters) this is not identically zero. So there is a θ 2 L such that 2 2 n−2 α := θ + βg(θ) + β1+gg2(θ) + ··· + β1+g+g g3(θ) + ··· + β1+g+g +···+g gn−1(θ) doesn't vanish. Check that β · αg = α to conclude the proof! 5 Continuation of Kummer Theory for cyclic extensions Theorem 2 Let K be a field. Let n be a positive integer prime to the characteristic of K. Assume that K contains µn := µn(K¯ ) all the n-th roots of unity. Any cyclic Galois extension L=K of degree n is obtained by the extraction of an n-th root of an element in K. Proof: Let ζn 2 µn be a primitive n-th root of unity. Since ζn 2 K, we have that NL=K (ζn) = n ∗ ζn = 1 so by HT90 there is an α 2 L such that ζn = g(α)/α or equivalently, g(α) = ζn · α: n Put a := NL=K (α). Then the minimal polynomial satisfied by α over K is X − a which tells us that α is a primitive element for L=K and that the conclusion of our theorem holds. 3 6 Continuation of Kummer Theory for all abelian extensions of exponent n Theorem 3 Let K be a field. Let n be a positive integer prime to the characteristic of K. Assume that K contains µn := µn(K¯ ) all the n-th roots of unity. Any finite Galois abelian field extension L=K of exponent n is of the form L = LA (as in section 2) for a unique finite subgroup A ⊂ K∗=(K∗)n. In particular, all such fields are generated over K by the extraction of (finitely many) n-th roots of elements in K. 4.