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 = {m ∈ L | h(m) = m for all h ∈ H}.

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 ∈ V and α, α0 ∈ 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 ∈ 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 {ai}i=1 form √ √ n n L = KA := K( ai; i = 1, 2, . . . , s) = K( a; a ∈ A).

Deﬁne the bilinear pairing: G × A −→ µn by setting (for any g ∈ G and a ∈ A) √ √ n n hg, ai := g( a)/ a ∈ 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 := |G| = [L : K].

∗ ∗ Theorem 1 If β ∈ L has the property that that NL/K (β) = 1 then there exists an α ∈ 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 ∈ Gal(L/K) and x ∈ L , xγ will denote γ(x). You’ll see why that’s convenient. Also if γ, γ0 ∈ 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 ∈ 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 ∈ µn be a primitive n-th root of unity. Since ζn ∈ K, we have that NL/K (ζn) = n ∗ ζn = 1 so by HT90 there is an α ∈ 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.