Bourbaki

Exercise 18, § 8

Commutative Algebra, VI, 468

Bourbaki marks this one as harder than the rest.

Let K be a Henselian field with a valuation v, k the residue field of v and suppose that every finite algebraic extension of k is cyclic over k (which is for example the case when k is finite). Let A be the ring of v and f ∈A[X]\in A[X] a monic polynomial such that, iff (X)=(X−α1)…(X−an)(X) = (X - \alpha_1) \ldots (X - a_n), where the αi\alpha_i belong to an algebraic closure of K, the element D=∏i<j(αi−αj)2D = \prod_{i < j} (\alpha_i - \alpha_j)^2 of A ("discriminant" off) is such that v(D)=0v(D) = 0. Let f~j\tilde{f}_j (1≤j≤s1 \leq j \leq s) be the irreducible factors of the canonical image f~\tilde{f} of f in k[X]k[X] and let rjr_j be the degree of f~j\tilde{f}_j. Show that the Galois over K of the splitting field of f, considered as a group of permutations of the a, is generated by a permutation σ\sigma which decomposes into cycles of respective lengths r1,r2,…,rsr_1, r_2, \ldots, r_s (observe that there exists only a single extension of k of given degree, up to a k-isomorphism). Deduce that, for D to be a square in A, it is necessary and sufficient that n−sn - s be even ("Stickelberger's Theorem"; examine under what condition D is invariant under σ\sigma).

Solution

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