Bourbaki

Exercise 19, § 1

Commutative Algebra, V, 358

(a) Let A be an integrally closed domain, K its field of fractions, L a separable algebraic extension of K, x an element of L integral over A, K' the field K(x) ⊂ L and F the minimal polynomial of x over K. If x is of degree n over K, show that the integral closure of A in K' is contained in the A-module generated by the elements 1/F'(x), x/F'(x), ..., x^{n-1}/F'(x) of K'. (If z ∈ K' is integral over A, write

zF'(x) = g(x), where g(X) = b_0 + b_1X + ... + b_{n-1}X^{n-1} ∈ K[X].

Show with the aid of Lagrange's interpolation formula that

g(X)=Tr⁡K′[X]/K[X](z⋅F(X)X−x). g(X) = \operatorname{Tr}_{K'[X]/K[X]} \left( z \cdot \frac{F(X)}{X - x} \right).

(b) Let AA be an integrally closed domain, KK its field of fractions and pp the characteristic exponent of KK. Suppose that the subring A1/pA^{1/p} of K1/pK^{1/p} consisting of the pp-th roots of the elements of AA is a finitely generated AA-module. Show that for every finite extension EE of KK the integral closure of AA in EE is a finitely generated AA-module (reduce to the case where EE is a quasi-Galois extension).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.