Bourbaki

Exercise 16, § 2

Commutative Algebra, V, 366

Let AA be an integral domain, KK its field of fractions, K′K' a finite extension of KK of degree nn and A′A' a subring of K′K' containing AA, with K′K' as field of fractions and integral over AA. Let mm be a maximal ideal of AA and BB the local ring A[X]mA[X]A[X]_{mA[X]} (Exercise 15).

(a) The subring B′=B[A′]B' = B[A'] of K′(X)K'(X) is integral over BB; its field of fractions is K′(X)K'(X) and [K′(X):K(X)]=n[K'(X):K(X)] = n.

(b) For every maximal ideal m′m' of A′A' containing mm, n′=m′B′n' = m'B' is a maximal ideal of B′B'; if we write n=mBn = mB, then

[(A′/m′):(A/m)]=[(B′/n′):(B/n)] [(A'/m'):(A/m)] = [(B'/n'):(B/n)]
and
[(A′/m′):(A/m)]s=[(B′/n′):(B/n)]s. [(A'/m'):(A/m)]_s = [(B'/n'):(B/n)]_s.
(Observe that (A′/m′)⊗AB(A'/m') \otimes_A B is isomorphic to the field (A′/m′)(X)(A'/m')(X) and deduce that B′/n′B'/n' is isomorphic to (A′/m′)(X)(A'/m')(X).)

(c) The mapping m′↦m′B′m' \mapsto m'B' is the bijection of the set of maximal ideals of AA containing mm onto the set of maximal ideals of B′B'; the inverse bijection is n′↦n′∩A′n' \mapsto n' \cap A'.

Solution

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Transcribed from the printed volume. Not translated.