Bourbaki

Exercise 2, § 2

Commutative Algebra, II, 123

Let AA be a ring and SS a multiplicative subset of AA. Show that the set nn of x∈Ax \in A such that sx=0sx = 0 for some s∈Ss \in S is an ideal of AA. Set An=A/nA_n = A/n and denote by S1S_1 the canonical image of SS in AnA_n. Show that no element of S1S_1 is a divisor of 0 in AA, and that the canonical homomorphism S−1A→S1−1A1S^{-1}A \to S_1^{-1}A_1 is bijective.

Deduce that, for S−1AS^{-1}A to be a finitely generated AA-module, it is necessary and sufficient that all the elements of S1S_1 be invertible in AA, in which case S−1AS^{-1}A is identified with AA.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.