Bourbaki

Exercise 18, § 1

Commutative Algebra, VII, 550

Bourbaki marks this one as harder than the rest.

Let A=⨁n∈ZAnA = \bigoplus_{n \in \mathbf{Z}} A_n be a graded Krull domain with positive degrees, such that A0A_0 is a field and let m=⨁n≫1Anm = \bigoplus_{n \gg 1} A_n a maximal ideal of A. Let S be the multiplicative set of homogeneous elements ≠0\neq 0 of A, so that the ring S−1AS^{-1}A is a principal ideal domain (Exercise 16 (b)). (a) Let pp be a prime ideal of A of height 1 which meets A−mA - m; then pp is not graded (otherwise p=Ap = A) and the ideal S−1pS^{-1}p of S−1AS^{-1}A is principal, generated by an element of the form x=1+x1+…+xrx = 1 + x_1 + \ldots + x_r, where xj∈(S−1A)jx_j \in (S^{-1}A)_j. Writing an element of p of the form 1+a1+…+ar1 + a_1 + \ldots + a_r, (ai∈Aia_i \in A_i) as a multiple of xx in S−1AS^{-1}A and using Chapter V, § 1, no. 3, Proposition 11, show that x∈px \in p; writing finally every element of p as a multiple of xx in S−1AS^{-1}A, show that p=Axp = Ax. (b) Deduce from (a) that C(A)C(A) and C(Am)C(A_m) are canonically isomorphic (use Proposition 17 of no. 10).

Solution

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Transcribed from the printed volume. Not translated.