Bourbaki

Exercise 31, § 2

Commutative Algebra, IV, 299

Bourbaki marks this one as harder than the rest.

(a) Let A be a Noetherian ring, m a maximal ideal of A and p a prime ideal contained in m; show that, if the Hausdorff completion \hat{A} of A with respect to the m-adic topology is an integral domain, the filter base of symbolic power p^{(n)} of p tends to 0 with respect to the m-adic topology on A. (Reduce it to the case where A is local; let q be a prime ideal of A^\hat{A} whose trace on A is p and for all n>0n > 0 let cn∉pc_n \notin p be such that cnp(n)⊂pnc_n p^{(n)} \subset p^n; show that cnp(n)A^⊂q(n)c_n p^{(n)} \hat{A} \subset q^{(n)}. Use Exercise 29 (e) and Chapter III, § 2, no. 7, Proposition 8.)

(b) Let A be a Noetherian ring, n an ideal of A, p a minimal prime ideal in Ass(A/n) and m0,mm_0, m two maximal ideals of A containing p; let A(n), A(m_0), A(m) denote the Hausdorff completions of A with respect to the n-adic, m0m_0-adic, m-adic topologies respectively. Let z be an element of A(n) whose canonical image in A(m_0) is zero. Show that if A(m) is an integral domain, the canonical image of z in A(m) is also zero. (Take z as the limit of a sequence (xn)(x_n) of elements of A such that xn−xm∈nmx_n - x_m \in n^m for n≥mn \geq m and xn∈m0nx_n \in m_0^n; deduce that xmx_m belongs to the closure of nmn^m with respect to the m0m_0-adic topology; using Exercise 8, conclude that xm∈p(m)x_m \in p^{(m)} for all m and complete with the aid of (a).)

(c) Let A be a Zariski ring and r a defining ideal of A; suppose that Spec(A) is connected and that for every maximal ideal m of A the Hausdorff completion A(m) of A with respect to the m-adic topology is an integral domain. Show then that the completion A^\hat{A} of A with respect to the r-adic topology is an integral domain. (Show that for every maximal ideal m of A, the canonical homomorphism A^+A(m)\hat{A} + A(m) is injective. For this, we may assume that r is the intersection of a finite number of prime ideals which are minimal in Ass(A/r), let r=p1∩p2∩⋯∩psr = p_1 \cap p_2 \cap \cdots \cap p_s; show that the hypothesis on Spec(A) implies for each i the existence of a maximal ideal m, containing (p1∩…∩pi−1)+pi(p_1 \cap \ldots \cap p_{i-1}) + p_i. Using (b), show that, if the canonical image of z∈A^z \in \hat{A} in A(m) is zero and for example p1⊂mp_1 \subset m, then the canonical image of z in each of the A(m_i) is zero, then conclude that the canonical image of z in A(m') is zero for every maximal ideal m' of A.)

Solution

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Transcribed from the printed volume. Not translated.