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 whose trace on A is p and for all let be such that ; show that . 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 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, -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 of elements of A such that for and ; deduce that belongs to the closure of with respect to the -adic topology; using Exercise 8, conclude that 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 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 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 ; show that the hypothesis on Spec(A) implies for each i the existence of a maximal ideal m, containing . Using (b), show that, if the canonical image of in A(m) is zero and for example , 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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