Bourbaki

Exercise 31, § 2

Commutative Algebra, IV

Bourbaki marks this one as harder than the rest.

a) Let AA be a noetherian ring, mm a maximal ideal of AA, pp a prime ideal contained in mm; now prove that if the separated completion A^\hat{A} of AA for the mm-adic topology is integral, the filter basis of the symbolic powers p(n)p^{(n)} of pp tends to 00 for the mm-adic topology in AA. (Reduce to the case where AA is local; let qq be a prime ideal of A^\hat{A} whose trace on AA is pp, and for every n>0n > 0, let cn∉pc_n \notin p such that cnp(n)⊂pnc_n p^{(n)} \subset p^n; now prove that one has cnp(n)A^⊂q(n)c_n p^{(n)} \hat{A} \subset q^{(n)}. Use Exercise 29 e), as well as Chapter III, § 2, No. 7, Proposition 8.)

b) Let AA be a noetherian ring, nn an ideal of AA, pp a minimal prime ideal in Ass⁡(A/n)\operatorname{Ass}(A/n), m0,mm_0, m two maximal ideals of AA containing pp; denote by A(n),A(m0),A(m)A(n), A(m_0), A(m) the separated completions of AA for the nn-adic, m0m_0-adic, mm-adic topologies respectively. Let zz be an element of A(n)A(n) whose canonical image in A(m0)A(m_0) is zero. Now prove that if A(m)A(m) is integral, the canonical image of zz in A(m)A(m) is zero also. (Take zz as the limit of a sequence (xn)(x_n) of elements of AA 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 for the m0m_0-adic topology; using Exercise 8, conclude that xm∈p(m)x_m \in p^{(m)} for every mm; complete using a.)

c) Let AA be a Zariski ring, rr a defining ideal of AA; suppose that Spec⁡(A)\operatorname{Spec}(A) is connected and that for every maximal ideal mm of AA, the separated completion A(m)A(m) of AA for the mm-adic topology is integral. Now prove that the completion A^\hat{A} of AA for the rr-adic topology is integral. (Now prove that for every maximal ideal mm of AA, the canonical homomorphism A^→A(m)\hat{A} \to A(m) is injective. For this, one may suppose that rr is the intersection of a finite number of prime ideals, minimal in Ass⁡(A/r)\operatorname{Ass}(A/r), namely r=p1∩p2∩…∩psr = p_1 \cap p_2 \cap \ldots \cap p_s; now prove that the assumption on Spec⁡(A)\operatorname{Spec}(A) implies for each ii the existence of a maximal ideal mim_i containing (p1∩…∩pi−1)+pi(p_1 \cap \ldots \cap p_{i-1}) + p_i. Using bb, now prove that if the canonical image of z∈A^z \in \hat{A} in A(m)A(m) is zero and if, for example, p1⊂mp_1 \subset m, then the canonical image of zz in each of the A(mi)A(m_i) is zero, then conclude that the canonical image of zz in A(m′)A(m') is zero for every maximal ideal m′m' of AA.)

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.