Lemma 38.14.2. Let R be a ring. Let I \subset R be an ideal. Let R \to S be a ring map. Let N be an S-module. If N is I-adically complete, then for any R-module M and for any prime \mathfrak q \subset S which is an associated prime of N \otimes _ R M we have \mathfrak q + I S \not= S.
Proof. Let S^\wedge denote the I-adic completion of S. Note that N is an S^\wedge -module, hence also N \otimes _ R M is an S^\wedge -module. Let z \in N \otimes _ R M be an element such that \mathfrak q = \text{Ann}_ S(z). Since z \not= 0 we see that \text{Ann}_{S^\wedge }(z) \not= S^\wedge . Hence \mathfrak q S^\wedge \not= S^\wedge . Hence there exists a maximal ideal \mathfrak m \subset S^\wedge with \mathfrak q S^\wedge \subset \mathfrak m. Since IS^\wedge \subset \mathfrak m by Algebra, Lemma 10.96.6 we win. \square
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