Lemma 10.63.13. Let \varphi : R \to S be a ring map. Let M be an S-module. If S is Noetherian, then \mathop{\mathrm{Spec}}(\varphi )(\text{Ass}_ S(M)) = \text{Ass}_ R(M).
Proof. We have already seen in Lemma 10.63.11 that \mathop{\mathrm{Spec}}(\varphi )(\text{Ass}_ S(M)) \subset \text{Ass}_ R(M). For the converse, choose a prime \mathfrak p \in \text{Ass}_ R(M). Let m \in M be an element such that the annihilator of m in R is \mathfrak p. Let I = \{ g \in S \mid gm = 0\} be the annihilator of m in S. Then R/\mathfrak p \subset S/I is injective. Combining Lemmas 10.30.5 and 10.30.7 we see that there is a prime \mathfrak q \subset S minimal over I mapping to \mathfrak p. By Proposition 10.63.6 we see that \mathfrak q is an associated prime of S/I, hence \mathfrak q is an associated prime of M by Lemma 10.63.3 and we win. \square
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