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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