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. $\square$

Comment #2825 by Dario Weißmann on

I think we should mention that we can choose $\mathfrak{q}$ to be a minimal prime of $S/I$, thus a minimal prime in the support of $S/I$. Then we can apply 10.62.6.

There are also:

• 13 comment(s) on Section 10.63: Associated primes

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).