Lemma 10.62.13. 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 10.62.11 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 10.29.5 and 10.29.7 we see that there is a prime $\mathfrak q \subset S$ minimal over $I$ mapping to $\mathfrak p$. By Proposition 10.62.6 we see that $\mathfrak q$ is an associated prime of $S/I$, hence $\mathfrak q$ is an associated prime of $M$ by Lemma 10.62.3 and we win.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2825 by Dario Weißmann on

Comment #2926 by Johan on

There are also: