The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.65.13. Let $\varphi : R \to S$ be a ring map. Let $M$ be an $S$-module. Denote $f : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ the associated map on spectra. If $\varphi $ is a finite ring map, then

\[ \text{WeakAss}_ R(M) = f(\text{WeakAss}_ S(M)). \]

Proof. One of the inclusions has already been proved, see Remark 10.65.12. To prove the other assume $\mathfrak q \in \text{WeakAss}_ S(M)$ and let $\mathfrak p$ be the corresponding prime of $R$. Let $m \in M$ be an element such that $\mathfrak q$ is a minimal prime over $J = \{ g \in S \mid gm = 0\} $. Thus the radical of $JS_{\mathfrak q}$ is $\mathfrak qS_{\mathfrak q}$. As $R \to S$ is finite there are finitely many primes $\mathfrak q = \mathfrak q_1, \mathfrak q_2, \ldots , \mathfrak q_ l$ over $\mathfrak p$, see Lemma 10.35.21. Pick $x \in \mathfrak q$ with $x \not\in \mathfrak q_ i$ for $i > 1$, see Lemma 10.14.2. By the above there exists an element $y \in S$, $y \not\in \mathfrak q$ and an integer $t > 0$ such that $y x^ t m = 0$. Thus the element $ym \in M$ is annihilated by $x^ t$, hence $ym$ maps to zero in $M_{\mathfrak q_ i}$, $i = 2, \ldots , l$. To be sure, $ym$ does not map to zero in $S_{\mathfrak q}$.

The ring $S_{\mathfrak p}$ is semi-local with maximal ideals $\mathfrak q_ i S_{\mathfrak p}$ by going up for finite ring maps, see Lemma 10.35.22. If $f \in \mathfrak pR_{\mathfrak p}$ then some power of $f$ ends up in $JS_{\mathfrak q}$ hence for some $n > 0$ we see that $f^ t ym$ maps to zero in $M_{\mathfrak q}$. As $ym$ vanishes at the other maximal ideals of $S_{\mathfrak p}$ we conclude that $f^ t ym$ is zero in $M_{\mathfrak p}$, see Lemma 10.22.1. In this way we see that $\mathfrak p$ is a minimal prime over the annihilator of $ym$ in $R$ and we win. $\square$


Comments (0)


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05E1. Beware of the difference between the letter 'O' and the digit '0'.