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 10.64.3. Let $R \to S$ be a ring map. Let $M$ be an $R$-module, and let $N$ be an $S$-module. If $N$ is flat as $R$-module, then

\[ \text{Ass}_ S(M \otimes _ R N) \supset \bigcup \nolimits _{\mathfrak p \in \text{Ass}_ R(M)} \text{Ass}_ S(N/\mathfrak pN) \]

and if $R$ is Noetherian then we have equality.

Proof. If $\mathfrak p \in \text{Ass}_ R(M)$ then there exists an injection $R/\mathfrak p \to M$. As $N$ is flat over $R$ we obtain an injection $R/\mathfrak p \otimes _ R N \to M \otimes _ R N$. Since $R/\mathfrak p \otimes _ R N = N/\mathfrak pN$ we conclude that $\text{Ass}_ S(N/\mathfrak pN) \subset \text{Ass}_ S(M \otimes _ R N)$, see Lemma 10.62.3. Hence the right hand side is contained in the left hand side.

Write $M = \bigcup M_\lambda $ as the union of its finitely generated $R$-submodules. Then also $N \otimes _ R M = \bigcup N \otimes _ R M_\lambda $ (as $N$ is $R$-flat). By definition of associated primes we see that $\text{Ass}_ S(N \otimes _ R M) = \bigcup \text{Ass}_ S(N \otimes _ R M_\lambda )$ and $\text{Ass}_ R(M) = \bigcup \text{Ass}(M_\lambda )$. Hence we may assume $M$ is finitely generated.

Let $\mathfrak q \in \text{Ass}_ S(M \otimes _ R N)$, and assume $R$ is Noetherian and $M$ is a finite $R$-module. To finish the proof we have to show that $\mathfrak q$ is an element of the right hand side. First we observe that $\mathfrak qS_{\mathfrak q} \in \text{Ass}_{S_{\mathfrak q}}((M \otimes _ R N)_{\mathfrak q})$, see Lemma 10.62.15. Let $\mathfrak p$ be the corresponding prime of $R$. Note that

\[ (M \otimes _ R N)_{\mathfrak q} = M \otimes _ R N_{\mathfrak q} = M_{\mathfrak p} \otimes _{R_{\mathfrak p}} N_{\mathfrak q} \]

If $\mathfrak pR_{\mathfrak p} \not\in \text{Ass}_{R_{\mathfrak p}}(M_{\mathfrak p})$ then there exists an element $x \in \mathfrak pR_{\mathfrak p}$ which is a nonzerodivisor in $M_{\mathfrak p}$ (see Lemma 10.62.18). Since $N_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ we see that the image of $x$ in $\mathfrak qS_{\mathfrak q}$ is a nonzerodivisor on $(M \otimes _ R N)_{\mathfrak q}$. This is a contradiction with the assumption that $\mathfrak qS_{\mathfrak q} \in \text{Ass}_ S((M \otimes _ R N)_{\mathfrak q})$. Hence we conclude that $\mathfrak p$ is one of the associated primes of $M$.

Continuing the argument we choose a filtration

\[ 0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M \]

such that each quotient $M_ i/M_{i-1}$ is isomorphic to $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$, see Lemma 10.61.1. (By Lemma 10.62.4 we have $\mathfrak p_ i = \mathfrak p$ for at least one $i$.) This gives a filtration

\[ 0 = M_0 \otimes _ R N \subset M_1 \otimes _ R N \subset \ldots \subset M_ n \otimes _ R N = M \otimes _ R N \]

with subquotients isomorphic to $N/\mathfrak p_ iN$. If $\mathfrak p_ i \not= \mathfrak p$ then $\mathfrak q$ cannot be associated to the module $N/\mathfrak p_ iN$ by the result of the preceding paragraph (as $\text{Ass}_ R(R/\mathfrak p_ i) = \{ \mathfrak p_ i\} $). Hence we conclude that $\mathfrak q$ is associated to $N/\mathfrak pN$ as desired. $\square$


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 0312. Beware of the difference between the letter 'O' and the digit '0'.