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 10.64.1. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Let $A$, $A'$, $A_{fin}$, $B$, and $B_{fin}$ be the subsets of $\mathop{\mathrm{Spec}}(S)$ introduced above.

  1. We always have $A = A'$.

  2. We always have $A_{fin} \subset A$, $B_{fin} \subset B$, $A_{fin} \subset A'_{fin} \subset B_{fin}$ and $A \subset B$.

  3. If $S$ is Noetherian, then $A = A_{fin}$ and $B = B_{fin}$.

  4. If $N$ is flat over $R$, then $A = A_{fin} = A'_{fin}$ and $B = B_{fin}$.

  5. If $R$ is Noetherian and $N$ is flat over $R$, then all of the sets are equal, i.e., $A = A' = A_{fin} = A'_{fin} = B = B_{fin}$.

Proof. Some of the arguments in the proof will be repeated in the proofs of later lemmas which are more precise than this one (because they deal with a given module $M$ or a given prime $\mathfrak p$ and not with the collection of all of them).

Proof of (1). Let $\mathfrak p$ be a prime of $R$. Then we have

\[ \text{Ass}_ S(N \otimes _ R \kappa (\mathfrak p)) = \text{Ass}_{S/\mathfrak pS}(N \otimes _ R \kappa (\mathfrak p)) = \text{Ass}_{S \otimes _ R \kappa (\mathfrak p)}(N \otimes _ R \kappa (\mathfrak p)) \]

the first equality by Lemma 10.62.14 and the second by Lemma 10.62.16 part (1). This prove that $A = A'$. The inclusion $A_{fin} \subset A'_{fin}$ is clear.

Proof of (2). Each of the inclusions is immediate from the definitions except perhaps $A_{fin} \subset A$ which follows from Lemma 10.62.16 and the fact that we require $\mathfrak p = R \cap \mathfrak q$ in the formulation of $A_{fin}$.

Proof of (3). The equality $A = A_{fin}$ follows from Lemma 10.62.16 part (3) if $S$ is Noetherian. Let $\mathfrak q = (g_1, \ldots , g_ m)$ be a finitely generated prime ideal of $S$. Say $z \in N \otimes _ R M$ is an element whose annihilator is $\mathfrak q$. We may pick a finite submodule $M' \subset M$ such that $z$ is the image of $z' \in N \otimes _ R M'$. Then $\text{Ann}_ S(z') \subset \mathfrak q = \text{Ann}_ S(z)$. Since $N \otimes _ R -$ commutes with colimits and since $M$ is the directed colimit of finite $R$-modules we can find $M' \subset M'' \subset M$ such that the image $z'' \in N \otimes _ R M''$ is annihilated by $g_1, \ldots , g_ m$. Hence $\text{Ann}_ S(z'') = \mathfrak q$. This proves that $B = B_{fin}$ if $S$ is Noetherian.

Proof of (4). If $N$ is flat, then the functor $N \otimes _ R -$ is exact. In particular, if $M' \subset M$, then $N \otimes _ R M' \subset N \otimes _ R M$. Hence if $z \in N \otimes _ R M$ is an element whose annihilator $\mathfrak q = \text{Ann}_ S(z)$ is a prime, then we can pick any finite $R$-submodule $M' \subset M$ such that $z \in N \otimes _ R M'$ and we see that the annihilator of $z$ as an element of $N \otimes _ R M'$ is equal to $\mathfrak q$. Hence $B = B_{fin}$. Let $\mathfrak p'$ be a prime of $R$ and let $\mathfrak q$ be a prime of $S$ which is an associated prime of $N/\mathfrak p'N$. This implies that $\mathfrak p'S \subset \mathfrak q$. As $N$ is flat over $R$ we see that $N/\mathfrak p'N$ is flat over the integral domain $R/\mathfrak p'$. Hence every nonzero element of $R/\mathfrak p'$ is a nonzerodivisor on $N/\mathfrak p'$. Hence none of these elements can map to an element of $\mathfrak q$ and we conclude that $\mathfrak p' = R \cap \mathfrak q$. Hence $A_{fin} = A'_{fin}$. Finally, by Lemma 10.62.17 we see that $\text{Ass}_ S(N/\mathfrak p'N) = \text{Ass}_ S(N \otimes _ R \kappa (\mathfrak p'))$, i.e., $A'_{fin} = A$.

Proof of (5). We only need to prove $A'_{fin} = B_{fin}$ as the other equalities have been proved in (4). To see this let $M$ be a finite $R$-module. By Lemma 10.61.1 there exists a filtration by $R$-submodules

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

such that each quotient $M_ i/M_{i-1}$ is isomorphic to $R/\mathfrak p_ i$ for some prime ideal $\mathfrak p_ i$ of $R$. Since $N$ is flat we obtain a filtration by $S$-submodules

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

such that each subquotient is isomorphic to $N/\mathfrak p_ iN$. By Lemma 10.62.3 we conclude that $\text{Ass}_ S(N \otimes _ R M) \subset \bigcup \text{Ass}_ S(N/\mathfrak p_ iN)$. Hence we see that $B_{fin} \subset A'_{fin}$. Since the other inclusion is part of (2) 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 05GB. Beware of the difference between the letter 'O' and the digit '0'.