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*}

10.66 Embedded primes

Here is the definition.

Definition 10.66.1. Let $R$ be a ring. Let $M$ be an $R$-module.

  1. The associated primes of $M$ which are not minimal among the associated primes of $M$ are called the embedded associated primes of $M$.

  2. The embedded primes of $R$ are the embedded associated primes of $R$ as an $R$-module.

Here is a way to get rid of these.

Lemma 10.66.2. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Consider the set of $R$-submodules

\[ \{ K \subset M \mid \text{Supp}(K) \text{ nowhere dense in } \text{Supp}(M) \} . \]

This set has a maximal element $K$ and the quotient $M' = M/K$ has the following properties

  1. $\text{Supp}(M) = \text{Supp}(M')$,

  2. $M'$ has no embedded associated primes,

  3. for any $f \in R$ which is contained in all embedded associated primes of $M$ we have $M_ f \cong M'_ f$.

Proof. Let $\mathfrak q_1, \ldots , \mathfrak q_ t$ denote the minimal primes in the support of $M$. Let $\mathfrak p_1, \ldots , \mathfrak p_ s$ denote the embedded associated primes of $M$. Then $\text{Ass}(M) = \{ \mathfrak q_ j, \mathfrak p_ i\} $. There are finitely many of these, see Lemma 10.62.5. Set $I = \prod _{i = 1, \ldots , s} \mathfrak p_ i$. Then $I \not\subset \mathfrak q_ j$ for any $j$. Hence by Lemma 10.14.2 we can find an $f \in I$ such that $f \not\in \mathfrak q_ j$ for all $j = 1, \ldots , t$. Set $M' = \mathop{\mathrm{Im}}(M \to M_ f)$. This implies that $M_ f \cong M'_ f$. Since $M' \subset M_ f$ we see that $\text{Ass}(M') \subset \text{Ass}(M_ f) = \{ \mathfrak q_ j\} $. Thus $M'$ has no embedded associated primes.

Moreover, the support of $K = \mathop{\mathrm{Ker}}(M \to M')$ is contained in $V(\mathfrak p_1) \cup \ldots \cup V(\mathfrak p_ s)$, because $\text{Ass}(K) \subset \text{Ass}(M)$ (see Lemma 10.62.3) and $\text{Ass}(K)$ contains none of the $\mathfrak q_ i$ by construction. Clearly, $K$ is in fact the largest submodule of $M$ whose support is contained in $V(\mathfrak p_1) \cup \ldots \cup V(\mathfrak p_ t)$. This implies that $K$ is the maximal element of the set displayed in the lemma. $\square$

Lemma 10.66.3. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. For any $f \in R$ we have $(M')_ f = (M_ f)'$ where $M \to M'$ and $M_ f \to (M_ f)'$ are the quotients constructed in Lemma 10.66.2.

Proof. Omitted. $\square$

Lemma 10.66.4. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module without embedded associated primes. Let $I = \{ x \in R \mid xM = 0\} $. Then the ring $R/I$ has no embedded primes.

Proof. We may replace $R$ by $R/I$. Hence we may assume every nonzero element of $R$ acts nontrivially on $M$. By Lemma 10.39.5 this implies that $\mathop{\mathrm{Spec}}(R)$ equals the support of $M$. Suppose that $\mathfrak p$ is an embedded prime of $R$. Let $x \in R$ be an element whose annihilator is $\mathfrak p$. Consider the nonzero module $N = xM \subset M$. It is annihilated by $\mathfrak p$. Hence any associated prime $\mathfrak q$ of $N$ contains $\mathfrak p$ and is also an associated prime of $M$. Then $\mathfrak q$ would be an embedded associated prime of $M$ which contradicts the assumption of the lemma. $\square$

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 02M4. Beware of the difference between the letter 'O' and the digit '0'.