Lemma 10.67.2. 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

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

$M'$ has no embedded associated primes,

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

**Proof.**
We will use Lemma 10.63.5 and Proposition 10.63.6 without further mention. 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\} $. Let

\[ K = \{ m \in M \mid \text{Supp}(Rm) \subset \bigcup V(\mathfrak p_ i)\} \]

It is immediately seen to be a submodule. Since $M$ is finite over a Noetherian ring, we know $K$ is finite too. Hence $\text{Supp}(K)$ is nowhere dense in $\text{Supp}(M)$. Let $K' \subset M$ be another submodule with support nowhere dense in $\text{Supp}(M)$. This means that $K_{\mathfrak q_ j} = 0$. Hence if $m \in K'$, then $m$ maps to zero in $M_{\mathfrak q_ j}$ which in turn implies $(Rm)_{\mathfrak q_ j} = 0$. On the other hand we have $\text{Ass}(Rm) \subset \text{Ass}(M)$. Hence the support of $Rm$ is contained in $\bigcup V(\mathfrak p_ i)$. Therefore $m \in K$ and thus $K' \subset K$ as $m$ was arbitrary in $K'$.

Let $M' = M/K$. Since $K_{\mathfrak q_ j}=0$ we know $M'_{\mathfrak q_ j} = M_{\mathfrak q_ j}$ for all $j$. Hence $M$ and $M'$ have the same support.

Suppose $\mathfrak q = \text{Ann}(\overline{m}) \in \text{Ass}(M')$ where $\overline{m} \in M'$ is the image of $m \in M$. Then $m \not\in K$ and hence the support of $Rm$ must contain one of the $\mathfrak q_ j$. Since $M_{\mathfrak q_ j} = M'_{\mathfrak q_ j}$, we know $\overline{m}$ does not map to zero in $M'_{\mathfrak q_ j}$. Hence $\mathfrak q \subset \mathfrak q_ j$ (actually we have equality), which means that all the associated primes of $M'$ are not embedded.

Let $f$ be an element contained in all $\mathfrak p_ i$. Then $D(f) \cap \text{supp}(K) = 0$. Hence $M_ f = M'_ f$ because $K_ f = 0$.
$\square$

## Comments (2)

Comment #7153 by Xuande Liu on

Comment #7301 by Johan on