
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. $\square$

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