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

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. 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 $K_ f = 0$. $\square$

Comment #7153 by Xuande Liu on

I think the proof is not very intuitive and it actually does not prove the third conclusion. And there is a better way to rewrite the proof:

Let $K=\\{m\in M| \text{supp(Rm)}\subset \bigcup V(p_i)\\}$. Obviously it is a submodule. And since $M$ is finitely generated over a noetherian ring, we know $K$ is also generated by finitely many such elements. Hence $\text{supp}(K)$ is nowhere dense in $\text{supp}(M)$. Let $K'$ be another submodule with support nowhere dense in $\text{supp}(M)$. Then $K'$ can be generated by finitely elements $m_i$ with support nowhere dense in $\text{supp}(M)$. Notice that $Rm_i\subset M$. Hence $Ass(Rm_i)\subset Ass(M)$. But $(m_i)_{q_j}=0$. Hence the support of $m_i$ is contained in $\bigcup V(p_i)$. Therefore we know $m_i\in K$ and thus $K'\subset K$.

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

Suppose $q=Ann(\overline{m})\in Ass(M')$. Let $m$ be an element in $M$ lying in the inverse image of $m$. Then the support of $Rm$ must contain one of $q_j$. Since $M_{q_j}=M'_{q_j}$, we know $\overline{m}_{q_j}\neq 0$. Hence $q\subset q_j$, which means that all the associated primes of $M'$ are not embedded.

Let $f$ be an element contained in all $p_i$. Then $D(f)\cap \text{supp}(K)=0$. Hence $M_f=M'_f$. Now let $f$ be an element contained in all $p_i$ but not contained in any $q_j$.

In this proof, we avoid to check that $K$ is independent of the choice of $f$. And $K$ is obviously the unique maximal element in that set of submodules.

Comment #7301 by on

OK, thanks. I made the changes you suggested. See this commit. I hope others will comment further to improve the exposition even more.

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