## 10.66 Embedded primes

Here is the definition.

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

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

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

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

$M'$ 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.**
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)