Lemma 10.67.4. 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 10.40.5 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.
$\square$

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