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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