Lemma 10.66.9. Let R be a ring. Let M be an R-module. Let \mathfrak p be a prime ideal of R which is finitely generated. Then
In particular, if R is Noetherian, then \text{Ass}(M) = \text{WeakAss}(M).
Lemma 10.66.9. Let R be a ring. Let M be an R-module. Let \mathfrak p be a prime ideal of R which is finitely generated. Then
In particular, if R is Noetherian, then \text{Ass}(M) = \text{WeakAss}(M).
Proof. Write \mathfrak p = (g_1, \ldots , g_ n) for some g_ i \in R. It is enough the prove the implication β\Leftarrow β as the other implication holds in general, see Lemma 10.66.6. Assume \mathfrak p \in \text{WeakAss}(M). By Lemma 10.66.2 there exists an element m \in M_{\mathfrak p} such that I = \{ x \in R_{\mathfrak p} \mid xm = 0\} has radical \mathfrak pR_{\mathfrak p}. Hence for each i there exists a smallest e_ i > 0 such that g_ i^{e_ i}m = 0 in M_{\mathfrak p}. If e_ i > 1 for some i, then we can replace m by g_ i^{e_ i - 1} m \not= 0 and decrease \sum e_ i. Hence we may assume that the annihilator of m \in M_{\mathfrak p} is (g_1, \ldots , g_ n)R_{\mathfrak p} = \mathfrak p R_{\mathfrak p}. By Lemma 10.63.15 we see that \mathfrak p \in \text{Ass}(M). \square
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