Lemma 10.64.2. Let $R$ be a Noetherian ring. Let $\mathfrak p$ be a prime ideal. Let $n > 0$. Then $\text{Ass}(R/\mathfrak p^{(n)}) = \{ \mathfrak p\}$.

Proof. If $\mathfrak q$ is an associated prime of $R/\mathfrak p^{(n)}$ then clearly $\mathfrak p \subset \mathfrak q$. On the other hand, any element $x \in R$, $x \not\in \mathfrak p$ is a nonzerodivisor on $R/\mathfrak p^{(n)}$. Namely, if $y \in R$ and $xy \in \mathfrak p^{(n)} = R \cap \mathfrak p^ nR_{\mathfrak p}$ then $y \in \mathfrak p^ nR_{\mathfrak p}$, hence $y \in \mathfrak p^{(n)}$. Hence the lemma follows. $\square$

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