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Let $R$ be a Noetherian ring, $P$ be a prime ideal, and $Q$ an ideal of $R$. How to prove that $$ \text{Ass}(R/Q)=\{P\} $$

if and only if $Q$ is $P$-primary?


Update

In fact, I have proved that if $Q$ is primary, then Ann$(R/Q)$ is primary. Let $P=\text{rad}(\text{Ann}(R/Q))$, we have $Q$ is $P$-primary and $\text{Ass}'(R/Q))=\{P\}$ (where $\text{Ass}'$ consists of all primes occur as the radical of some $\text{Ann}(x)$ )

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    $\begingroup$ Decompose $\mathfrak q$ as an intersection of primary ideals $\bigcap \mathfrak q_i$. If the only associated prime of $\mathfrak q$ is $\mathfrak p$, every $\mathfrak q_i$ is $\mathfrak p$-primary, and thus is the intersection $\mathfrak q$. Conversely, if $\mathfrak q$ is $\mathfrak p$-primary, you already have a decomposition, and thus $\mathfrak p$ is the only associated ideal. $\endgroup$
    – Pedro Tamaroff
    May 21 '14 at 20:48
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Here's a relatively elementary proof, which is (in my opinion) one of many extremely beautiful proofs in the theory of associated primes and primary decomposition:

An ideal $Q$ is primary iff every zerodivisor in $R/Q$ is nilpotent, i.e. the set of zerodivisors is equal to the nilradical. Since zerodivisors are a union of associated primes in a Noetherian ring, the nilradical is the intersection of all minimal primes, and every minimal prime is an associated prime, one sees that $Q$ is primary iff $\DeclareMathOperator{\Ass}{Ass}$ $$\bigcup_{p \in \Ass(R/Q)} p = \bigcap_{p \in \text{Min}(R/Q)} p = \bigcap_{p \in \Ass(R/Q)} p$$ which occurs iff $|\Ass(R/Q)| = 1$ (since neither side is empty). It follows that $\sqrt{Q} = P$ for the single associated prime $\{P\} = \Ass(R/Q)$.

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