How to prove Ass$(R/Q)=\{P\}$ if and only if $Q$ is $P$-primary when $R$ is Noetherian? [duplicate]

Solution 1:

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 $\{P\} = \Ass(R/Q)$.