Let $R$ be a reduced ring. Show that $\operatorname{Ass}R$ is the set of minimal prime ideals of $R$.
I think that the first inclusion must come from using $\operatorname{Ass}R \subseteq\operatorname{Supp}R$, assuming the ideal is not minimal, and then showing some contradiction. No idea how to prove every minimal prime is associate, since the ring is not necessarily noetherian.
Best Answer
If $R$ is reduced and $P=\operatorname{Ann}(x)$ is an associated prime, suppose $Q$ is a prime properly contained in $P$. Then $(x)P\subseteq Q$ implies $x\in Q$. But then $x^2=0$, a contradiction. So $P$ was already minimal.
The other direction isn't clear to me. In this paper they talk about necessary and sufficient conditions for a reduced ring to have the property that all finitely generated ideals of zero divisors to have a nonzero annihilator, so presumably both cases can happen.
I see here that "weakly associated primes" are exactly the minimal primes in a reduced ring though. In that case, minimal primes are obviously weakly associated, because a minimal prime is minimal over $\operatorname{Ann}(1)=\{0\}$.