Associated primes in a reduced ring 
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.

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 $\operatorname{Ann}(1)=\{0\}$.
