# 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.

• Would you mind adding the definition of $\text{Ass}(R)$ that you're using to the question? Also this post might help: stacks.math.columbia.edu/tag/0546 especially 10.65.3 Jul 16, 2019 at 19:29
• Ass $R$ meaning the associate primes of $R$, that is, $\{P\in \text{Spec }R |P \text{ is the annihilator of an element } r \in R\}$. No mention of $R$ being noetherian.
– José
Jul 16, 2019 at 19:54
• The converse doesn't hold. If $R=K[X_1,\dots,X_n,\dots]/(X_1X_2,X_3X_4,\dots)$, then $\mathfrak p=(x_1,x_3,\dots)$ is a minimal prime which is not associated. Jul 17, 2019 at 21:05

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