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.

  • $\begingroup$ 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 $\endgroup$
    – desiigner
    Jul 16, 2019 at 19:29
  • $\begingroup$ 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. $\endgroup$
    – José
    Jul 16, 2019 at 19:54
  • 1
    $\begingroup$ 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. $\endgroup$
    – user26857
    Jul 17, 2019 at 21:05

1 Answer 1


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


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