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Today I tried to prove that if a Noetherian ring $A$ satisfies Serre's $R_0$ and $S_1$ conditions, then $A$ is reduced. Now we recall that $R_0$ means the localization at any minimal prime is a field (Nakayama's lemma) and $S_1$ means that $\operatorname{depth} A_{\mathfrak{p}} \geq \min\{ 1, \dim A_{\mathfrak{p}}\}$.

Is it true that the $S_1$ condition is equivalent to saying that $A$ has no embedded primes, i.e. every associated prime is minimal?

If this is true, then it would allow us to give the following quick proof that $R_0 + S_1 \implies$ reduced:

We know $A$ injects into the product $\prod_{\mathfrak{p} \in \operatorname{Ass} A} A_{\mathfrak{p}}$. However, every associated prime is minimal and the localization at a minimal prime is field. It follows $A$ injects into a finite direct product of fields and thus is reduced.

user26857
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David Benjamin Lim
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1 Answers1

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Yes, $S_1$ is equivalent to saying that every associated prime is minimal for Noetherian rings. This follows directly from the following facts:

i) $\text{Ass}(S^{-1}A) = \text{Ass}(A) \cap \text{Spec}(S^{-1}A)$. In particular $p \in \text{Ass}(A) \iff pA_p \in \text{Ass}(A_p)$.
ii) if $(A,m)$ is local, then $\text{depth}(A) = 0$ iff $m \in \text{Ass}(A)$. In particular $pA_p \in \text{Ass}(A_p) \iff \text{depth}(A_p) = 0$.

Thus, if $A$ is $S_1$, then for any associated prime $p$, $0 = \text{depth}(A_p) \ge \min(1, \dim A_p) \implies \dim A_p = 0$, so $p$ is minimal. For the converse, if $p$ is nonminimal, then $p$ is not associated, so $\text{depth}(A_p) \ge 1$, hence $A$ is $S_1$.

zcn
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