Lemma 10.157.3. Let $R$ be a Noetherian ring. The following are equivalent:

$R$ is reduced, and

$R$ has properties $(R_0)$ and $(S_1)$.

** Reduced equals R0 plus S1. **

Lemma 10.157.3. Let $R$ be a Noetherian ring. The following are equivalent:

$R$ is reduced, and

$R$ has properties $(R_0)$ and $(S_1)$.

**Proof.**
Suppose that $R$ is reduced. Then $R_{\mathfrak p}$ is a field for every minimal prime $\mathfrak p$ of $R$, according to Lemma 10.25.1. Hence we have $(R_0)$. Let $\mathfrak p$ be a prime of height $\geq 1$. Then $A = R_{\mathfrak p}$ is a reduced local ring of dimension $\geq 1$. Hence its maximal ideal $\mathfrak m$ is not an associated prime since this would mean there exists an $x \in \mathfrak m$ with annihilator $\mathfrak m$ so $x^2 = 0$. Hence the depth of $A = R_{\mathfrak p}$ is at least one, by Lemma 10.63.9. This shows that $(S_1)$ holds.

Conversely, assume that $R$ satisfies $(R_0)$ and $(S_1)$. If $\mathfrak p$ is a minimal prime of $R$, then $R_{\mathfrak p}$ is a field by $(R_0)$, and hence is reduced. If $\mathfrak p$ is not minimal, then we see that $R_{\mathfrak p}$ has depth $\geq 1$ by $(S_1)$ and we conclude there exists an element $t \in \mathfrak pR_{\mathfrak p}$ such that $R_{\mathfrak p} \to R_{\mathfrak p}[1/t]$ is injective. Now $R_\mathfrak p[1/t]$ is contained in the product of its localizations at prime ideals, see Lemma 10.23.1. This implies that $R_{\mathfrak p}$ is a subring of a product of localizations of $R$ at $\mathfrak p \supset \mathfrak q$ with $t \not\in \mathfrak q$. Since these primes have smaller height by induction on the height we conclude that $R$ is reduced. $\square$

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