
Reduced equals R0 plus S1.

Lemma 10.151.3. Let $R$ be a Noetherian ring. The following are equivalent:

1. $R$ is reduced, and

2. $R$ 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 10.24.1. 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 a $x \in \mathfrak m$ with annihilator $\mathfrak m$ so $x^2 = 0$. Hence the depth of $A = R_{\mathfrak p}$ is at least one, by Lemma 10.62.9. 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. This implies that $R_{\mathfrak p}$ is a subring of localizations of $R$ at primes of smaller height. Thus by induction on the height we conclude that $R$ is reduced. $\square$

Comment #929 by correction_bot on

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