Reduced equals R0 plus S1.

Lemma 10.157.3. 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 10.25.1. 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. $\square$

Comment #929 by correction_bot on

Comment #7081 by Yuto Masamura on

I think that the sentence "This implies that $R_{\mathfrak p}$ is a subring of localizations of $R$ at primes of smaller height" (in the 2nd paragraph of proof) is not trivial. We can find a prime $\mathfrak q\subsetneq\mathfrak p$ with $t\notin\mathfrak q$ (here I assume $t\in R$, i.e., $t\in\mathfrak p$ without loss of generality) and thus maps $R_{\mathfrak p}\to R_{\mathfrak p}[1/t]\to R_{\mathfrak q}$, but I do not know how to verify that the map $R_{\mathfrak p}\to R_{\mathfrak q}$ is injective. Or did you mean 'the product of localizations...', i.e., that we have $R_{\mathfrak p}\to\prod_{\mathfrak q\subseteq\mathfrak p\setminus\lbrace t\rbrace}R_{\mathfrak q}$?

Comment #7258 by on

Yes, I did mean product of localizations. Thanks very much for catching this. I have made some edits in this commit.

Comment #8783 by Mateo on

The inclusion of prime ideals is reversed in the second paragraph.

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• 4 comment(s) on Section 10.157: Serre's criterion for normality

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