Lemma 10.157.3 (031R)—The Stacks project

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. $\square$

Comments (7)

Comment #929 by correction_bot on August 22, 2014 at 18:44

Hah, bad slogan!

Comment #939 by Johan on August 22, 2014 at 22:33

Thanks very much. Fixed here.

Comment #7081 by Yuto Masamura on February 26, 2022 at 04:56

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 Johan on May 01, 2022 at 18:47

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 January 30, 2024 at 19:20

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

Comment #9296 by Stacks project on May 22, 2024 at 13:49

Thanks and fixed here.

Comment #10001 by Shubhankar on February 22, 2025 at 07:30

Perhaps I am missing something (in which case apologies) but the forward direction seems like it can be simplified (in a minor sense). We want to show that reduced implies $(S_1)$ which by the previous lemma is equivalent to non-existence of embedded primes. Suppose $R$ had an embedded prime $\mathfrak{p}$ then $R_{\mathfrak{p}}$ is a dimension $\geq 1$ reduced ring. On the other hand $\mathfrak{p} R_{\mathfrak{p}}$ is an associated prime of $R_{\mathfrak{p}}$ therefore contains an element which is squre zero (so no appeal to depth).

Also in the forward direction the notation $A=R_{\mathfrak{p}}$ is introduced but never used.

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