Lemma 10.30.6 (00FL)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.30: More on images
Lemma 10.30.6 (cite)

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Lemma 10.30.6. Let $R \to S$ be a ring map. The following are equivalent:

The kernel of $R \to S$ consists of nilpotent elements.
The minimal primes of $R$ are in the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ .
The image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is dense in $\mathop{\mathrm{Spec}}(R)$ .

Proof. Let $I = \mathop{\mathrm{Ker}}(R \to S)$ . Note that $\sqrt{(0)} = \bigcap _{\mathfrak q \subset S} \mathfrak q$ , see Lemma 10.17.2. Hence $\sqrt{I} = \bigcap _{\mathfrak q \subset S} R \cap \mathfrak q$ . Thus $V(I) = V(\sqrt{I})$ is the closure of the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ . This shows that (1) is equivalent to (3). It is clear that (2) implies (3). Finally, assume (1). We may replace $R$ by $R/I$ and $S$ by $S/IS$ without affecting the topology of the spectra and the map. Hence the implication (1) $\Rightarrow$ (2) follows from Lemma 10.30.5. $\square$

Comments (2)

Comment #9517 by Goodluckthere on July 21, 2024 at 23:15

The equality $\sqrt{I}=\cap_{q\in S}R\cap q$ is better to be written as "the preimage of the nilradical $\sqrt{(0)}$ of $S$ equals $\sqrt{I}$ ."

Comment #9518 by Goodluckthere on July 22, 2024 at 07:30

Also in the penultimate line, it's strange to take $S/IS$ since $I$ is not inside $S$ and its image in $S$ is zero. I think what it means is that the map $R/I\to S$ is an injective ring map, therefore by lemma $\ref{https://stacks.math.columbia.edu/tag/00FK}$ , the corresponding spectra map $\operatorname{Spec}(S)\to \operatorname{Spec}(R/I)$ hits all minimal prime ideals of $R/I$ . A minimal prime ideal of $R$ will contain the nilpotent elements of $R$ , therefore also $I$ by (1) thus corresponding to a minimal prime ideal of $R/I$ which in turn will have a preimage in $\operatorname{Spec}(S)$ .

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