The Stacks project

Lemma 10.41.5. Let $R \to S$ be a ring map. Let $T \subset \mathop{\mathrm{Spec}}(R)$ be the image of $\mathop{\mathrm{Spec}}(S)$. If $T$ is stable under specialization, then $T$ is closed.

Proof. We give two proofs.

First proof. Let $\mathfrak p \subset R$ be a prime ideal such that the corresponding point of $\mathop{\mathrm{Spec}}(R)$ is in the closure of $T$. This means that for every $f \in R$, $f \not\in \mathfrak p$ we have $D(f) \cap T \not= \emptyset $. Note that $D(f) \cap T$ is the image of $\mathop{\mathrm{Spec}}(S_ f)$ in $\mathop{\mathrm{Spec}}(R)$. Hence we conclude that $S_ f \not= 0$. In other words, $1 \not= 0$ in the ring $S_ f$. Since $S_{\mathfrak p}$ is the directed colimit of the rings $S_ f$ we conclude that $1 \not= 0$ in $S_{\mathfrak p}$. In other words, $S_{\mathfrak p} \not= 0$ and considering the image of $\mathop{\mathrm{Spec}}(S_{\mathfrak p}) \to \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ we see there exists a $\mathfrak p' \in T$ with $\mathfrak p' \subset \mathfrak p$. As we assumed $T$ closed under specialization we conclude $\mathfrak p$ is a point of $T$ as desired.

Second proof. Let $I = \mathop{\mathrm{Ker}}(R \to S)$. We may replace $R$ by $R/I$. In this case the ring map $R \to S$ is injective. By Lemma 10.30.5 all the minimal primes of $R$ are contained in the image $T$. Hence if $T$ is stable under specialization then it contains all primes. $\square$

Comments (5)

Comment #3001 by David Holmes on

"Since S_p is the directed limit of the rings S_f..." Here the "limit" should be "colimit" I think.

Comment #7873 by Ryo Suzuki on

It can be proved considering constructible topology.

First, is quasi-compact in constructible topology by Lemma 5.23.2. Hence is also quasi-compact in constructible topology. Now is Hausdorff in constructible topology, again by Lemma 5.23.2. So is closed in constructible topology. By Lemma 5.23.6, is closed in usual topology.

There are also:

  • 7 comment(s) on Section 10.41: Going up and going down

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