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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #1583 by Alex Torzewski on

Comment #3001 by David Holmes on

Comment #3124 by Johan on

There are also: