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 (7)
Comment #1583 by Alex Torzewski on
Comment #3001 by David Holmes on
Comment #3124 by Johan on
Comment #7873 by Ryo Suzuki on
Comment #8144 by Aise Johan de Jong on
Comment #10038 by Zhouhang Mao on
Comment #10543 by Stacks Project on
There are also: