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
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