Lemma 10.40.6. Let $R \to S$ be a ring map. The following are equivalent:

Going up holds for $R \to S$, and

the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is closed.

Lemma 10.40.6. Let $R \to S$ be a ring map. The following are equivalent:

Going up holds for $R \to S$, and

the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is closed.

**Proof.**
It is a general fact that specializations lift along a closed map of topological spaces, see Topology, Lemma 5.19.7. Hence the second condition implies the first.

Assume that going up holds for $R \to S$. Let $V(I) \subset \mathop{\mathrm{Spec}}(S)$ be a closed set. We want to show that the image of $V(I)$ in $\mathop{\mathrm{Spec}}(R)$ is closed. The ring map $S \to S/I$ obviously satisfies going up. Hence $R \to S \to S/I$ satisfies going up, by Lemma 10.40.4. Replacing $S$ by $S/I$ it suffices to show the image $T$ of $\mathop{\mathrm{Spec}}(S)$ in $\mathop{\mathrm{Spec}}(R)$ is closed. By Topology, Lemmas 5.19.2 and 5.19.6 this image is stable under specialization. Thus the result follows from Lemma 10.40.5. $\square$

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