Lemma 10.29.4. Let $R$ be a ring and let $T \subset \mathop{\mathrm{Spec}}(R)$ be constructible. Then there exists a ring map $R \to S$ of finite presentation such that $T$ is the image of $\mathop{\mathrm{Spec}}(S)$ in $\mathop{\mathrm{Spec}}(R)$.

Proof. The spectrum of a finite product of rings is the disjoint union of the spectra, see Lemma 10.21.2. Hence if $T = T_1 \cup T_2$ and the result holds for $T_1$ and $T_2$, then the result holds for $T$. By Lemma 10.29.3 we may assume that $T = D(f) \cap V(g_1, \ldots , g_ m)$. In this case $T$ is the image of the map $\mathop{\mathrm{Spec}}((R/(g_1, \ldots , g_ m))_ f) \to \mathop{\mathrm{Spec}}(R)$, see Lemmas 10.17.6 and 10.17.7. $\square$

Comment #4937 by awllower on

I think the last ring map should be $R_f\rightarrow R_f/(g_1,\ldots,g_m)$?

Comment #4938 by Laurent Moret-Bailly on

The proof contains the result that the constructible sets of $\mathrm{Spec}(R)$ are the finite unions of sets of the form $D(f)\cap V(g_1,\dots,g_m)$. This fact is useful by itself, so why not make it a separate statement (or incorporate it in 00F6)? It would make 00F8 immediate, and probably simplify other proofs, too.

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• 6 comment(s) on Section 10.29: Images of ring maps of finite presentation

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