Lemma 10.29.6. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $S = R/I$. Then the image of a constructible subset of $\mathop{\mathrm{Spec}}(S)$ is constructible in $\mathop{\mathrm{Spec}}(R)$.

Proof. If $I = (f_1, \ldots , f_ m)$, then we see that $V(I)$ is the complement of $\bigcup D(f_ i)$, see Lemma 10.17.2. Hence it is constructible, by Lemma 10.29.1. Denote the map $R \to S$ by $f \mapsto \overline{f}$. We have to show that if $\overline{U}, \overline{V}$ are retrocompact opens of $\mathop{\mathrm{Spec}}(S)$, then the image of $\overline{U} \cap \overline{V}^ c$ in $\mathop{\mathrm{Spec}}(R)$ is constructible. By Lemma 10.29.1 we may write $\overline{U} = \bigcup D(\overline{g_ i})$. Setting $U = \bigcup D({g_ i})$ we see $\overline{U}$ has image $U \cap V(I)$ which is constructible in $\mathop{\mathrm{Spec}}(R)$. Similarly the image of $\overline{V}$ equals $V \cap V(I)$ for some retrocompact open $V$ of $\mathop{\mathrm{Spec}}(R)$. Hence the image of $\overline{U} \cap \overline{V}^ c$ equals $U \cap V(I) \cap V^ c$ as desired. $\square$

Comment #6455 by Jonas Ehrhard on

The statement should be:

"The image of a constructible subset of $\operatorname{Spec}(S)$ is constructible in $\operatorname{Spec}(R)$."

Comment #7355 by Hao Peng on

The proof has a gap that the $f(U\cap V)$ may not be equal to $f(U)\cap f(V)$. But this can be easily solved using explict form in tag0G1P.

Comment #7356 by Hao Peng on

Nevermind, it is a closed embedding...

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

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