Lemma 10.28.3. 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.**
Let $T \subset \mathop{\mathrm{Spec}}(R)$ be constructible. The spectrum of a finite product of rings is the disjoint union of the spectra, see Lemma 10.20.2. Hence if $T = T_1 \cup T_2$ and the result holds for $T_1$ and $T_2$, then the result holds for $T$. In particular we may assume that $T = U \cap V^ c$, where $U, V \subset \mathop{\mathrm{Spec}}(R)$ are retrocompact open. By Lemma 10.28.1 we may write $T = (\bigcup D(f_ i)) \cap (\bigcup D(g_ j))^ c = \bigcup \big (D(f_ i) \cap V(g_1, \ldots , g_ m)\big )$. In fact we may assume that $T = D(f) \cap V(g_1, \ldots , g_ m)$ (by the argument on unions above). In this case $T$ is the image of the map $R \to (R/(g_1, \ldots , g_ m))_ f$, see Lemmas 10.16.6 and 10.16.7.
$\square$

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