Lemma 10.28.4. 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 10.20.2. Hence if $T = T_1 \cup T_2$ and the result holds for $T_1$ and $T_2$, then the result holds for $T$. By Lemma 10.28.3 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 10.16.6 and 10.16.7.
$\square$

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