Lemma 33.7.8. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $A$ be a $k$-algebra. Let $V \subset X_ A$ be a quasi-compact open. Then there exists a finitely generated $k$-subalgebra $A' \subset A$ and a quasi-compact open $V' \subset X_{A'}$ such that $V = V'_ A$.

**Proof.**
We remark that if $X$ is also quasi-separated this follows from Limits, Lemma 32.4.11. Let $U_1, \ldots , U_ n$ be finitely many affine opens of $X$ such that $V \subset \bigcup U_{i, A}$. Say $U_ i = \mathop{\mathrm{Spec}}(R_ i)$. Since $V$ is quasi-compact we can find finitely many $f_{ij} \in R_ i \otimes _ k A$, $j = 1, \ldots , n_ i$ such that $V = \bigcup _ i \bigcup _{j = 1, \ldots , n_ i} D(f_{ij})$ where $D(f_{ij}) \subset U_{i, A}$ is the corresponding standard open. (We do not claim that $V \cap U_{i, A}$ is the union of the $D(f_{ij})$, $j = 1, \ldots , n_ i$.) It is clear that we can find a finitely generated $k$-subalgebra $A' \subset A$ such that $f_{ij}$ is the image of some $f_{ij}' \in R_ i \otimes _ k A'$. Set $V' = \bigcup D(f_{ij}')$ which is a quasi-compact open of $X_{A'}$. Denote $\pi : X_ A \to X_{A'}$ the canonical morphism. We have $\pi (V) \subset V'$ as $\pi (D(f_{ij})) \subset D(f_{ij}')$. If $x \in X_ A$ with $\pi (x) \in V'$, then $\pi (x) \in D(f_{ij}')$ for some $i, j$ and we see that $x \in D(f_{ij})$ as $f_{ij}'$ maps to $f_{ij}$. Thus we see that $V = \pi ^{-1}(V')$ as desired.
$\square$

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