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I would like to know if the following claim is true. I found this claim in a paper without proof :(.

Let $k$ an uncountable algebraically closed field of characteristic $0$.

Let $S=\operatorname{Spec}(k[X_1,..., X_m]/I(S))=\operatorname{Spec}(k[X_1,..., X_m]/\langle f_1,...,f_n\rangle)$ be an integral affine scheme of finite type over $k$.

Claim: We can choose a countable algebraically closed subfield $k_0\subset k$ such that there is an irreducible quasi projective scheme $S_0$ over $k_0$ with $S=S_0\times_{\operatorname{Spec}(k_0)}\operatorname{Spec}(k)$.

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Let $T$ be the finite set of coefficients appearing in the $f_i$. Then $k_0=\overline{\Bbb Q(T)}$ and $S_0=\operatorname{Spec} k_0[X_1,\cdots,X_m]/(f_1,\cdots,f_m)$ suffice. The verification that this has all the properties you request is straightforward; if you are stuck on any part, please leave a comment.

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  • $\begingroup$ Thank you @KReiser! Another question that I have is: Why $S_0$ over $k_0$ should be quasi-projective (notice that they initially say that $S$ over $k$ is affine)?. $\endgroup$
    – Roxana
    Mar 4, 2021 at 9:51
  • $\begingroup$ This is probably because they're gearing up to apply some result for quasi-projective $k_0$-schemes (it would be easier to say for sure if you included the source). Being affine, $S_0$ is obviously quasi-projective, and in fact as affineness descends along fpqc morphisms, any $S_0$ you find satisfying $S=S_0\times_{k_0} k$ will be affine. $\endgroup$
    – KReiser
    Mar 4, 2021 at 11:03
  • $\begingroup$ Ohhh I see @KReiser. Now, since $k_0$ is countable algebraically closed, Can I say that there are only countably many closed subsets in $S_0$? $\endgroup$
    – Roxana
    Mar 4, 2021 at 13:24
  • $\begingroup$ Yes, that's true, but it doesn't require $k_0$ to be algebraically closed, just countable. Any closed subset of $S_0$ is given by an ideal of $k_0[X_1,\cdots,X_m]/(f_1,\cdots,f_m)$, every ideals is finitely generated as this ring is noetherian, and the ring has countably many elements therefore the collection of ideals is countable. $\endgroup$
    – KReiser
    Mar 4, 2021 at 20:40

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