Countable algebraically closed field inside an incountable algebraically closed of characteristic zero

Question

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 $S=S_0\times_{\operatorname{Spec}(k_0)}\operatorname{Spec}(k)$.

Answer 1

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.