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I apologize if this question is "too elementary" - I am not an algebraic geometer. Are the following statements true?

Let $k\subset K$ an extension of algebraically closed fields of characteristic 0, and $f_{1},\ldots, f_{m}\in k[x_{0},\ldots, x_{n}]$ homogeneous polynomials. Consider the subschemes $S_{k}\subset\mathbb{P}^n_{k}$ and $S_{K}\subset\mathbb{P}^n_{K}$ given by $f_{1}=\cdots=f_{n}=0$ (to define $S_{K}$ we just use $k[x_{0},\ldots, x_{n}]\subset K[x_{0},\ldots, x_{n}]$).

Then (1) $\dim S_{k} = \dim S_{K}$, (2) if $\dim S_{k} = \dim S_{K}=0$ then $S(k)=S(K)$.

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    $\begingroup$ Yes, these are both true. For the first, a citable reference is Tag 02NK, but that's way more general. The key point is that the dimension is computed (at least for affine varieties) as the unique $n$ such that there exists a finite map $X \to \mathbf A^n$, and if you have this over $k$, then also over $K$. See also Tag 00P3. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 27 at 1:14
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    $\begingroup$ For the second, they agree set-theoretically (but not scheme-theoretically: their rings of functions differ by $-\otimes_k K$). I'm not sure what is the best way to prove this. Algebraically, what you do is extend the ideal to be radical (which does not change the points over $k$ or $K$), and then prove that $S \cong \coprod_{i=1}^n \operatorname{Spec} k$ for some $n$ (equivalently, $\operatorname{Spec} k^n$, by the Chinese remainder theorem for rings), which after base change becomes $\coprod_{i=1}^n \operatorname{Spec} K$ (for the same $n$). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 27 at 1:27
  • $\begingroup$ Sorry, there surely must be more elementary ways to see these, without involving scheme theory or tensor products. But this is the language in which these types of questions are usually addressed (especially when the fields are not algebraically closed, which is of course more subtle). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 27 at 1:31
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    $\begingroup$ A reformulation of the above comment for (1) is to consider the transcendence degree of the function field. This is one definition of the dimension. It doesn't change after making an algebraic extension. $\endgroup$
    – Daniel Loughran
    Commented Aug 27 at 13:13
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    $\begingroup$ @YCor both fields are assumed algebraically closed. The OP almost certainly means whether $S(k) = S(K)$, which is true for zero-dimensional finite type $k$-schemes. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 27 at 14:35

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