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Let $k$ be a field and let $X$ be a finite type scheme over $k$, explicitly given by finitely many affine patches which are $\mathrm{Spec}$ of finitely generated $k$-algebras, glued along other affine opens which are $\mathrm{Spec}$ of finitely generated $k$-algebras. Is there an algorithm to check whether $X$ is proper over $k$?

I assume the answer is yes, because I've never found it hard to do. But I just realized that both definitions of "proper" involve quantifying over very infinite sets: All $k$-schemes $Y$, if one is using universal closeness, or all maps from $\mathrm{Frac}$ of a valuation ring to $X$ if we use the valuative criterion. (We can make the valuation ring into a dvr, but that's still pretty darn infinite.)

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    $\begingroup$ You can use Chow's Lemma to replace your original finite type, separated $k$-scheme by an (explicitly) quasi-projective $k$-scheme. For a given locally closed subvariety of projective space, it is fairly explicit to work out whether or not it is closed. You do not need to use valuation rings. $\endgroup$
    – Jason Starr
    Commented Feb 10, 2015 at 14:52
  • $\begingroup$ @JasonStarr I was confused by your statement at first, since I thought Chow's lemma had as a hypothesis that $X$ is proper. I imagine what you mean is something like Exercise 18.9.H in Ravi's notes math.stanford.edu/~vakil/216blog/FOAGjan2915public.pdf ? Thanks, I think that works. $\endgroup$
    – David E Speyer
    Commented Feb 10, 2015 at 17:08
  • $\begingroup$ I think any approach (including Jason's, which is quite natural) would involve deciding first whether $X$ is separated. $\endgroup$
    – Laurent Moret-Bailly
    Commented Feb 10, 2015 at 17:29
  • $\begingroup$ @LaurentMoret-Bailly Right, but that I do more or less know how to make computable. If $X = \bigcup U_i$, then the diagonal $\Delta$ is closed in $X \times X$ if and only if $\Delta \cap (U_i \times U_j) \cong U_i \cap U_j$ is closed in $U_i \times U_j$. First check that the given $U_i \cap U_j$ are affine, then check whether $\mathcal{O}(U_i) \otimes \mathcal{O}(U_j)$ generates $\mathcal{O}(U_i \cap U_j)$. $\endgroup$
    – David E Speyer
    Commented Feb 10, 2015 at 17:51

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