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Let $X, Y, S$ be noetherian schemes, $X$ flat and quasi-projective over $S$, $Y$ projective over $S$.

  • Is the hom-functor $T\mapsto\text{Hom}_T(X_T, Y_T)$ representable?

If $X$ is flat and projective, it is, by a standard graph argument to deduce representability from that of the Hilbert functor of $X\times_SY$. If $X$ is proper and flat, then the hom-functor is an algebraic space.

  • If so, is the representing object quasi-projective/projective?
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    $\begingroup$ No, that is not representable. This is worked out explicitly in Claim 3.1 of the following article: arxiv.org/pdf/math/0602646.pdf If the functor is representable for $Y$ projective, then for every Zariski open $U$ in $Y$, the functor is also representable with $Y$ replaced by $U$. Now let $X$ and $Y$ both equal $\mathbb{A}^1_k$. Then the Zariski tangent space of the functor at the identity morphism is a $k$-vector space of countably infinite dimension. If the functor were representable, then the dimension would be either finite or uncountable. $\endgroup$ Jan 12, 2018 at 8:56
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    $\begingroup$ @JasonStarr: A slightly more refined way to get a contradiction is to use Grothendieck's criterion for being locally of finite presentation (which also applies to algebraic spaces) to see that the tangent spaces have to be finite-dimensional in representable cases. $\endgroup$
    – nfdc23
    Jan 12, 2018 at 13:17

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