Let $\mathbb{K}$ be any field and let $X$ be a proper $\mathbb{K}$-scheme. Does it exist a smooth, proper $\mathbb{K}$-scheme $Y$ and a closed immersion from $X$ to $Y$? This is tautological if $X$ is projective, but what about the general case?
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5$\begingroup$ Francesco's answer is completely correct. It does raise another question, that has been much studied: does there exist an immersion from $X$ into a smooth algebraic space? There are still counterexamples if $X$ is allowed to be non-separated, cf. Edidin-Hassett-Kresch-Vistoli. However, to the best of my knowledge, this question is open for separated $X$ (there is a lovely article of Totaro on this question). $\endgroup$– Jason StarrCommented Sep 30, 2015 at 14:40
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3$\begingroup$ Also, this new question has strong motivation: such immersions are used in the main construction of Todd homomorphisms in the singular Grothendieck-Riemann-Roch theorem of Baum-Fulton-MacPherson. $\endgroup$– Jason StarrCommented Sep 30, 2015 at 14:42
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1 Answer
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For a counterexample over $\mathbb{C}$, it sufficies to take a proper scheme $X$ with trivial Picard group (there are examples among toric varieties).
Such a scheme cannot be embedded in any smooth variety $Y$; in particular, $X$ is necessarily singular.
In fact, assume the contrary and take a dense affine subset $U$ of $Y$. Then we can choose an effective divisor $D_U$ on $U$, whose closure would give a non-trivial Cartier (since $Y$ is smooth) divisor $D_Y$ on $Y$, hence a non-trivial Cartier divisor $D_X$ on $X$, against the assumption $\textrm{Pic}\,X=\{\mathcal{O}_X \}$.
answered Sep 30, 2015 at 14:21
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$\begingroup$ How does $D_Y$ give a non-trivial divisor on $X$ ? Couldn't $D_Y$ be disjoint from $X$ ? $\endgroup$ Commented Mar 17, 2016 at 9:40
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$\begingroup$ Well, we can always take an affine chart $U$ of $Y$ containing a point $x \in X$ and an effective divisor $D_U$ passing through $x$. Then the closure of $D_U$ in $X$ cannot be trivial. I'm missing something? $\endgroup$ Commented Mar 17, 2016 at 11:05
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$\begingroup$ Thanks for your very clear answer, Francesco: as usual you definitely do not miss anything! $\endgroup$ Commented Mar 17, 2016 at 11:25
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$\begingroup$ Dear Georges, you are welcome. $\endgroup$ Commented Mar 17, 2016 at 12:28