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Does every proper algebraic space (over a field, say) admit a closed immersion into a smooth proper algebraic space?

Remark: Of course, if we say "projective" instead of "proper" then the answer is tautologically "yes": we can take the ambient variety to be some projective n-space. But I'm curious about the general case.

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    $\begingroup$ No : there are proper normal surfaces $S$ with trivial Picard group (see for instance [Stefan Schröer, On non-projective normal surfaces]). If you could embed $S$ in a smooth and proper $X$, you could choose a Weil (hence Cartier) effective divisor in $X$ intersecting $S$ but not containing it, showing that the Picard group of $S$ is not trivial. $\endgroup$
    – Olivier Benoist
    Nov 7, 2014 at 15:04
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    $\begingroup$ Are you sure that this exclude the possibility of embedding $S$ into a smooth algebraic space? It seems to me that this argument only works if $X$ is a scheme, or I am missing something? $\endgroup$
    – Francesco Polizzi
    Nov 7, 2014 at 15:12
  • $\begingroup$ @FrancescoPolizzi Ah you are absolutely right ! Sorry ! $\endgroup$
    – Olivier Benoist
    Nov 7, 2014 at 15:24

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I do not know the general answer to this interesting question.

However, there are surely examples of proper algebraic varieties that admit no embeddeding into smooth schemes.

Look for instance to this paper by Roth and Vakil, page 11.

Note that in the paper the authors claim that their example admits no embedding into a smooth algebraic space; however, their proof actually only work for smooth schemes, see the corrigendum in Vakil's webpage here.

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  • $\begingroup$ Thanks for the comments and pointer to the literature. I guess the answer may not be known -- too bad! $\endgroup$
    – Dustin Clausen
    Nov 10, 2014 at 9:45

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