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I just asked a similar question here: Do there exist manifolds which cannot be smoothly embedded in a compact manifold?

The answer there was pretty topological, so I think this merits its own question. (Though it would be interesting to me if a similar construction could be done in the algebraic category.)

Let $M$ be a smooth affine variety (over an algebraically closed field). Does there exist a smooth projective variety $N$ into which it embeds as an open subset?

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  • $\begingroup$ @JohnMa I don't see any guarantee that the "completion" (closure) will remain smooth. In fact it is easy to build examples where it is not - for example, take the nodal cubic and remove the line containing the singular point. The result is an affine variety whose projective closure is singular. $\endgroup$
    – Elle Najt
    Commented Nov 11, 2015 at 14:41
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    $\begingroup$ In characteristic zero, resolution of singularities will do the trick. $\endgroup$
    – Mohan
    Commented Nov 11, 2015 at 14:44
  • $\begingroup$ @Mohan In the sense that you resolve the singularities of the projective closure? So the general statement is that away from the locus of singular points being resolved the resolution is an isomorphism? $\endgroup$
    – Elle Najt
    Commented Nov 11, 2015 at 14:45
  • $\begingroup$ Yes, that is part of the yoga. You only need to blow up along singular loci to resolve. $\endgroup$
    – Mohan
    Commented Nov 11, 2015 at 14:48
  • $\begingroup$ @Mohan Thanks. It's not obvious to me that this question is equivalent to resolution of singularities, so I will hope that there is an easier argument somewhere. $\endgroup$
    – Elle Najt
    Commented Nov 11, 2015 at 14:58

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It is already mentioned by Mohan in the comments, but the problem is essentially equivalent to resolution of singularities.

If we have resolution (e.g. if $\operatorname{char} k = 0$, or $\dim X \leq 2$), then we can just start with any projective model (e.g. take an affine model and take the closure in the associated projective space). Blowing up will make the variety smooth. (Note: it's not obvious that you can do this without affecting the given variety $X$. You would need to dive into the proof to see whether you may assume that you don't change the stuff that is already smooth. I have not done this.)

On the other hand, if you could prove your statement, then a weak version of resolution would follow. Indeed given any $X$ over $k$ projective, let $X^0$ be the smooth locus (which is nonempty when $k$ is perfect, e.g. algebraically closed). Then your result implies that we can find $Y \supseteq X^0$ smooth projective. This at least gives $$\phi \colon Y \dashrightarrow X$$ birational from the smooth variety $Y$ to $X$. The usual way to extend to an actual morphism from a smooth projective onto $X^0$ is by resolving the graph of $\phi$, which we cannot do (unless we already know resolution).

However, it seems to me that proving your statement could well be a major breaktrough already! (I hope someone will correct me if I'm wrong.)

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