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Consider the following fact from algebraic geometry:

Any (complex) smooth algebraic variety can be embedded into a complete smooth variety as a locally closed set.

I know how to prove this fact using Hironaka's and Nagata's theorems. However it looks much more elementary (at least more elementary then Hironaka's theorem). In particular, it is obvious for quasi-projective varieties.

So here are my questions:

  1. Does this fact have an elementary proof?

  2. Does anybody know the analogous fact for ($C^\infty$) Nash manifolds?

Thank you very much.

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How does this follow from Hironaka's and Nagata's theorems? I can only imagine the following. Take a compactification of your smooth variety (Nagata). If this compactification turns out to be smooth, we're done. If not, take a desingularization. But how do you get an open immersion into the desingularization? I must be missing something very elementary.... – Harry Apr 9 '12 at 20:13
Harry, if $\bar{X}$ is the compactification of $X$, the desingularization only modifies ${\bar X} \setminus X$. Thus $X$ embeds into it as an open set! – Karl Schwede Apr 9 '12 at 20:39
Dear Rami, I don't think 1. has an elementary answer (at least, it can be no more elementary than Nagata's theorem over $\mathbb{C}$). Indeed, suppose that you had a new way to prove 1., then you could also prove Nagata's theorem as follows. Embed $X$ into $Y$ as a locally closed set, take the closure of $X$ in $Y$ and you are done. – Karl Schwede Apr 9 '12 at 20:43
What is a Nash manifold? – Igor Rivin Apr 9 '12 at 21:39
Hi Karl, Thank you very much. Does Nagata's theorem for smooth varieties is as complicated as for general varieties? If not can one at list avoid Hironaka? e.g. Can embed any complete singular variety to a complete smooth one? Can one do it in an elementary way? Thanks again – Rami Apr 10 '12 at 21:41

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