Let $U$ be a smooth quasiprojective variety. Does there always exist a smooth compactification of $U$? If not always when can we have smooth compactification? In particular, suppose $X$ is a singular projective variety and $U$ is the smooth locus. The question is does there always exist a smooth projective variety $Y$ containing $U$? If we add the condition that $U$ is open in $Y$ is $Y$ uniquely determined upto isomorphism?

9$\begingroup$ You should search for "resolution of singularities" to answer the question about existence. About uniqueness: in dimension at least 2, no, because of blowups. $\endgroup$ – user5117 Mar 11 '13 at 21:14
The main idea is this: You can always find an $X$ such that $X\supseteq U$ and $X\setminus U\supseteq \Sigma := \mathrm{Sing} X$. Then in characteristic zero apply Hironaka's resolution theorem, which says that there exists a resolution of singularities $\pi:Y\to X$ such that $\pi$ is an isomorphism over $X\setminus \Sigma\supseteq U$. In particular, $\pi^{1}: U\hookrightarrow Y$ gives an embedding.
In positive characteristic resolution is not known in general and similarly this embedding result is not known either (although I am not saying that knowing this would prove the existence of resolutions).
There is actually a newer, expanded version of Kollár's paper in book form: Lectures on Resolution of Singularities.
And of course it is not unique as long as $U$ itself is not projective, since as you can always blowup $Y$ outside of $U$.

1$\begingroup$ A remark: though $U$ doesnotdetermine $Y$, for any cohomology theory that factorizes through Voevodsky′s motives the image $H(Y)\to H(U)$ is canonical and functorial (this is the zeroth level of the weight filtration). One can also look at weight complexes. In characteristic $p$ one can prove a somewhat similar result for any cohomology whose target is a $Z[1/p]$linear category. $\endgroup$ – Mikhail Bondarko Mar 12 '13 at 8:59

$\begingroup$ @Mikhail: I'm sorry, but I don't see your point. Could you explain? $\endgroup$ – Sándor Kovács Mar 12 '13 at 16:03

$\begingroup$ So, smooth compactifications are certainly not unique; yet this nonuniqueness can be controlled to a certain extent. $\endgroup$ – Mikhail Bondarko Mar 12 '13 at 17:35

$\begingroup$ @Mikhail, I'm sorry, but I still don't understand how this is more control. Perhaps my problem is what Artie mentions that I can't read half of your comment... $\endgroup$ – Sándor Kovács Mar 12 '13 at 21:12

In characteristic 0, yes, yes, no. Check out Kollar's paper:
http://arxiv.org/abs/math/0508332
In characteristic p, unknown.