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Let $X$ be a normal algebraic variety defined over $\mathbb{C}$. Is it possible to compactify $X$ to a normal projective variety? And if $X$ is smooth, is it possible to compactify $X$ to a smooth (or normal) projective variety?

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The answer is no. For example, $X$ could be a normal(or smooth) complete (and thus compact) algebraic variety which is not projective. Not too many options for compactifications there. – Mohan Sep 16 '10 at 17:15

With regards to the first question, the answer is yes as long as the variety is quasi-projective, meaning, it has some projective compactification (for example, if you started with a complete non-projective variety, for example, Hartshorne, Chapter II, Exercise 7.13, then the answer would be no, as Mohan Kumar pointed out above):

Here's the solution. Suppose that $X$ has some compactification inside projective space $\bar{X}$. Let $X^N$ be the normalization of $\bar{X}$ with normalization map $f$ (which is finite), and we need to show that $X^N$ is also projective. We will show that $X^N$ has a ample line bundle. Take $L$ to be an ample line bundle on $\bar{X}$, by Hartshorne, Chapter III, exercise 5.7(d), $f^* L$ is also ample and thus ${X}^N$ is projective.

With regards to the second question (smooth), the answer is also yes (with the same caveat).

Here's the solution. Take any embedding $\bar{X}$ into projective space. We know that $\bar{X}$ is only (possibly) singular on $\bar{X} \setminus X$. Now, apply virtually any resolution of singularities algorithm (from Hironaka, to Bierstone-Milman, to Villamyor, to Wlodarczyk, to ...). This will only blow-up over the singular points of $\bar{X}$ and thus gives you the desired compactification (note, blow-ups take projective varieties to projective varieties).

For many other classes of singularities (like rational singularities), this is an open problem. For some discussion of the rational singularities question, see chapter 12 of Koll\'ars "Shaferavich Maps and Automorphic Forms."

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What about if $X$ is not quasi-projective, but we only ask for our compactification to be proper, not projective? – David Speyer Sep 16 '10 at 17:42
David, you are asking if the normalization of a proper variety is still proper? Normalization in this setting is a proper map so composition of proper = proper. Likewise with resolutions of singularities. – Karl Schwede Sep 16 '10 at 17:46
Thank you very much for your answer. – Yemon Dai Sep 16 '10 at 17:50
Ah, sorry, I misunderstood. What you describe is an old theorem of Nagata. I think the following is the right reference (although perhaps the toric approach might be simpler?).… – Karl Schwede Sep 16 '10 at 18:07
Brian Conrad wrote up a modern account of Nagata's theorem (based on notes of Deligne), that are available at his web-site: Just glancing over his web-site, I also noticed that there is a paper (joint with Lieblich and Olsson) on Nagata compactification for algebraic spaces. – Emerton Sep 17 '10 at 3:31

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