# A variety always has a compactification. Is there an easy proof?

This is of course a special case of the more difficult question on compactifiable morphisms.

I would like to know if there is an easier proof of the following fact:

Every algebraic variety $X$ admits an open embedding into a proper variety.

It may be just an easy homework exercise, but I'm not sure how to proceed.

-
Even this special case of the compactification theorem is quite deep, so may as well just deduce it from the general case. For the generalization to algebraic spaces, the case over a field was addressed by Raoult a long time ago, and the general case (for separated finite type maps between quasi-compact quasi-separated alg. spaces) was settled just in the past 2 or 3 years. – BCnrd Apr 24 '10 at 16:28
Where does the subtlety lie? For quasiprojective ones one can simply take the closure in some embedding, no? – Mariano Suárez-Alvarez Apr 24 '10 at 16:37
Mariano: Presumeably, the definition "variety" at work here the intrinsic definition: a variety is a separated reduced [integral] scheme of finite type over an [algebraically closed] field. The terms in brackets, as I understand it, are stronger requirements imposed by some authors, including Mumford and Hartshorne. For an excellent account of the (more restrictive) definition that does not invoke schemes, see Chapter I of Mumford's Red Book. – Charles Staats Apr 24 '10 at 16:47
@unknown(google): it isn't easy beyond q-proj case. Proof becomes simpler in minor ways for integral schemes of f.t. over field. (Base scheme doesn't change in argument, and method is characteristic-free, so I don't think working over C adds simplicity.) I don't remember if significant simplifications occur, or if non-reduced or reducible schemes still intervene, but to sort that out just sit down with the proof in general and follow it through in the special case you care about. (My exposition of Deligne's notes separates off the non-noetherian aspects, so you can ignore that if you prefer.) – BCnrd Apr 24 '10 at 17:23
Martin, Nagata wrote an entire first paper on the case over a field, and already that paper was tough going. I suggest you think more carefully about the meaningfulness of the strategy you are suggesting to use beyond the affine case. – BCnrd Apr 24 '10 at 19:12