While reading Andreas Gathmann's notes on Algebraic Geometry, I stumbled upon this statement: "Projective varieties form a large class of “compact” varieties that do admit such a unified global description. In fact, the class of projective varieties is so large that it is not easy to construct a variety that is not (an open subset of) a projective variety.".

I know that we can sometimes glue affine varieties together and create compact spaces (in fact, Gathmann constructs $\mathbb{P^1}(\mathbb{C})$ as a compactification of $\mathbb{A}^1$). Also affine varieties are not compact unless they are single points. But my question is: is there an example of a variety which is "compact" but not projective?

Gathman does not provide such an example, so maybe someone here can help.

  • $\begingroup$ Well, I think you first have to find out what Gathmann means by "compact" variety. I am pretty sure he is secretly talking about complete varieties. $\endgroup$ Commented Aug 1, 2013 at 11:03
  • $\begingroup$ @NilsMatthes He is in fact surreptitiously talking about complete varieties. So maybe my question can be paraphrased as: is there a complete variety which is not projective? $\endgroup$
    – user39280
    Commented Aug 1, 2013 at 11:08
  • $\begingroup$ Yes, that is exactly what I thought you were asking. $\endgroup$ Commented Aug 1, 2013 at 11:20

1 Answer 1


Such example does not exist in dimension 1. For dimension 3, see the appendix B in Hartshorne, Example 3.4.1

As Liu pointed out below, there is a list discussing related questions in Hartshorne Chapter II Beneath Remark 4.10.2.

  • $\begingroup$ Thank you, I will have a look at it and see whether I understand it. $\endgroup$
    – user39280
    Commented Aug 1, 2013 at 11:10
  • 1
    $\begingroup$ Dear @LiYutong, I remember that Hartshorne mentioned that there is an example of a complex complete singular surface which is not projective in II.4, so your answer has a small mistake. $\endgroup$
    – Yuchen Liu
    Commented Aug 2, 2013 at 16:53
  • $\begingroup$ Thank you! I have edited the answer. $\endgroup$
    – Li Yutong
    Commented Aug 7, 2013 at 2:00
  • 1
    $\begingroup$ There is also an exercise to "construct" one (actually, you merely prove that such a thing must exist as a certain deformation) in Chapter III, but I don't have the book in front of me to give the exact number. $\endgroup$
    – Matt
    Commented Aug 7, 2013 at 2:28

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