Despite the obvious differences in definitions, I have a very strong impression that any theorem whose assumption requiring the morphism to be projective can be replaced by the morphism to be proper.

I learn algebraic geometry theorems mainly from the book by Hartshorne, but for the limited times I consult EGA for a general statement, I realize that the projective condition can always be weakened by properness.

My questions are rather soft:

Is this impression always correct? Or, are there any perticular situations/theorems which I must be very careful about projective assumption? Or, what is on earth the reason make these two concepts so close such that one can usually replaced one by the other.

The followings are something I am aware of:

(1)EGA and Hartshorne have incompatible definitions of projective morphism.

(2)Proper morphism is closed to projective morphism by Chow's lemma. -- However, I had never seen an application of this lemma in a non-conceptual way.

(3)From algebraic geometry perspective, I could understand the projective condition: it benefits people to start proof from a projective space. From complex geometry perspective, I could understand proper condition: properness is a very important tool in analysis -- however, I do not know how does it directly benefit the proof algebraically.

  • $\begingroup$ You bring up Chow's lemma, and say you've never seen it used. It's often time used to effectively carry out your desire. Namely, you reduced from projective to proper. A good example is the adaptation of Serre's proof of GAGA for projective varieties over $\mathbb{C}$ to proper varieties oveR $\mathbb{C}$. $\endgroup$ Feb 7, 2014 at 18:25
  • $\begingroup$ Thank you for pointing out this to me! I did read GAGA (mainly for learning French) and remember Serre mentioned Chow's results somewhere, but it seems that result was more complicated than what I mentioned here... But I might be wrong, and I will keep your comment in mind... $\endgroup$
    – Li Yutong
    Feb 7, 2014 at 22:15
  • 1
    $\begingroup$ The quotient of a projective variety by a finite group always exists, but this fails for proper varieties in general. A projective variety of positive dimension has a non trivial Picard group, this fails for proper varieties in general. You have Bertini type theorem for projective varieties, but not for proper varieties etc... $\endgroup$
    – Cantlog
    Feb 7, 2014 at 22:31
  • $\begingroup$ Are you sure there is no Bertini type theorem for proper varieties? I guess there is a whole bunch of variations of that theorems, if neither of them valid in proper case, then it might the place I should be very careful... $\endgroup$
    – Li Yutong
    Feb 9, 2014 at 1:49

1 Answer 1


It's true that in many cases projective assumptions can be weakened to properness without changing results, but not always. The key difference to my mind is that projective varieties have ample line bundles, but proper varieties need not.

One consequence of this, for example, is that on a projective variety, no effective cycle is numerically trivial; on a proper variety, this need not be the case. (Hartshorne Appendix B has an example.)

  • 1
    $\begingroup$ Thank you, I see what you mean, this is a big difference. $\endgroup$
    – Li Yutong
    Feb 7, 2014 at 16:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .