Suppose $X$ is a projective variety and $f:X\to Y$ a morphism, is the image $f(X)$ projective?(the schematic image is well-defined in this case, or the induced reduced structure on the closed subset $f(X)$) Hartshorne says this property holds for properness. Usually how can we claim a variety(scheme) is projective?
Asked
Active
Viewed 808 times
6
-
1Let me supplement Georges' excellent answer by mentioning the paper Conducteur, descente et pincement http://www.math.jussieu.fr/~dferrand/Conducteur.pdf by D. Ferrand. In Section 6 he gives examples of nonprojective proper schemes obtained by glueing together two rational curves in $\mathbf P^3$. – May 06 '14 at 11:13
-
Thanks for the interesting link, Asal. Ferrand's pinchings ("pincements") look like a good way to construct schemes. I wonder if they appear elsewhere in the literature. Did you notice that the article was published 32 years after its conception ? – Georges Elencwajg May 06 '14 at 11:33
-
@GeorgesElencwajg: Indeed I did. That gives me hope for some of my own papers! – May 06 '14 at 11:35
-
Dear @Asal: I'm certainly looking forward to reading you (but hopefully in less than 32 years) ! – Georges Elencwajg May 06 '14 at 11:37
1 Answers
10
No, the image of a projective variety is not always projective.
Indeed Chow's lemma states that given any complete irreducible variety $X$, there exists a projective variety $X'$ and a birational surjective morphism $f:X'\to X$.
Choosing for $X$ a complete irreducible variety that is not projective thus gives an example of a projective variety $X'$ whose image $X$ is not projective.
To be complete (!), let me mention that an example due to Hironaka of a complete but not projective smooth $3$-dimensional variety is given in Shafarevich's Basic Algebraic Geometry, volume 2, page 75.
Georges Elencwajg
- 150,790