Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f:X\to Y$ be a morphism of varieties over a field $k,$ such that $X(\overline{k})\to Y(\overline{k})$ is bijective. Is $f$ necessarily an affine morphism?

share|improve this question
    
Dear shenghao: Perhaps clarify that for this question, "variety" means "geometrically irreducible", right? –  BCnrd Nov 8 '10 at 15:49
    
Dear Brian: is there an obvious counter-example if I don't assume geometric irreducibility? –  shenghao Nov 12 '10 at 16:29
add comment

1 Answer

up vote 9 down vote accepted

No. Here is an example: Let $g:\mathbb A^2\to Y$ be a morphism that glues two closed points $P$ and $Q$ together and otherwise it is an isomorphism. Now let $X=\mathbb A^2\setminus \{P\}$ and $f$ the restriction of $g$ to $X$.

If you add to the conditions that $f$ is projective, then the statement is true, because then $f$ is finite and hence affine.

EDIT: A minute ago there was another question asking how to define $Y$. It has now disappeared, but perhaps it is still interesting to include some references.

1) See this MO question and the discussion.

2) See this paper of Karl Schwede. Especially, Theorem 3.4 in general and Corollaries 3.6 and 3.9 in particular.

3) Try to construct it directly. If you get stuck, look at Karl's paper and try to carry out the computation in this special case. It might be a worthy exercise if you have never done anything like this before.

share|improve this answer
2  
You beat me to it. –  Karl Schwede Nov 8 '10 at 16:04
1  
Me too (and by a few more minutes than Karl). –  Torsten Ekedahl Nov 8 '10 at 16:11
1  
@Karl&Torsten :) –  Sándor Kovács Nov 8 '10 at 16:14
    
@Karl: just so you look at this... Do you have anything to add to the EDIT above? –  Sándor Kovács Nov 9 '10 at 4:21
2  
I can add the explicit formula in my (aborted) answer: If one glues $(0,\pm1)$, then the affine algebra of $Y$ is $k[x,y^2+1,y^3+y]$. –  Torsten Ekedahl Nov 9 '10 at 6:36
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.