# affine morphism

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?

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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

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.

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You beat me to it. – Karl Schwede Nov 8 '10 at 16:04
Me too (and by a few more minutes than Karl). – Torsten Ekedahl Nov 8 '10 at 16:11
@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
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