Welcome to MathOverflow

Question

2

1

What is an interesting example of that? Things like $Spec(K) \to Spec(L)$ do not count cause they are not interesting.

edited Sep 24 2010 at 4:59

asked Sep 24 2010 at 4:47

Dung Nguyen
157●6

2

An open embedding that is not closed. I don't how "interesting" that is. – Angelo Sep 24 2010 at 4:54

5

Dear Dung: Are you aware of Zariski's Main Theorem, which in one form says that Angelo's example and your example are nearly the only examples? To be precise, a quasi-finite separated map $X \rightarrow S$ with $S$ quasi-compact and quasi-separated (e.g., noetherian) always factors as an open immersion into a finite $S$-scheme. So by the transitive property of "not interesting", can we say that all quasi-finite separated maps are "not interesting" (at least locally on the base)? I'd hope not, because I can think of interesting ones (e.g., torsion in semi-abelian schemes). – BCnrd Sep 24 2010 at 5:37

I do not understand, what this ``$\Spec(K)\to\Spec(L)'' example should be. I assume, that $L$ is a field and $K/L$ a field extension. But then we do have: If $\Spec(K)\to\Spec(L)$ is quasi-finite, then $K/L$ is a finite extension and $\Spec(K)\to\Spec(L)$ is a finite morphism. Right? – Sebastian Petersen Sep 28 2010 at 7:30

Sorry, something went wrong above. I cannot delete / edit the comment. I do not understand, what this $Spec(K)\to Spec(L)$ example should be. I assume $L$ is a field and $K/L$ a field extension. But then we do have: If $Spec(K)\to Spec(L)$ is quasi-finite, then $K/L$ is finite and $Spec(K)\to Spec(L)$ is a finite morphism. Right? – Sebastian Petersen Sep 28 2010 at 7:33

It does not have to be a finite extension. – Dung Nguyen Oct 9 2010 at 2:44

Georges Elencwajg · Answer 1 · 2010-09-24 22:32:05Z UTC

1

Dear Dung, a pleasantly geometric example of a quasi-finite, separated, but not finite morphism is the projection of the hyperbola $xy=1$ in the affine $x,y$ plane on the $x$-axis. Its image is the affine line minus the origin. It is clearly quasi-finite (even injective) but not finite, since its image is not closed . (Also, morphisms between affine schemes are separated)

answered Sep 24 2010 at 22:32

Georges Elencwajg
24.3k●65●132

My only excuse for posting an example subsumed under BCnrd's comment is that I am, alas, completely unable to think spontaneously at his level of generality and that only simple pictures like the above come to my mind. – Georges Elencwajg Sep 24 2010 at 22:40

1

Dear Georges: Since your example is an isomorphism onto the punctured affine line, it is an open immersion (as we can also see algebraically). In that sense, it meshes better with Angelo's comment than with mine. – BCnrd Sep 28 2010 at 2:13

I should have added "surjective", but BCnrd's comment answered my question. – Dung Nguyen Oct 9 2010 at 2:41

Quasi-finite + separated but not finite morphism

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.