# Quasi-finite + separated but not finite morphism

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

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An open embedding that is not closed. I don't how "interesting" that is. –  Angelo Sep 24 '10 at 4:54
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 '10 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)$is a finite morphism. Right? – Sebastian Petersen Sep 28 '10 at 7:30 Sorry, something went wrong above. I cannot delete / edit the comment. I do not understand, what this$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)$is a finite morphism. Right? – Sebastian Petersen Sep 28 '10 at 7:33 It does not have to be a finite extension. – John Doe Oct 9 '10 at 2:44 ## 1 Answer Dear Dung, a pleasantly geometric example of a quasi-finite, separated, but not finite morphism is the projection of the hyperbola$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)

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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 '10 at 22:40
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 '10 at 2:13
I should have added "surjective", but BCnrd's comment answered my question. –  John Doe Oct 9 '10 at 2:41