Example of inclusion which is not a finite morphism - MathOverflow [closed]most recent 30 from http://mathoverflow.net2013-06-19T20:50:52Zhttp://mathoverflow.net/feeds/question/17658http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/17658/example-of-inclusion-which-is-not-a-finite-morphismExample of inclusion which is not a finite morphismPaul Yuryev2010-03-09T21:41:54Z2010-03-10T01:00:42Z
<p>Every closed immersion is a finite morphism. Can you give an example of quasi-projective varieties $X\subset Y$ such that inclusion $X\hookrightarrow Y$ is not finite? Same with Y projective?</p>
<p>Thanks!</p>
<p><strong>Edit:</strong> Sorry this question is very simple, I made a mistake asking the question. For a corrected version, check out <a href="http://mathoverflow.net/questions/17678/example-of-restriction-of-a-finite-morphism-which-is-not-finite" rel="nofollow">this one.</a></p>
http://mathoverflow.net/questions/17658/example-of-inclusion-which-is-not-a-finite-morphism/17660#17660Answer by Qing Liu for Example of inclusion which is not a finite morphismQing Liu2010-03-09T21:54:59Z2010-03-09T21:54:59Z<p>An open immersion is never finite unless it is also a closed immersion (for finite morphisms are proper). So you just need to take a non-empty open subset $X$ which is not a connected component in $Y$. </p>