Does finite+reduced fibers+connected fibers imply isomorphism? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:02:00Z http://mathoverflow.net/feeds/question/114382 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114382/does-finitereduced-fibersconnected-fibers-imply-isomorphism Does finite+reduced fibers+connected fibers imply isomorphism? HNuer 2012-11-24T23:08:21Z 2012-11-25T21:39:02Z <p>Suppose I have a morphism of Noetherian schemes over a field $k$ (if one needs this then assume $k$ is algebraically closed) $f:C'\rightarrow S$ which is finite with geometrically connected and reduced fibers. Is $f$ an isomorphism?</p> http://mathoverflow.net/questions/114382/does-finitereduced-fibersconnected-fibers-imply-isomorphism/114385#114385 Answer by Jason Starr for Does finite+reduced fibers+connected fibers imply isomorphism? Jason Starr 2012-11-24T23:39:15Z 2012-11-24T23:39:15Z <p>No: <code>$\text{Spec} k[\epsilon]/\langle \epsilon \rangle$</code> mapping to <code>$\text{Spec} k[\epsilon]/\langle \epsilon^2 \rangle$</code>.</p> http://mathoverflow.net/questions/114382/does-finitereduced-fibersconnected-fibers-imply-isomorphism/114425#114425 Answer by Mohan for Does finite+reduced fibers+connected fibers imply isomorphism? Mohan 2012-11-25T16:36:09Z 2012-11-25T16:36:09Z <p>Though any closed embedding will give a counter example, it is true if you ask whether it is an isomorphism to the (scheme theoretic) image. In other words, the map is an isomorphism from $C'\to f(C')$ where $f(C')$ is thought of as the scheme theoretic image (which makes sense since the map is assumed to be finite), with $k$ algebraically closed.</p> http://mathoverflow.net/questions/114382/does-finitereduced-fibersconnected-fibers-imply-isomorphism/114455#114455 Answer by Ray Hoobler for Does finite+reduced fibers+connected fibers imply isomorphism? Ray Hoobler 2012-11-25T21:39:02Z 2012-11-25T21:39:02Z <p>Or, perhaps more to the point, the map from the normalization to a cusp on a curve. Note that such a map is an isomorphism on points without being an isomorphism of schemes.</p>