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?
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$\begingroup$ Note that this is true if $S$ is normal, by Zariski's main theorem. $\endgroup$– Damian RösslerCommented Nov 26, 2012 at 8:07
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$\begingroup$ Also note that if you do not assume that $S$ is surjective, then any closed immersion gives a counterexample (Jason Starr's answer is a closed immersion but you can see that he assumed that you wanted the morphism to be surjective). $\endgroup$– Damian RösslerCommented Nov 26, 2012 at 8:10
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3 Answers
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No: $\text{Spec} k[\epsilon]/\langle \epsilon \rangle$ mapping to $\text{Spec} k[\epsilon]/\langle \epsilon^2 \rangle$.
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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.
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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.
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2$\begingroup$ Fiber above the cusp is not reduced. $\endgroup$– R.P.Commented Nov 26, 2012 at 9:39
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$\begingroup$ Yes. You are right. I was thinking only about the isomorphism on points, not the fiber! $\endgroup$ Commented Nov 28, 2012 at 17:24