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Let $f: X \to Y$ be a flat map of algebraic varieties or of complex analytic spaces which is bijective on closed points (or just bijective in the secnond case). Suppose both $X$ and $Y$ are reduced. Is it true that $f$ has reduced fibres?

If it is true, I would be most grateful for a reference.

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    Is $Y$ reduced?2012-05-13
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    yes, $Y$ is reduced2012-05-13
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    I haven't tried fleshing this out into a proof, but my first thought is that there seems to be enough conditions to argue that $f$ must induce isomorphisms on local rings. Since Y reduced, its local rings have no zero divisors. Since iso, X local rings have no zero divisors, but now relate this to the local rings of the fibers to get that the fibers are reduced. Maybe?2012-05-13
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    one must use flatness at some point, because the normalisation of a cuspidal curve is a bijection but with a non-reduced fibre at the singularity2012-05-13
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    I assumed it will come up in the argument that you have an iso on local rings (which is equivalent to flat and unramified). The cuspidal curve example doesn't violate this since you don't have such an iso. Georges example points out nicely that ramification might happen, so I'm not sure still ...2012-05-13
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    why would you have such an isomorphism? are you speaking abouth the algebraic or analytic case?2012-05-13
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    My idea was purely intuitive off of the following thoughts: If f is unramified, then you do have such an iso. Geometrically, ramification happens when "fibers come together". Your conditions force the map to be of relative dimension 0, and moreover all closed fibers to be singletons, so with that picture in mind fibers can't come together. Thus it seems to me the map is unramified. This could be a situation where my intuition is leading me astray.2012-05-13

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Let $f: X\to Y$ be a bijective flat morphism of reduced algebraic varieties over $\mathbb C$ (or any algebraically closed field $k$ of characteristic $0$), then $f$ is an isomorphism.

First $f$ is quasi-finite, hence finite and étale (because characteristic $0$) above some dense open subset $V$ of $Y$. As we work over an algebraically closed field, $f^{-1}(V)\to V$ is then an isomorphism.

Let $x\in X$ and $y=f(x)$. Then $O_{Y,y}\to O_{X,x}$ is flat, hence faithfully flat, therefore injective. This implies that the quotient $O_{X,x}/O_{Y,y}$ is flat over $O_{Y,y}$. But the total rings of fractions of $O_{Y,y}$ and $O_{X,x}$ coincide because $X\to Y$ is birational by the above. So $O_{X,x}/O_{Y,y}$ is of torsion over $O_{Y,y}$, hence equal to $0$. So $f$ is an open immersion. But $f$ is surjective, it is an isomorphism.

The proof should work for reduced complex analytic spaces.

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If $D=Spec(\mathbb C[T]/T^2)$ is the double point and $P$ is the simple point, the morphism of analytic spaces $D\to P$ has reduced base, is flat and bijective but has non reduced fiber.

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    ok, I am sloppy again. Let both $X$ and $Y$ be reduced.2012-05-13
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    Then consider a ramified cover of Riemann surfaces.2012-05-13
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    QiL, $f$ is bijective.2012-05-13
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    let me put it this way: I know of no example of a ramified bijective flat cover, but do not know how to prove that no such example exists.2012-05-13
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    @DimaSustretov, ah sorry. Consider the Frobenius morphism $x\mapsto x^p$ on the affine line over a characteristic $p>0$ field.2012-05-13