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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? – Matt May 13 '12 at 15:27
yes, $Y$ is reduced – Dima Sustretov May 13 '12 at 15:33
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? – Matt May 13 '12 at 15:40
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 singularity – Dima Sustretov May 13 '12 at 15:45
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 ... – Matt May 13 '12 at 16:08
up vote 3 down vote accepted

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. – Dima Sustretov May 13 '12 at 15:58
Then consider a ramified cover of Riemann surfaces. – user18119 May 13 '12 at 16:37
QiL, $f$ is bijective. – Dima Sustretov May 13 '12 at 16:40
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. – Dima Sustretov May 13 '12 at 16:42
@DimaSustretov, ah sorry. Consider the Frobenius morphism $x\mapsto x^p$ on the affine line over a characteristic $p>0$ field. – user18119 May 13 '12 at 17:01

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