are fibres of a flat bijective map reduced?

Let $f: X \to Y$ be a level map of algebraic selections or of intricate analytic rooms which is bijective on shut factors (or simply bijective in the secnond instance). Intend both $X$ and also $Y$ are lowered. Is it real that $f$ has lowered fibers?

If it holds true, I would certainly be most happy for a reference.

2022-07-25 20:45:52
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Answers: 1

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 ├ętale (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.

2022-07-25 21:21:54