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The question arises when I am considering something which is similar to the case of isomorphism of fiber bundles.

Assume working over an algebraically closed field $k$. Let $E$ be a locally free sheaf of rank $N+1$ over Y, which is nonsingular, and hence $\mathbb{P}(E)$ be a projective space bundle over $Y$. Let $X$ be a nonsingular variety with a flat morphism $\pi: X\rightarrow Y$.

Now if there is a morphism $f: X\rightarrow \mathbb{P}(E)$ over Y, which induces an isomorphism on each fiber $f_y: X_y \rightarrow \mathbb{P}^N, \forall y \in Y$, then is the morphism $f$ itself also an isomorphism?

I think the statement should be true. In fact, if I can get a compatible local trivialization of $X$ over $Y$ like that of $\mathbb{P}(E)$, then this statement is actually the definition of an isomorhpism between fiber bundles.

I am also considering a more general question. Let $X$, $Y$ and $Z$ be nonsingular varieties, and there is a morphism $f: X\rightarrow Y$ over $Z$ which induces isomorphism on each fiber $f_z: X_z \rightarrow Y_z$. Then when is $f$ an isomorphism itself. Is it enough for me to assume $X\rightarrow Z$ and $Y\rightarrow Z$ are flat (or smooth to be stronger)? or maybe the question is easier when $Z$ is a curve?

I am not sure if there is any direct interpretation of this question in algebra.

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Ziwen
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1 Answers1

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We start with the "critère de platitude par fibre" (Section Tag 039A) which assures you that if $f_z$ is flat and $X$ and $Y$ are flat over $Z$, then $f$ is flat in an open neighbourhood of $X_z$.

Thus if $X$ and $Y$ are flat over $Z$ and everything is of finite type over an algebraically closed field $k$, and if all fibre morphisms $f_z$ for $k$-points $z$ of $Z$ are isomorphisms, then $f$ is a bijective, unramified, flat morphism. An unramified flat morphism of Noetherian schemes is étale. Then $f : X \to Y$ is an étale morphism of finite type $k$-schemes which is injective on $k$-points. and hence an open immersion. A bijective open immersion is an isomorphism. So, yes, $X$, $Y$ flat over $Z$ is enough.

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Count Dracula
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  • Comment deleted as your comment got changed – Count Dracula Apr 26 '14 at 11:30
  • Thank you for your answer, which seems perfect to me. By the way, by $k$-points, do you mean points whose residue field is $k$? – Ziwen Apr 26 '14 at 13:12
  • If $k$ is not algebraically closed, the statement is also true if we take all closed points,right ? –  Apr 26 '14 at 13:15
  • @cant_log It seems so because I don't see where algebraically closedness is used here. Also this reminds me maybe we have to assume $char k$ is 0 in order that $f$ is unramified. Not sure if it is necessary. – Ziwen Apr 26 '14 at 13:39
  • @cant_log Yes, mutatis mutandis. – Count Dracula Apr 26 '14 at 13:45