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For simplicity let's assume we're working over an algebraically closed field $k$, and all schemes considered are smooth curves over $k$ and the morphisms are nonconstant.

Let's say I have a family $f_t:Y\rightarrow X$ (for $t\in T$) and a fixed map $Z\rightarrow X$. My question is, can we say for an open subset of $t\in T$ the schemes $Y\times_X Z$ taken along $f_t$ are isomorphic?


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Dear Randy, What do you mean by a flat family $f_t$? I guess you are supposing that there exists a morphism $f:Y \times T \to X$, but now what is assumed to be flat? Regards, – Matt E Sep 22 '11 at 20:44
@Matt: Yes you're of course right, my kneejerk reaction to add flat didn't quite make sense. I've edited the question accordingly. – user16544 Sep 22 '11 at 21:20
up vote 3 down vote accepted

The anwer is no. To see this, imagine that we have that $X = Y\times T$ (everything taken over a field, say), and $f_t$ is just the inclusion of the fibre $Y\times \{t\} \hookrightarrow Y\times T$. Take $Z$ to be some closed subscheme of $Y\times T$. We are then asking if the intersection $Z\cap (Y\times \{t\})$ is constant for an open subset of $t$.

To get a counterexample, let $Y$ be $\mathbb P^n$ (for some $n$), let $T$ be the $\mathbb P^N$ that parameterizes degree $d$ hypersurfaces in $Y$ (for some $d$), and let $Z \hookrightarrow Y\times T$ be the universal family of degree $d$ hypersurfaces. Then $Z \cap (Y\times \{t\})$ is the particular degree $d$ hypersurface corresponding to the parameter $t$, and (if $d$ is large enough) the isomorphism class of this hypersurface won't be constant on any open subset of $t$s. (One could take $n = 2$ and $d = 3$ to get a concrete example.)

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Okay thanks. char limit – user16544 Sep 22 '11 at 22:58

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