Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?


share|improve this question
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. –  Ryan Sep 22 '11 at 21:20
add comment

1 Answer

up vote 2 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.)

share|improve this answer
Okay thanks. char limit –  Ryan Sep 22 '11 at 22:58
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.