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Suppose that $f\colon X\to Y$ is a flat morphism of varieties over an algebraically closed field $k$. Let $E\subseteq X$ and $F\subseteq Y$ be closed subvarieties such that $f(E) = F$. Is it true that the restricted morphism $f|_E\colon E\to F$ is also flat? If not, are there some additional conditions on $f$ which would make this true?

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+1 Nice question: I had never asked myself that, and I should have! –  Georges Elencwajg Jun 24 '12 at 19:18
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No, the restriction $f:E\to F$ needn't be flat.

Take $Y=\mathbb A^2, X=\mathbb A^2 \times \mathbb P^1$ and for $f:X\to Y $ take the first projection, which is flat.
Now inside $X$ lies the blow-up $B\subset X$ of $Y$ at the origin $(0,0)\in \mathbb A^2=Y$.
The restricted map $f\mid B: B\to \mathbb A^2$ is well known not to be flat, because all its fibers outside the origin are single points, whereas the fiber at the origin is the one-dimensional projective space $\mathbb P^1$: flat maps do not tolerate such dimension jumps.

As to your second question, I am pessimistic about a general criterion ensuring that the restriction of a flat map will remain flat.
Flatness is a subtle relation between the fibers of a morphism, and I have the feeling that restricting a flat morphism to a subvariety of its domain will usually destroy this relation.

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That is a great answer! But in Gr\"obner degeneration over the affine line, the fiber over $0$ (all the irreducible components over $0$) cannot have dimension larger or smaller than the fiber over a nonzero point, correct? –  math-visitor Jun 24 '12 at 18:27
    
Thanks a lot! If we were very restrictive about $f$, say $f\colon X\to Y$ is a finite surjective morphism between smooth varieties (and hence flat) so that all fibers are dimension $0$, would it still be false that $f\colon E\to F$ is necessarily flat? –  froggie Jun 24 '12 at 18:49
    
@math-visitor: I dont know Gröbner degenerations.You might ask this as a new question. –  Georges Elencwajg Jun 24 '12 at 19:12
    
Dear froggie, you would have to assume that $E$ is smooth too but, even then, why would $F$ be smooth? –  Georges Elencwajg Jun 24 '12 at 19:16
    
No, I guess I didn't mean that either $E$ or $F$ is smooth, just $X$ and $Y$. –  froggie Jun 24 '12 at 19:17
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