# Morphism of varieties with all fibers isomorphic

Let $$k$$ be an algebraically closed field. Let $$f:X \to Y$$ be a morphism of $$k$$-algebraic schemes, that is separated schemes of finite type over $$k$$.

Furthermore:

1) $$Y$$ is a smooth variety (that is, it is integral too) over $$k$$ resp. regular.

2) All fibers $$X_y$$ for $$y$$ a closed point of $$Y$$ are isomorphic to a smooth/regular variety $$F$$.

Is it true then, that $$X$$ is a smooth/regular variety? If not in general, is it true if $$F$$ is a group variety over $$k$$?

It would be enough of course, to show, that $$X$$ is smooth over $$Y$$ or even only that $$f$$ is flat, or that $$X$$ is Cohen-Macaulay but at the moment I find no way to prove even this.

• I think you can construct some counterexample the following way. Take $X=\operatorname{Spec}k[x,y]/(xy) \sqcup \mathbb{G}_m\times\mathbb{A}^1\sqcup\{(0,0)\}$ and $Y=\mathbb{A}^1$. Consider the map induced by the first projection. Then each fiber is isomorphic to a copy of $\mathbb{A}^1\sqcup\{0\}$. So each fiber is smooth and isomorphic to each other (but not irreducible and the irreducible components do not have the same dimension) but $X$ is not smooth. – Roland Feb 9 at 19:15