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 $k$.


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.

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    $\begingroup$ 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. $\endgroup$ – Roland Feb 9 at 19:15

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