## Does smooth target space and smooth fibers imply smooth total space?

Suppose that $f: X \rightarrow Y$ is a morphism between algebraic varieties. If $Y$ is smooth, and the fibers of $f$ over closed points of $Y$ are proper and nonsingular, does it follow that $X$ is smooth?

Update: The answer to the question as posed, is NO. See a comment by Karl Schwede below for a counterexample.

Modified question: Let $f$ be a surjective morphism of algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field). Let $x \in X$ be a closed point and let $y = f(x)$. Just because the fiber $f^{-1}(y)$ is smooth does not mean $X$ is smooth at $x$. What if $X \times_Y Spec \mathcal{O}_y/m^n$ is smooth over $Spec \mathcal{O}_y/m^n$ for every positive integer n - is $X$ smooth at $x$? Here $m$ is the maximal ideal of the local ring at $y$.

Is there any condition on $f$ or the fibers which will guarantee smoothness of the total space? Flatness plus smooth fibers is one, is there anything weaker?

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 In the modified question, the scheme-theoretic fibre of f over $\mathcal{O}_{y}/\mathfrak{m}^n$ will be nonreduced for n>1, and hence not a regular scheme --- am I misinterpreting the question? – Mike Roth Jun 1 2010 at 19:53 No, it wasn't carefully thought out. Thanks - now it is more open ended. – unknown Jun 1 2010 at 20:31

An even more basic example: take $X$ to be any singular affine variety, and $f$ to be the inclusion of $X$ into the affine space $\mathbb{A}^N$.

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 That is a good point, I should probably include surjetive. – unknown Jun 1 2010 at 19:14

Here is an example where all the spaces involved are irreducible.

Let Y = variety of nilpotent 2 by 2 matrices.

X = variety of pairs (N, F) where N is in Y and F is a line preserved by N.

Let f : X -> Y be the natural projection. Now X is certainly smooth (as the projection to P^1 is a smooth morphism) and the fibres of f are points or P^1's. But Y is not regular.

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 That's backwards. He wants an example where Y is smooth but X is not. – David Speyer Jun 1 2010 at 18:08

I think the answer is no. Consider the case where $X$ is the two coordinate axes in $\mathbb{A}^2$ (corresponding to the ring $\mathbb{C}[x,y]/(xy)$) and $f$ is the projection onto the first axis (corresponding to $\mathbb{C}[x] \to \mathbb{C}[x,y]/(xy)$). Then the fibers of this map are a point, except over zero where the fiber is an $\mathbb{A}^1.$

I realize that this map is not proper, but I'm sure you could modify this example so that the map is proper.

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 Nice example, but for me, an algebraic variety includes irreducible... – unknown Jun 1 2010 at 15:37

No. The blow up of a point on the plane provides a counterexample. You need to add flatness.

Added: It seems I answered something different from what was asked. Perhaps someone can answer the actual question, which isn't so clear to me.

10 seconds later: It looks like Karl Schwede has a counterexample below.

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The blowup of a smooth variety along a smooth subvariety is still smooth though? – unknown Jun 1 2010 at 14:41
I'm asking about the domain X, not the morphism f. – unknown Jun 1 2010 at 15:00
OK, sorry. I misread your question. – Donu Arapura Jun 1 2010 at 15:06
Modifying Donu's example very slightly, blow up the ideal $(x^2, y^2)$. You have two charts, one is $k[x, y, (y/x)^2] = k[a,b,c]/(a^2*c - b)$ and the other is the other obvious (and symmetric) one. This chart isn't smooth, the fiber over the closed point $(x, y)$ is $k[c]$. – Karl Schwede Jun 1 2010 at 15:11
Ah, I think you are right (though you mean b^2 in your formula). Thanks, is there any other condition on the situation (besides flatness) which would guarantee that the domain is smooth or is it just hopeless? – unknown Jun 1 2010 at 15:33