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

Question

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?

Answer 1

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.

Answer 2

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 $\mathbb{A}^N$.

Answer 3

I apologize for answering such an old question, but it seems fundamental. A classical counterexample occurs for the abel map of a Prym variety with exceptional singularities on the theta divisor. The point is that the fibers of the abel prym map X-->Y of the double cover C'-->C are included among those for the abel map of C', hence are all smooth. (A map obtained by restricting another map over a subvariety of the target has the same fibers.)

Nonetheless X is singular at any exceptional divisor. (see lemma 2.13 of A Riemann singularities theorem for Prym theta divisors, with applications).

The point of the previous paper was that generalizing the Riemann - Kempf singularity theorem to prym varieties is easy when X is smooth. But when X is singular it is considerably harder:

A necessary and sufficient condition for Riemann's singularity theorem to hold on a Prym theta divisor

Singularities of the Prym theta divisor

For a detailed discussion of the case of the abel prym map for a prym variety isomorphic to the intermediate jacobian of the cubic threefold, see:

On parametrizing exceptional tangent cones to Prym theta divisors

The answer is yes however if the target Y is a smooth curve, since X is smooth at any point lying on a smooth cartier divisor, (compare Mumford, chap.7, Prop. 2, redbook.)

Answer 4

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.

Answer 5

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.