# 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?

• 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? Jun 1, 2010 at 19:53
• No, it wasn't carefully thought out. Thanks - now it is more open ended. Jun 1, 2010 at 20:31

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.

• The blowup of a smooth variety along a smooth subvariety is still smooth though? Jun 1, 2010 at 14:41
• I'm asking about the domain X, not the morphism f. Jun 1, 2010 at 15:00
• OK, sorry. I misread your question. 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 $k[c]$. 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? Jun 1, 2010 at 15:33

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 $\mathbb{A}^N$.

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

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.)

• Cool stuff! Cheers Jan 22, 2019 at 21:25

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.

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

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.

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