most recent 30 from http://mathoverflow.net 2013-06-19T06:24:28Z http://mathoverflow.net/feeds/question/26723 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26723/does-smooth-target-space-and-smooth-fibers-imply-smooth-total-space unknown 2010-06-01T14:32:27Z 2010-06-02T15:03:43Z

<p>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?</p> <p>Update: The answer to the question as posed, is NO. See a comment by Karl Schwede below for a counterexample. </p> <p>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 $y$. </p> <p>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?</p>

http://mathoverflow.net/questions/26723/does-smooth-target-space-and-smooth-fibers-imply-smooth-total-space/26724#26724 Donu Arapura 2010-06-01T14:35:57Z 2010-06-01T15:15:39Z

<p>No. The blow up of a point on the plane provides a counterexample. You need to add flatness.</p> <p>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.</p> <p>10 seconds later: It looks like Karl Schwede has a counterexample below.</p>

http://mathoverflow.net/questions/26723/does-smooth-target-space-and-smooth-fibers-imply-smooth-total-space/26728#26728 Mike Skirvin 2010-06-01T15:19:55Z 2010-06-01T15:19:55Z

<p>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 $\mathbb{A}^1.$</p> <p>I realize that this map is not proper, but I'm sure you could modify this example so that the map is proper.</p>

http://mathoverflow.net/questions/26723/does-smooth-target-space-and-smooth-fibers-imply-smooth-total-space/26745#26745 JT 2010-06-01T17:48:01Z 2010-06-01T17:48:01Z

<p>Here is an example where all the spaces involved are irreducible.</p> <p>Let Y = variety of nilpotent 2 by 2 matrices.</p> <p>X = variety of pairs (N, F) where N is in Y and F is a line preserved by N.</p> <p>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.</p>

http://mathoverflow.net/questions/26723/does-smooth-target-space-and-smooth-fibers-imply-smooth-total-space/26749#26749 David Speyer 2010-06-01T18:46:58Z 2010-06-01T18:46:58Z

<p>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 $\mathbb{A}^N$. </p>