For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:36:37Z http://mathoverflow.net/feeds/question/32083 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32083/for-a-morphism-f-from-a-regular-scheme-should-there-exist-an-open-subscheme-u-of For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular. Mikhail Bondarko 2010-07-15T23:15:18Z 2010-08-25T14:37:43Z <p>For a finite type morphism $f:X\to S$, $X$ is a regular scheme, should there always exist an open (dense) subscheme $U\subset S$ such that the fibre of $f$ at each Zariski point of $U$ is regular? All schemes are excellent.</p> <p>If the anwser is 'yes', then: could one choose such an $U$ such that the preimage of any regular subscheme of $U$ is regular? Are these conditions on $U$ equivalent?</p> http://mathoverflow.net/questions/32083/for-a-morphism-f-from-a-regular-scheme-should-there-exist-an-open-subscheme-u-of/32096#32096 Answer by David Speyer for For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular. David Speyer 2010-07-16T00:39:29Z 2010-08-25T14:37:43Z <p>Over a field of characteristic zero, your result is true. This is Corollary III.10.7 in Hartshorne. </p> <hr> <p>In characteristic $p$ no. The simplest example is to take $k$ an algebraically closed field and map $\mathbb{A}^1_k$ to itself by $x \mapsto x^p$. For every $t \in k$, the fiber above $t$ is $\mathrm{Spec}\ k[x]/(x^p-t) \cong \mathrm{Spec} \ k[y]/y^p$ where the isomorphism is $x-t^{1/p} = y$. So every fiber is singular.</p> <p>There is a more interesting counter-example due to Serre: Let $k$ be an algebraically closed field of characteristic $2$. Consider the planar cubic <code>$$F_{a,b,c}(x,y,z) :=a (y^2 z + y z^2) + b (x^2 z + x z^2) + c (x^2 y + x y^2) \quad (*)$$</code> We leave it to the reader to check that $F=0$ is singular at $(\sqrt{a}:\sqrt{b}:\sqrt{c})$. Generically, the singularity is a <strike>node</strike> cusp. Choose $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ in $k^3$ and try to map $\mathbb{P}^2_k \to \mathbb{P}^1_k$ by <code>$$(x:y:z) \mapsto (F_{(a_1,\ b_1,\ c_1)}(x,y,z) : F_{(a_2,\ b_2,\ c_2)}(x,y,z))$$</code> Then the fiber over $(t_1:t_2)$ is <code>$F_{(t_1 a_1+t_2 a_2,\ t_1 b_1+t_2 b_2,\ t_1 c_1+t_2 c_2)}=0$</code> which, as we just computed, is singular. More precisely, the above map is undefined at the $9$ points where $F_{(a_1,\ b_1,\ c_1)} = F_{(a_2,\ b_2,\ c_2)} =0$. But, if we take $X$ to be $\mathbb{P}^2$ blown up at those $9$ points, then we get a map from the regular $X$ to the regular $S$, where every fiber is a <strike>nodal</strike> cuspidal cubic or worse.</p> <p>Remark: Both of these counter-examples are still counter-examples if you replace "algebraically closed" by "perfect", but it would make my exposition messier.</p>