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.

If the answer 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?

  • 9
    $\begingroup$ Assume $X$ connected, hence irred. Assume $f$ dominant, or else the complement of closure of the image does the trick. Thus, $S$ is irred (and presumably you assume quasi-compact, so noetherian since excellent). By generic flatness, can assume $f$ flat by passing to dense open in $S$. Generic fiber is regular, so if generic point of $S$ is char. 0 then generic fiber is smooth. Then by flatness, $f$ is smooth over dense open (EGA IV$_4$, sec. 17 somewhere), so win without excellence. False without generic char. 0 (Speyer's example). Final equivalence unclear (I guess false); feels useless. $\endgroup$
    – BCnrd
    Jul 16, 2010 at 0:54

1 Answer 1


Over a field of characteristic zero, your result is true. This is Corollary III.10.7 in Hartshorne.

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.

There is a more interesting counter-example due to Serre: Let $k$ be an algebraically closed field of characteristic $2$. Consider the planar cubic $$ F_{a,b,c}(x,y,z) :=a (y^2 z + y z^2) + b (x^2 z + x z^2) + c (x^2 y + x y^2) \quad (*)$$ 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 node 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 $$(x:y:z) \mapsto (F_{(a_1,\ b_1,\ c_1)}(x,y,z) : F_{(a_2,\ b_2,\ c_2)}(x,y,z))$$ Then the fiber over $(t_1:t_2)$ is $F_{(t_1 a_1+t_2 a_2,\ t_1 b_1+t_2 b_2,\ t_1 c_1+t_2 c_2)}=0$ 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 nodal cuspidal cubic or worse.

Remark: Both of these counter-examples are still counter-examples if you replace "algebraically closed" by "perfect", but it would make my exposition messier.


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