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?



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




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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.


