Let us work over an algebraically closed field $k$ and suppose $\pi:S\rightarrow C$ where $S$ is a surface, $C$ is a smooth curve, and the fibers over closed points are all isomorphic to a fixed smooth projective curve $D$. Must there exist an open subscheme $U\subseteq C$ such that $\pi^{-1}(U)\cong U\times D$? More strongly can I choose $U$ to contain any 2 points of $C$ of my choice?
If $D=\mathbb{P}^1$, then by Prop V.2.2 in Hartshorne, $S$ is isomorphic over $C$ to the projectivization of $\mathcal{E}$, a locally free sheaf of rank 2. Taking an affine open $U$ containing two points of your choice trivializes the vector bundle, and hence $\pi^{-1}(U)\cong U\times\mathbb{P}^1$. Does the same hold (although clearly by a different argument) for $D$ an arbitrary smooth projective curve?
Thanks
This doesn't hold in higher genus.
Consider the $C=\mathbb A^1\setminus \{0\}$ with coordinate $t$, let $S\to C$ be the familly of elliptic curves defined by $$y^2z=x^3+tz^3$$ (suppose $k=\mathbb C$). Each fiber is isomorphic to $D: y^2z=x^3+z^3$ (by the obvious change of variables), but, as you can't take the $6$-th root of $t$ in $O_C(C)$, you see easily that $S\times_C k(C)$ is not isomorphism $D\times_k k(C)$.
What happens in general is there exists a finite étale cover $C'\to C$ such that $S\times_C C'\to D\times_k C'$ ($C'$ can be obtained by adjoining the $n$-torsions of $\mathrm{Pic}^0_{S/C}$ for some $n\ge 3$ prime to the characteristic of $k$). In the above example, $C'=\mathrm{Spec}k[t^{1/3}]$.
The relative curves $S\to C$ under your hypothesis are classified by $H^1_{ét}(C, \mathrm{Aut}(D))$. The point is that the curve $S\times_C k(C)$ over $k(C)$ is a twist of $D\times_k k(C)$ for some finite extension $k(C')/k(C)$ with finite étale $C'\to C$.