Let $X$ be a surface and $C $ a smooth connected curve. Assume the is a flat and proper map $\pi: X \to C$ such that every fiber is an irreducible curve and every geometric fiber has arithmetic genus one: $\dim H^1(X_c, \mathcal{O}_{X_c})=1$ for every geometric point $c \in C$.
Does this assumption suffice to conclude that the derived direct image sheaves $R^i\pi_* \mathcal{O}_X$, $i \ge 0$ are locally free?
I think I found an argument why it should be true, could somebody check if the argument really works and eventually if it's possible to fill to gaps: The key is the Grauert's Theorem which says that $ R^i \pi_* \mathcal{O}_X $ is locally free if the function $ c \mapsto \dim H^i(X_c, \mathcal{O}_{X_c})$ is constant. The main question is if it suffice to require the constancy only in all geometric points? I think this suffice, but I'm not sure:
We assumed that for all geometric points $\overline{c} \in C$ we have $\dim H^i(X_c, \mathcal{O}_{X_c})=1 $ if $i=0,1$ and zero else, that's the irreducibility of fibers and genus one condition. And I'm asking if it suffice to make this assumption for geometric points only to conclude that it's already true that $\dim H^i(X_c, \mathcal{O}_{X_c})=1 $ if $i=0,1$ and else $0$ for all points $c \in C$, not only geometric ones.
I think the answer should be yes, because the function $ c \mapsto \dim H^i(X_c, \mathcal{O}_{X_c})$ is upper semicontinuous, what meant that for any $n \ge 0$ the set $\{c \in C \ \vert \ \dim H^i(X_c, \mathcal{O}_{X_c}) \ge n \}$ is closed. But every non empty closed subset of $C$ should contain a geometric point. This should prove that there is no non empty closed subset in $C$ with \dim $H^i(X_c, \mathcal{O}_{X_c}) \ge 2$ in $i=0,1$ or non zero in $i \ge 2$ cince otherwise it would contain a geometric point for which this not holds.
Is the argument correct? If yes, then under assumption of flatness and properness of $\pi$ we could state that $ R^i \pi_* \mathcal{O}_X $ are locally free iff $\dim H^i(X_c, \mathcal{O}_{X_c})$ are constant in geometric points, so other points not matter. Ater all, I can't find a flaw in my argument, on the other hand the statement seems to be to strong, even though I haven't a conterexample.