Local-freeness of the Hodge bundle Suppose $X\xrightarrow{f}Y$ is a proper flat morphism of locally noetherian schemes, with all of its fibers being geometrically integral regular curves of genus $g$ (so $h^{0}(X|_{q},\mathscr{O}_{X|_{q}})=1$ for all $q\in Y$). In particular $f$ is smooth of relative dimension $1$, and $\mathscr{O}_{Y} = f_{\ast}\mathscr{O}_{X}$. I would like to verify that the pushforward $f_{\ast}\Omega_{X/Y}$, the so called Hodge bundle, is a locally free sheaf (of rank $g$). Cf. R.Vakil's note, exercise 28.1.N. 
Using cohomology and base change theorem ("for $p=2$"), one can reduce to the fact that $R^{1}f_{\ast}\Omega_{X/Y}$ is locally free. In fact as $H^{2}(X_{q},\Omega_{X_{q}/q})=0$ by dimensional vanishing, $R^{2}f_{\ast}\Omega_{X/Y}=0$ is forced (it is hence locally free), and hence by cohomology and base change theorem, $R^{1}f_{\ast}\Omega_{X/Y}|_{q}\to H^{1}(X_{q},\Omega_{X_{q}/q})$ is surjective (in fact bijective) for all $q$. To proceed further, I think one must prove $R^{1}f_{\ast}\Omega_{X/Y}$ is locally free (at least to use the base change theorem for "$p=1$"). (As $h^{1}(X_{q},\Omega) = h^{0}(X_{q},\mathscr{O})=1$ we know $R^{1}f_{\ast}\Omega_{X/Y}$ must be a line bundle.) But I don't have any idea on proving this fact by hand yet. Are there any ways of tackling this (perhaps, without quoting too much, like Grothendieck-Serre duality)?

As Alex Youcis have mentioned in the comment above, H.Hida's [Geometric Modular Forms and Elliptic Curves] contains a proof of the fact that $R^{1}f_{\ast}\Omega_{X/Y}\approx \mathscr{O}_{Y}$ in 2.1.2 Grothendieck-Serre duality, pp. 96-97. The rest of the proof is straightforward---the local freeness implies $R^{0}f_{\ast}\Omega_{X/Y}|_{q}\to H^{0}(X_{q},\Omega_{X_{q}})$ are isomorphisms, and as "$p=-1$ case" fiber maps are trivially surjective, we know $f_{\ast}\Omega_{X/Y}$ is locally free of rank $g$.  
