Obtain locally freeness of a sheaf from direct image sheaf The question comes from the book "Arithmetic moduli of elliptic curves" by Katz and Mazur. In somewhere of the review of elliptic curves, we want to prove the following results:
Let $E$ be an elliptic curve over a scheme $S$ (affine and noetherian) with $f: E\to S$, $\mathscr{L}$ be an invertible sheaf on E, degree one. Then 
1) $f_{*}\mathscr{L}$ is invertible sheaf on $E$, of formation compatible with any base change.
2)$R^{1}f_{*}(\mathscr{L}) = 0$
In the book it says we only need to prove 2), because we can apply a theorem and 1) is automatic. That theorem is from Mumford "Abelian Varieties", here I cite the theorem:
$f:X \to Y$ a proper morphism of noetherian schemes with $Y$ affine, $\mathscr{F}$ a coherent sheaf. Assume for some $p$ that $H^{p}(X_{y},\mathscr{F}_{y})=0$, for all $y \in Y$. Then $R^{p-1}f_{*}(\mathscr{F}) \otimes_{\mathcal{O}_{y}}k(y) \to H^{p-1}(X_{y},\mathscr{F}_{y})$ is an isomorphism for all $y \in Y$.
My question is how to use the above theorem on 2) to imply 1)?
Any comments are welcome!
 A: As $f: E\to S$ is a proper morphism of relative dimension one, we have $R^if_*\mathcal{F} = 0$ for any quasicoherent sheaf $\mathcal{F}$ on $E$ and any $i>1$. In particular, adding the assumption of 2), we then have that $R^if_*\mathscr{L}=0$ for $i>0$. 
By Grothendieck vanishing, for all $p>1$ we have that $H^p(X_y,\mathscr{L}_y)=0$ because each fiber is one dimensional. Applying the theorem as stated, $H^2(X_y,\mathscr{L}_y) = 0$ for all $y$, and therefore $R^{1}f_*(\mathscr{L}) \otimes_{\mathcal{O}_y} k(y) \to H^{1}(X_y,\mathscr{L}_y)$ is an isomorphism for all $y$. But $R^1f_*\mathscr{L}=0$, so in fact $H^1(X_y,\mathscr{L}_y)=0$ for all $y$, and applying the theorem again, we have that $f_*(\mathscr{L}) \otimes_{\mathcal{O}_y} k(y) \to H^{0}(X_y,\mathscr{L}_y)$ is an isomorphism. 
This implies that $f_*\mathscr{L}$ is locally free on $S$ (for additional details about this step, consult chapter 3 section 12 of Hartshorne - this is essentially a consequence of the semicontinuity theorem). To show it is invertible, it is enough to compute the rank at each point. By our last formula, it is enough to show that $H^0(X_y,\mathscr{L}_y)$ is rank 1 for all $y$. But since $\mathscr{L}$ is degree one, restriction to closed subschemes preserves degree, and $H^i(X_y,\mathscr{L}_y)=0$ unless $i=0$, we have proven the claim.
Since $Y$ is affine and $f_*\mathscr{L}$ is an invertible sheaf, it is isomorphic to $\widetilde{P}$ where $P$ is a projective module of rank 1 over $\mathcal{O}_Y(Y)$. Thus it is a perfect object compatible with arbitrary base change.
