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!
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.