Reading Katz & Mazur's book "Arithmetic moduli of elliptic curves" (available here), I came across the following statement that I fail to understand (p.66).
Let $E$ be an elliptic curve over an algebraically closed field $k$. We denote by $f$ the morphism $E\rightarrow \operatorname{Spec}(k)$. Let $\mathcal{L}$ be an invertible sheaf over $E$, fiber-by-fiber of degree one. Then, $\operatorname{R}^1f_{\star}\mathcal{L}=0$.
The justification given in the book is the following:
For invertible sheaves $\mathcal{L}$ on $E$ of degree $> 2g-2=0$, we have $\operatorname{H}^1(E,\mathcal{L})=0$.
I have no problem with the last statement, which follows from Riemann-Roch theorem. However, I fail to see the connexion between the second and the first propositions. How does the vanishing of $\operatorname{H}^1(E,\mathcal{L})$ imply the vanishing of $\operatorname{R}^1f_{\star}\mathcal{L}$?
I thank you very much for your clarifications.
Best Answer
When $f : X \to \text{Spec(k)}$ is the morphism to the point, then the direct images are just cohomology groups of $F$, i.e $R^if_*F = H^i(X,F)$. For example, if $i=0$ this is just the tautological fact $H^0(X,F) = f_*F(pt)$.
In fact, it also works in a more general situation : if $f : E \to S$ is a family of elliptic curves (so $E_s := f^{-1}(s)$ is an elliptic curve for all $s \in S$, and $f$ is proper) and $L$ is an invertible sheaf so that $L_s$ is of degree $1$ on $E_s$, then $R^1f_*L = 0$. This is because its stalk at $s \in S$ is $H^1(E_s, L_s)$ by proper base change. As you said, Riemann-Roch implies that $H^1(E_s, L_s) = 0$.