Cohomology of an elliptic surface with section

Question

Let $\pi:S \to B$ be an elliptic surface with a section $\sigma$.

Let $f$ be the linear equivalence class of any fiber.

Then for all integers $a$

1. $h^0(\mathcal O_S(-\sigma + af)) = 0 $.
2. $R^0\pi_*\mathcal O_S(-\sigma + af) = 0 $.
3. $R^2\pi_*\mathcal O_S(-\sigma + af) = 0 $ since $\pi$ has relative dimension 1.

For part 1: $\mathcal O_S(-\sigma + af)(S) = \{g\in \mathcal O_S(S)| (g)-\sigma+af \ge 0\}$. So any global section $g$ should vanish on the curve $\sigma$, here I am inclined to say that such a $g$ should be zero, but I don't see the precise reason? For part 2: I guess it should be exactly the same argument as for part 1. For part 3: How does the relative dimension influence $R^2\pi_*\mathcal O_S(-\sigma + af) = 0 $?

Edit Using Cranium Clamp's suggestion, here is what I found.

Assuming that $X$ and $B$ are both projective , they are noetherian schemes and we can apply Grauert's theorem. If we denote $f$ the fiber above $b\in B$, the functions $b \to h^i(f,\mathcal{O}_S(-\sigma + af)) $ is clearly constant for $i=0,2$. Hence by Grauert's theorem, $R^i\pi_*\mathcal{O}_S(-\sigma + af) \cong H^i(f, \mathcal{O}_S(-\sigma + af)) $ for $i=0,2$. Hence part 2 follows from part 1. And since the fibers $f$ are elliptic curves (here we use the relative dimension one), obviously all cohomology of degree $i>1$ vanishes, hence part 3. It remains to understand why part 1 is true.

Thank you for any help/corrections!

Answer 1

Because $ f \cdot f = 0 $ and $ \sigma \cdot f = 1 $, the line bundle $ \mathcal{O}_S(-\sigma + af) $ restricts to every fibre with negative degree as $ (-\sigma + af)\cdot f = -1 $ and hence has no global sections. Therefore by Grauert's theorem, $ \pi_* \mathcal{O}_S( \sigma + af) = 0 $ which proves (1) and (2). For (3), you are right. The fibres are one-dimensional so have vanishing $ H^2 $ so the same theorem shows $ R^2 \pi_* \mathcal{O}_S( -\sigma + af) = 0 $.