This is a problem from chapter 6 in Beauville's book Complex Algebraic Surfaces. I have a smooth projective surface $S$ whose smooth hyperplane sections $H$ are elliptic curves. I want to show that $S$ is either a del Pezzo surface or is an elliptic ruled surface.
I want to proceed by the hint in the book. I have managed to show that $q=h^1(S,\mathcal O_S)\le 1$ by looking at two holomorphic 1-forms, restricting them to the hyperplane sections, doing some stuff and getting that they have to be proportional everywhere. Now I want to take care of the cases $q=0$ and $q=1$ separately. When $q=0$, Beauville suggests that I show that $K\equiv -H$, i.e. that the canonical class is the negative of $H$ in $\mathrm{Pic}(S)$. This would solve this part of the problem because I have a result from an exercise a while ago saying if $K\equiv -H$ then $S$ is a del Pezzo surface. The problem is that I'm not sure how to go about showing this.
The best I can think of is to show that $H=-K$ in cohomology, which would suffice because the Chern class map $\mathrm{Pic}(S)\to H^2(S,\mathbb Z)$ is an injection since $q=0$. However I'm not sure how this helps. To be frank, I don't know how to go abut showing two divisor classes are the same if I don't have some explicit description for them.
For the case $q=1$, I'm hoping that the previous part will show me that $H.K<0$ in general or something, as this would imply that $S$ is ruled. I haven't thought too much about this part to be honest.
Any help is appreciated. I'm more looking for a hint or nudge that will get me thinking in the right direction, but even just ideas which are not fleshed out at all could be helpful.
I have a solution to this problem now. My original proof showing that $q\leq 1$ was false (or rather incomplete), so I'll present the whole thing here.
First, I had to blackbox the Lefschetz hyperplane theorem. It's unfortuante but I'm okay with it. Applying it here, I get that $H^1(S,\mathbb Z)\to H^1(H,\mathbb Z)$ is injective, and hence $b_1\le 2.$ Since $q=\frac{1}{2}b_1$ we have the result.
On to the rest of the proof. Suppose that $q=h^1(\mathcal O_S(K))=0.$ From the short exact sequence $$0\to \mathcal O_S(K)\to \mathcal O_S(K+H)\to \mathcal O_H(K+H)\to 0$$ we form the long exact sequence in cohomology and find that the map $$H^0(S,\mathcal O_S(K+H))\to H^0(H,\mathcal O_H(K+H))\cong H^0(H,\mathcal O_H)$$ is surjective (here we use the fact that $q=0$). Thus $\mathcal O_S(K+H)$ has a section, say $D\in |K+H|.$ If $D\neq 0,$ let $H$ be a smooth hyperplane section not contained in $D$ but which intersects the components of $D.$ Then $D$ restricts to $H$ nontrivially, contradicting the fact that $H+K|_H\equiv K_H\equiv 0.$ Thus $D\equiv 0$ and $K\equiv -H$ on $S,$ so $S$ is a del Pezzo surface by some previous exercise in the book.
Now suppose that $q=1$ and again consider the same long exact sequence in cohomology. The map $$H^0(S,\mathcal O_S(K+H))\to H^0(H,\mathcal O_H(K+H))$$ can no longer be surjective since then $S$ would be rational by the argument above, and since $H^1(S,\mathcal O_S(K))$ has rank $1,$ the above map must be zero. It follows that $H^0(S,\mathcal O_S(K))\cong H^0(S,\mathcal O_S(K+H)),$ so they have the same dimension: the geometric genus $p_g$ of $S.$ Now applying Riemann-Roch to $K+H$ (and noting $h^0(-H)=0$) we get $$p_g=h^0(K+H)+h^0(-H)\ge p_g+\frac{1}{2}(H^2+K.H)>p_g+\frac{1}{2}K.H.$$ It follows that $K.H<0,$ so by a lemma in the chapter, $S$ must be ruled. Finally, $S$ is ruled over an elliptic surve since the genus of the curve is exactly $q=1$ (this fact is somewhere in the chapter on ruled surfaces).
There you have it. For me this was a good lesson in remembering that all the of equalities $q=h^1(\mathcal O_S)=h^1(K_S)=\frac{1}{2}b_1$ hold. Also don't forget about $q=h^0(\Omega^1_S)$!