This is probably a silly question, but I'm a bit confused. Regarding exercises 6.11 and 6.12 of Chapter II of Hartshorne:
Let $X$ be a nonsingular projective curve over an algebraically closed field $k$, and let $\mathcal{F}$ be a coherent sheaf on $X$ with rank $r$ (at the generic point). Part 11(c) says there is a divisor $D$ and a short exact sequence:
$0 \rightarrow O_X(D)^{\oplus r} \rightarrow \mathcal{F} \rightarrow \mathcal{J} \rightarrow 0$
where $\mathcal{J}$ is a torsion sheaf. But $X$ is defined over a field, doesn't that imply that $\mathcal{J} = 0$? (Since $H^0(X,\mathcal{J}) = \bigoplus_{P \in X} \mathcal{J}_P$, and $H^0(X,\mathcal{J}) = 0$ being a torsion $k$-module, hence all the stalks of $\mathcal{J}$ are $0$).
That would mean all coherent sheaves on $X$ of rank $r$ are locally free, I don't quite believe that.
(Side note: What I was trying to think about originally is whether such a sheaf $\mathcal{F}$ has global sections or not by considering the degree of the divisor $D$. Any comment on that would also be quite helpful).
Best Answer
Torsion means torsion over $\mathcal O_X$. For a coherent sheaf on a curve, it is equivalent to having support a finite number of closed points.