Exercise II 6.11:
Let $X$ be a nonsingular curve over an algebraically closed field $k$.
(c) If ${\mathscr{F}}$ is any coherent sheaf of rank $r$ (means that its stalk at the generic point has dimension $r$ as a $K(X)$-vector space), show that there is a divisor $D$ on $X$ and an exact sequence $0 \rightarrow \mathscr{L}(D)^{\oplus r}\to \mathscr {F} \rightarrow \mathscr {J} \rightarrow 0$, where $ \mathscr {J}$ is a torsion sheaf (means that its stalk at the generic point is $0$. ).
The following are my attempts:
Let $\zeta$ be the generic point of $X$, then rank$\mathscr{F}=r$ implies that $\mathcal{F}_\zeta \simeq \mathcal{O}_{X,\zeta}^{\oplus r}$. Since $\mathscr{F}$ is a coherent sheaf, there is an open neighborhood $U$ of $\zeta$ such that $\mathscr{F}|_U\simeq \mathcal{O}_U^{\oplus r}$.
We can assume that $U=\operatorname{Spec} A$ is affine, and $\mathcal{F}|_U\simeq \widetilde{M}$ for some free $A$-module $M$ with generators $m_1,\cdots,m_r$. But how to extend these $m$ to the global section of $\mathscr{F}$?
Since $X$ is a curve, $X-U$ consists of finite points. For each $p\in X-U$, we can take $U_p=\operatorname{Spec}(B)$ containing $p$ and $f\in B$ such that $p=V(f)$. Furthermore, we can assume that $m\in \mathcal{F}(D(f))\simeq \mathcal{F}(U_p)_f$, then $f^{N_p} m\in \mathcal{F}(U_p)$ for some $N_p$. If we set $D=\Sigma N_p p$ be the effective Weil divisor, we can construct a morphism $\mathscr{L}(-D)\to \mathscr{F}$ locally by multiplying $f^{N_p}$ locally, but I do not think this is well defined on the intersection of $U_p$ and $U_q$ for distinct $p,q\in X-U$.
Best Answer
Let $\xi$ be the generic point of $X$. $\def\cF{\mathcal{F}}\def\cO{\mathcal{O}}\def\cL{\mathcal{L}}\def\Spec{\operatorname{Spec}}\def\cT{\mathcal{T}}\def\cI{\mathcal{I}}\def\g{\gamma}$ As $\cF_\xi \cong k(X)^r$ is a free $k(X)$-module, we have by exercise II.5.7(a) that $\cF$ is free in an open neighborhood $U$ of $\xi$, which implies that we have an isomorphism $\cO_X^r|_U\to \cF|_U$ on $U$. Suppose $X\setminus U = \{P_1,\cdots,P_n\}$, and let $D=\sum P_i$. As $\cL(eD)|_U$ is trivial for any $e$, we have $\cF|_U\cong (\cF\otimes \cL(eD))|_U$. I claim that for a sufficiently large integer $e$, the composite $\cO_X^{\oplus r}|_U\to (\cF\otimes \cL(eD))|_U$ lifts to a map $\cO_X^{\oplus r}\to \cF\otimes\cL(eD)$, which after tensoring by $\cL(-eD)$ gives an injection $\cL(-eD)\to\cF$ which is surjective on $U$, and thus the quotient $\cF/\cL(-eD)$ is supported on the finite set $X\setminus U$ and is torsion.
To show this, let $\{(U_i,f_i),(U,1)\}$ be the Cartier divisor associated to $D$, where we assume that each $U_i$ is actually $\Spec A_i$ after an appropriate shrinking and subdivision and the number of $U_i$ is finite by quasi-compactness. Now the map $\cO_X^{\oplus r}|_U\to \cF|_U$ is given by an $r$-tuple of sections in $\cF(U)$, and each section of $\cF(U)$ gives us a section of $\cF$ over $D(f_i) \subset U_i$. By lemma II.5.3, for each $i$ and each section over $D(f_i)$ there is an integer $n$ so that $f^n$ times that section extends to a section over $U_i$. Taking the maximum of these $n$ across all finitely many $i$ and the $r$ sections over each $D(f_i)$, we obtain $e$ so that $\cO_X^{\oplus r}|_U\to \cF|_U$ extends to $\cO_X^{\oplus r}\to\cF\otimes\cL(eD)$. The kernel of this map is a subsheaf of $\cO_X^{\oplus r}$ and is thus locally free as $X$ is a smooth curve, plus it's zero at the generic point by definition, so $\cO_X^{\oplus r}\to \cF\otimes\cL(eD)$ is injective. As tensoring with an invertible sheaf is an isomorphism, we get an injection $\cL(-eD)^{\oplus r}\to \cF$ with cokernel a torsion sheaf $\cT$ as requested.
The construction of $\cL(D)$ takes care of your worries about well-definedness. Recall that by construction, the transition maps for $\cL(D)$ from $(U,f)$ to $(V,g)$ take a section from $\cL(U)$ and multiply by $\frac{g}{f}$ - so in our case above, the transition maps from $\cL(U)$ to $\cL(U_i)$ are multiplication by $f_i$ and we really do find that for big enough $e$ we get a global section of $\cF\otimes\cL(eD)$.