TITLE: Lazarsfeld-Mukai bundles are stable on a K3 surface of picard number 1
QUESTION [6 upvotes]: Let $S$ be a K3 surface with $Pic(S) = \mathbb{Z}.[C]$, where $C$ is a smooth curve. Let $A$ be a complete, base point free $g_d^r$ on $C$. How to show that the Lazarsfeld-Mukai bundle $E_{C,A}$ associated to $(C,A)$ is $C$-stable on $S$?  (The bundle $E_{C,A}$ is as defined in Lazarsfeld's paper 'Brill-Noether-Petri without degenerations').

REPLY [3 votes]: You can argue as follows. Put $E:=E_{C,A}$. You need only to know : $c_1(E)$ is the positive generator of $\mathrm{Pic}(S)$,  there is a homomorphism $\mathscr{O}_S^r\rightarrow E$ which is generically surjective, and $H^0(S,E^*)=0$.
If $E$ is not stable,  it admits a torsion-free quotient $\mathscr{F}$ with $c_1(\mathscr{F})\leq 0$.  The bidual $F$ of $\mathscr{F}$ is a vector bundle with the same $c_1$, and a generically surjective homomorphism $\mathscr{O}_S^r\rightarrow F$. Let $s:=\rm{rk}( F)$; 
 we get a generically surjective homomorphism $\wedge^s\mathscr{O}_S^r\rightarrow \det(F)$. Therefore $\det(F)=\mathscr{O}_S$, and there is a direct factor $\mathscr{O}_S^s$ of $\mathscr{O}_S^r$ such that the induced homomorphism $\mathscr{O}_S^s\rightarrow F$ has nonzero determinant, hence is an isomorphism. But this contradicts 
$H^0(S,E^*)=0$.