# Why is any rank $1$ sheaf always stable?

Let $X$ be a projective scheme over $\mathbb{C}$. For a sheaf on $X$, $$p_E(d)=\chi(X,E(d))$$ be the Hilbert polynomial of $E$. A sheaf $E$ on $X$ is siad to be stable if for every proper subsheaf $F\subset E$, $$p_F(d)/rk(F)<p_E(d)/rk(E)$$ for sufficiently large $d>0$. Why is any rank $1$ sheaf always stable?

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What does $rk(F)$ mean in this context? Are you restricting to locally free sheaves, or are you allowing coherent sheaves but taking rank at the generic point (though that would need irreducibility, which isn't assumed here)? –  only Nov 11 '12 at 13:30
One must suppose $F\ne 0$. –  user18119 Nov 11 '12 at 15:54
@QiL You are right. Thank you for pointing this out. –  M. K. Nov 12 '12 at 1:04

If $F$ is a non-zero sub-coherent sheaf of $E$, then it also has rank 1. Let $S=E/F$. Then $p_E(d)=p_F(d)+p_S(d)$ with $p_S(d)>0$ when $d$ is big enough. Thus $$p_F(d)/\mathrm{rk}(F)=p_F(d) < p_E(d)=p_E(d)/\mathrm{rk}(E).$$