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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)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)?2012-11-11
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    One must suppose $F\ne 0$.2012-11-11
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    @QiL You are right. Thank you for pointing this out.2012-11-12

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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 $$p_F(d)/\mathrm{rk}(F)=p_F(d) < p_E(d)=p_E(d)/\mathrm{rk}(E).$$