# Degree of sub sheaf of locally free sheaf is bounded

Let $X$ be smooth projective curve over a field $k$.

Let $\mathcal F$ be a finite rank locally free sheaf on $X$. Now, consider the set $D = \{\operatorname {deg}(\mathcal G) : \mathcal G$ is a subsheaf of $\mathcal F \}$. Why is set $D$ bounded, i.e. $\operatorname {sup} D$ is less than $n$, $n \in \mathbb{Z}$?

• Can you do this if $\mathcal{F}$ is of rank one? – Mohan Feb 28 '18 at 19:37
• No, I don't how to do it for one. I guess if I could this for rank one that would imply it for the general case as the degree is the degree of the determinant bundle. – grok Feb 28 '18 at 20:09
• In the rank one case, if $G\subset F$, then $F\otimes G^{-1}$ has a section, so its degree which is $\deg F-\deg G\geq 0$. – Mohan Feb 28 '18 at 21:25