Saturated line subbundle on a smooth curve

A rank-one subsheaf of a vector bundle on a smooth algebraic curve is saturated when the quotient is torsion-free. Intersecting a one-dimensional subspace of the generic fibre with each stalk constructs a saturated subsheaf. The local ring of a smooth algebraic curve is a discrete valuation ring, so the subsheaf and its torsion-free quotient are locally free. Saturating a nonzero map from a line bundle enlarges that line bundle by an effective divisor, recording the map's zeros.

For a rank-two bundle $E$, the sufficiently positive twist $E(m{p}_0)$ is globally generated. The nowhere-vanishing section of a globally generated bundle on a curve gives a subbundle ${\mathcal{O}}_C\subset E(m{p}_0)$. Twisting back produces a genuine line subbundle of degree $-m$, with locally free quotient. Thus line-subbundle degrees are bounded above but not below. Merely inserting a negative divisor on an algebraic curve into a fixed subline would allow a torsion quotient and would not establish this stronger conclusion.

Choose a rational basis of the dual of a vector bundle and an effective divisor $D$ bounding all its poles. Evaluation gives a generically injective map $E\to {\mathcal{O}}_C(D{)}^{\oplus r}$. A line subbundle maps nontrivially to at least one summand, so ${\mathcal{O}}_C(D)\otimes L^{-1}$ has a nonzero regular section. Consequently $\mathrm{deg}L\le \mathrm{deg}D$. This bound depends on $E$ and the chosen rational frame, and does not require $E$ to be stable.