# Locally free sheaves and vector bundles over smooth connected projective curve

Let $$X$$ be a connected smooth projective curve over an algebraically closed field $$K$$. Let $$\mathcal{F}$$ be a locally free sheaf on $$X$$ and $$\mathcal{E}$$ a subsheaf of $$\mathcal{F}$$, which is again locally free since $$dim(X)=1$$. Let $$E,F$$ be the corresponding vector bundles associated with $$\mathcal{E},\mathcal{F}$$ respectively.

Definition: The vector subbundle of $$F$$ generically generated by $$E$$ is a vector subbundle $$\bar{E}\subset F$$, which is the vector bundle associated with the locally free sheaf $$\bar{\mathcal{E}}:=\pi^{-1}\bigg(\mathcal{T}\big(\mathcal{F}/\mathcal{E}\big)\bigg)$$, where $$\pi:\mathcal{F}\rightarrow\mathcal{F}/\mathcal{E}$$ is the projection and $$\mathcal{T}\big(\mathcal{F}/\mathcal{E}\big)$$ is the torsion subsheaf of the quotient.

## Question 1

Why is $$\mathcal{F}/\bar{\mathcal{E}}$$ torsion-free (and hence locally-free)?

By using this property we obtain that $$\bar{E}$$ is a vector subbundle of $$F$$

## Question 2

Why do the following equations: $$\quad rk(\bar{\mathcal{E}})=rk(\mathcal{E})\qquad deg(\bar{\mathcal{E}})\ge deg(\mathcal{E})\quad$$ hold? Do they hold in general when we are dealing with inverse image sheaf or they are a special case?

Question 1: this can be checked locally, on affine opens or local rings, and then becomes an exercise: if $$0 \to M' \to M \overset{\pi}{\to} M'' \to 0$$ is an exact sequence of $$R$$-modules, then the submodule of $$M$$ generated by $$M'$$ and $$\pi$$-preimage of torsion in $$M''$$ has torsion free quotient. (E.g. for finitely-generated $$\mathbb{Z}$$-modules a submodule which has torsion-free quotient is called primitive, and every submodule can be enlarged to become primitive one by this construction.)
Question 2: the rank claim can be checked locally again; it follows from the fact that torsion modules have rank zero, by definition. For the degree claim, we compute using additivity of degree on short exact sequences $$\deg(\overline{\mathcal{E}}) = \deg(\mathcal{F}) - \deg(\mathcal{(\mathcal{E}/\mathcal{F}})/\mathcal{T}) = \deg(\mathcal{F}) - \deg(\mathcal{\mathcal{E}/\mathcal{F}}) + \deg(\mathcal{T}) \ge \deg(\mathcal{E})$$ where we used that the degree of a torsion-sheaf is non-negative (in fact, positive if it's nonzero).