Hartshorne, chapter V, exercise 2.11, very ample divisors on ruled surfaces I have a problem with proving the following exercise:

Let $X$ be a ruled surface over the curve $C$, defined by a normalized bundle $\mathcal{E}$ and $\mathfrak{e}$ be the divisor on $C$ for which $\mathcal{O}_C(\mathfrak{e}) \simeq \bigwedge^2 \mathcal{E}$. Let $\mathfrak{b}$ be any divisor on $C$.
(a) If $|\mathfrak{b}|$ and $|\mathfrak{b} + \mathfrak{e}|$ have no basepoints and if $\mathfrak{b}$ is nonspecial, then there is a section $D \sim C_0 + \mathfrak{b}f$ and $|D|$ has no base points.
(b) If $\mathfrak{b}$ and $\mathfrak{b}+\mathfrak{e}$ are very ample on $C$, and for every point $P \in C$, we have $\mathfrak{b}-P$ and $\mathfrak{b}+\mathfrak{e}-P$ is nonspecial, then $C_0 + \mathfrak{b}f$ is very ample.

The notations are the ones used by Hartshorne, namely $C_0$ is a section of $C$ such that $\mathcal{O}_X(C_0) \simeq \mathcal{O}_X(1)$ and $f$ is a fibre. I'm starting off by proving that $C_0 + \mathfrak{b} f$ is a section. By Proposition 2.6 this is equivalent to the existence of a surjective morphism $\mathcal{E} \to \mathcal{O}_C(\mathfrak{b}+\mathfrak{e}) \to 0$. The kernel of this morphism would be isomorphic to $\mathcal{O}_C(-\mathfrak{b})$, so the existence of the section is equivalent to the existence of an injective morphism $0 \to \mathcal{O}_C(-\mathfrak{b}) \to \mathcal{E}$, which is equivalent to the fact that $\mathcal{E} \otimes \mathcal{O}_C(\mathfrak{b})$ has a nowhere vanishing section. And here I got stuck. By the hypothesis, I know that an injective morphism $0 \to \mathcal{O}_C \to \mathcal{O}_C(\mathfrak{e} + \mathfrak{b})$ exists, but I don't know how to use that. Any hints on how to prove this and the rest of the exercise would be much appreciated.
 A: For (a), first prove two claims:$\newcommand{\fb}{\mathfrak{b}}\newcommand{\fe}{\mathfrak{e}}\newcommand{\sL}{\mathcal{L}}$
Claims:

*

*If $\fb$ is nonspecial, then $h^i(\sL_X(C_0+\fb f))=h^i(\sL_C(\fb))+h^i(\sL_C(\fb+\fe))$.

*If $P\in C$, then $\lvert C_0+\fb f\rvert$ is base-point-free on the fibre $Pf$ over $P$ iff $h^0(\sL_X(C_0+\fb f-Pf))=h^0(C_0+\fb f)-2$
Proof: (1) is an easy exercise, (2) is essentially the same proof as Proposition 3.1(a) in Chapter IV.  $\checkmark$
In particular, $h^0(\sL_X(C_0+\fb f))\geq 2$, so there exists section $D\sim C_0+\fb f$.  Then we compute
$$
\begin{aligned}
      h^0(\sL_X(D-Pf))&=h^0(\sL_C(\fb-P))+h^0(\sL_C(\fb+\fe-P))\\
                      &=h^0(\sL_C(\fb))-1+h^0(\sL_C(\fb+\fe-P))\\
                      &=h^0(\sL_C(\fb))-1+h^0(\sL_C(\fb+\fe))-1\\
                      &=h^0(D)-2,
\end{aligned}
$$
(the second equality follows since $\fb$ is bpf, the third follows since $P$ is not a base point of $\fb+\fe$) so $\lvert D\rvert$ has no base points.
For part (b): We compute, for any $P,Q\in C$
$$h^0(\sL_X(D-Pf-Qf)) = h^0(\sL_C(\fb-P-Q))+h^0(\sL_C(\fb+\fe-P-Q))=h^0(\sL_X(D)-4$$
and apply
Claim
$\lvert H\rvert=\lvert C_0+\fb f\rvert$ is very ample if and only if $h^0(\sL_X(H-(P+Q)f))=h^0(\sL_X(H))-4$ for any $P,Q\in C$.
Proof: Essentially the same as Proposition 3.1(b) in Chapter IV.
