Is a base point free nonspecial invertible sheaf generated by two global sections?

Question

Let $X$ be a projective nonsingular integral curve of genus $g$ over an aglebraically closed field. In Hartshorne Chapter IV exercise 6.8, we know that:

1. If $d\ge g+1$ then there is an effective nonspecial base point free divisor $D$ of degree $d$.

My question is that: can such $\mathscr L(D)$ be generated by two global sections?(that is, there is a sheaf surjection: $\mathcal O_X\oplus \mathcal O_X\to \mathscr L(D)$) That's what I need to solve exercise V 2.4 (a), which can be reduced to the following question:

2. For every $d\ge g+1$, find an effective divisor $D$ of degree $d$ such that there is a sheaf surjection: $\mathcal O_X\oplus \mathcal O_X\to \mathscr L(D)$.

Edit:

Let me explain KReiser's idea more concretely and state my issues with it:

Of course the question $2$ above is equivalent to: find a degree $d$ morphism $g:X\to \mathbb P^1$. KReiser's idea is to find a nonconstant global section $f\in \Gamma(X,\mathscr L(D))$, and since $\Gamma(X,\mathscr L(D))\subset K(X)$, $f$ indeed induces a morphism $f':X\to \mathbb P^1$. But the question is: the degree of $f'$ may not be $d$.

Let me explain why. By the proof of Hartshorne's II proposition 7.7 (a), you will find that: the principal divisor $(f)=(f)_0-D$, here $(f)_0$ is the effective divisor of zeros as defined in page 157. Now consider the morphism $f'$ induced by $f$, we can see that the pull back divisor of $\infty$: $(f')^*(\infty)=(f)_{\infty}\subset D$. So $\deg(f')=d$ is equivalent to saying that: $$d=\deg (f')^*(\infty)=\deg (f)_{\infty}$$ That's equivalent to say: The whole $D$ is the pole set of $f$, that's equal to say: the divisor $(f)_0$ doesn't decrease the coefficients of points in $D$, which is also equivalent to say: $(f)_0$ doesn't intersect with $D$. But this is just equivalent to say that $f$ and $1$ in $\Gamma(X,\mathscr L(D))$ generate $\mathscr L(D)$ (note that $(1)_0=D$). So the question is sill the same...

Any help is appreciated, thanks!

Answer 1

Let me explain myself (or at least try to). The question we're really trying to solve here is the following:

Hartshorne Exercise V.2.4. Let $C$ be a curve of genus $g$, and let $X$ be the ruled surface $C\times\Bbb P^1$. We consider the question, for what integers $s\in \Bbb Z$ does there exist a section $D$ of $\pi:X\to C$ with $D^2=s$? First show that $s$ is an even integer, say $s=2r$.

(a) Show that $r=0$ and $r\geq g+1$ are always possible. Cf. (IV, Ex. 6.8)

It's quick to show that $s=2r$: by adjunction, $2g-2=D.(D+K)$; by corollary V.2.11, $K\equiv -2C_0+(2g-2)f$ where we can take $C_0=C\times\{\infty\}$; combining the two we see that $$2g-2=D^2+D.(-2C_0+(2g-2)f)$$ and as $D.f=1$ since $D$ is a section of $\pi:X\to C$, we find that $D^2=2D.C_0$. Thus $r=D.C_0$, and so if we can find a function $f:C\to \Bbb P^1$ which has a pole divisor consisting of $r$ points, the section $C\to X\cong C\times\Bbb P^1$ defined by $id_C\times f$ will be a section $D$ such that $D.C_0=r$.

To show that we can always find such a function, consider an effective nonspecial base-point free divisor $D=\sum d_ip_i$ of degree $r\geq g+1$: by nonspeciality, $l(K-D)=0$, so by Riemann-Roch we have $l(D)=r+1-g$. Since $r\geq g+1$, $l(D)\geq 2$ and therefore $l(D)$ contains nonconstant functions. By base-point freeness of $D$, for all $P\in C$ we have $l(D-P)=l(D)-1$ (proposition IV.3.1(a)), so as $P$ runs through the points $p_i$ in $D$ the union of $|D-P|$ in $|D|$ cannot be the whole space. Therefore we can find a function $f$ which has poles supported exactly on $D$ and not on any effective subdivisor of $D$ (where I mean an effective divisor $D'$ such that there exists another effective divisor $D''$ so that $D=D'+D''$).