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

algebraic-geometrydivisors-algebraic-geometry

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:

  1. 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!

Best Answer

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 $D=D'+D''$).

Related Question