For an effective divisor \(D\) on a curve \(X\) of genus \(g\), show that \(\dim \abs{D} \leq \deg D\). Furthermore, equality holds if and only if \(D=0\) or \(g=0\).
Solution
Write \[ d=\deg D, \qquad \ell(D)=h^0(X,\mco_X(D)). \] Since \(D\) is effective, \(d\ge0\) and \[ \dim\abs{D}=\ell(D)-1. \]
1. The divisor sequence gives\[ 0\to\mco_X\to\mco_X(D)\to\mco_X(D)|_D\to0, \]and\[ h^0(D,\mco_X(D)|_D)=d. \]
The first map is multiplication by the canonical section with zero divisor \(D\). The finite scheme \(D\) has length \(d\). Since \(\mco_X(D)|_D\) is locally free of rank one on \(D\), it also has length \(d\), so its space of global sections has dimension \(d\).
2. One has \(\dim\abs{D}\le\deg D\).
Global sections of step 1 give \[ 0\to H^0(X,\mco_X)\to H^0(X,\mco_X(D)) \to H^0(D,\mco_X(D)|_D). \] Since \(h^0(X,\mco_X)=1\), step 1 gives \(\ell(D)\le d+1\). Therefore \[ \dim\abs{D}=\ell(D)-1\le d=\deg D. \]
3. If \(D=0\), then equality holds.
Here \(\ell(0)=h^0(X,\mco_X)=1\), so \[ \dim\abs{0}=0=\deg0. \]
4. If \(g=0\), then equality holds for every effective \(D\).
3. If \(d>0\), Riemann–Roch gives\[ \ell(D)-\ell(K-D)=d+1. \]Now \(\deg K=-2\), so \(\deg(K-D)<0\) and hence\(\ell(K-D)=0\). Thus \(\ell(D)=d+1\), and therefore\[ \dim\abs{D}=d=\deg D. \]
5. Conversely, suppose equality holds and \(D\ne0\). Then\(\ell(K-D)=g\).
Equality means \(\ell(D)=d+1\). Riemann–Roch gives \[ \ell(D)-\ell(K-D)=d+1-g. \] Hence \(\ell(K-D)=g\). It also gives \(\ell(K)=g\).
6. Choose a point \(P\) in the support of \(D\).
Because \(D-P\) is effective, the section spaces for \(K-D\), \(K-P\), and \(K\) are nested. The first and third have dimension \(g\) by step 5, so the middle one also has dimension \(g\). Riemann–Roch for the degree-one divisor \(P\) then gives \[ \ell(P)=2. \]
7. The equality \(\ell(P)=2\) forces \(X\cong\PP^1\), hence \(g=0\).
Constants give one section of \(\mco_X(P)\), so a second linearly independent section gives a nonconstant rational function \[ f\in H^0(X,\mco_X(P)). \] Its pole divisor is bounded by \(P\). Since every global regular function on the proper integral curve \(X\) is constant, \(f\) must actually have pole divisor exactly \(P\).
Thus \(f\) defines a finite morphism \[ f:X\longrightarrow\PP^1 \] whose degree is the degree of its polar divisor, namely \(1\). A finite degree-one morphism between integral smooth projective curves is birational, and since \(\PP^1\) is normal it is an isomorphism. Hence \(X\cong\PP^1\) and \(g=0\).
8. Equality holds if and only if \(D=0\) or \(g=0\).
Steps <1>3–<1>4 prove the two sufficient cases. Steps <1>5–<1>7 show that equality with \(D\ne0\) forces \(g=0\).
9. Q.E.D.
Step 2 proves the inequality and step 8 characterizes equality.