Let \(X\) be a nonsingular projective variety over an algebraically closed field, and let \(D\) be a divisor on \(X\). For any \(n > 0\) consider the complete linear system \(\abs{nD}\). The Riemann-Roch problem is to determine \(\dim \abs{nD}\) as a function of \(n\), and in particular its behavior for large \(n\).
If \(\mcl\) is the corresponding invertible sheaf, then \(\dim \abs{nD} = \dim \Gamma(X, \mcl^n) - 1\), so an equivalent problem is to determine \(\dim \Gamma(X, \mcl^n)\) as a function of \(n\).
Show that if \(D\) is very ample, and if \(X \injects \PP^N_k\) is the corresponding embedding in projective space, then for all sufficiently large \(n\) we have \(\dim \abs{nD} = P_X(n) - 1\), where \(P_X\) is the Hilbert polynomial of \(X\) (I, §7). Thus in this case \(\dim \abs{nD}\) is a polynomial function of \(n\) for \(n\) large.
If \(D\) corresponds to a torsion element of \(\Pic X\) of order \(r\), then \(\dim \abs{nD} = 0\) if \(r \divides n\) and \(-1\) otherwise. In this case the function is periodic of period \(r\).
It follows from the general Riemann-Roch theorem that \(\dim \abs{nD}\) is a polynomial function for \(n\) large whenever \(D\) is an ample divisor; see (IV, 1.3.2), (V, 1.6), and Appendix A. In the case of algebraic surfaces, Zariski has shown for any effective divisor \(D\) that there is a finite set of polynomials \(P_1, \ldots, P_r\) such that for all sufficiently large \(n\), \(\dim \abs{nD} = P_{i(n)}(n)\), where \(i(n) \in \ts{1, \ldots, r}\) is a function of \(n\).
Solution
Put \(\mathcal L=\OO_X(D)\) and \(h^0(X,\mathcal F)=\dim_k\Gamma(X,\mathcal F)\). The complete linear system \(\abs{nD}\) has dimension \(h^0(X,\mathcal L^n)-1\), with the empty system assigned dimension \(-1\).
1. For the embedding given by a very ample \(D\), one has \(h^0(X,\mathcal L^n)=P_X(n)\) for all sufficiently large \(n\).
Let \(i:X\hookrightarrow\PP_k^N\) be the embedding and write \[ S=k[x_0,\ldots,x_N]/I_X \] for its homogeneous coordinate ring. The isomorphism \(i^*\OO(1)\cong\mathcal L\) identifies \(\mathcal L^n\) with \(\OO_X(n)\). Apply the high-degree comparison of P-AGH259GAMMASTAR, step 2, to the finite graded \(S\)-module \(S\). Its hypotheses hold because \(S_0=k\) and \(S\) is generated by finitely many degree-one elements. It gives \[ S_n\xrightarrow{\cong}\Gamma(X,\OO_X(n)) \] for every sufficiently large \(n\). The Hilbert polynomial satisfies \(P_X(n)=\dim_k S_n\) for every sufficiently large \(n\) [@Har10a, Chapter I, §7]. Taking \(n\) beyond both bounds proves the equality.
2. Part (a) gives the eventual polynomial\[ \boxed{\dim\abs{nD}=P_X(n)-1\qquad(n\gg0).} \]
Substitute the equality in step 1 into the dimension formula for the complete linear system. The result is a polynomial in \(n\) because \(P_X\) is a polynomial.
3. A torsion invertible sheaf on \(X\) with a nonzero global section is trivial.
First, \(\Gamma(X,\OO_X)=k\). A global regular function gives a morphism \(X\to\AA_k^1\subseteq\PP_k^1\). Since \(X\) is projective, its image in \(\PP^1\) is closed; it is irreducible and misses \(\infty\), so it is a point. The field \(k\) is algebraically closed, and \(X\) is reduced, so the function equals that point’s constant value as a section.
Let \(\mathcal N\) be invertible with \(\mathcal N^a\cong\OO_X\) for some \(a>0\), and suppose \(0\ne s\in\Gamma(X,\mathcal N)\). As \(X\) is integral, \(s\) is nonzero at the generic point: on each trivializing affine open, restriction of a regular function to the fraction field is injective. Thus \(s^{\otimes a}\) is also nonzero at the generic point. Under the chosen isomorphism it becomes a nonzero global regular function, hence a nonzero constant by the preceding argument.
At any point, write \(s=b e\) in a local frame for \(\mathcal N\). The image of \(e^{\otimes a}\) under the isomorphism \(\mathcal N^a\cong\OO_X\) is a unit, so the constancy of \(s^{\otimes a}\) makes \(b^a\) a unit and therefore makes \(b\) a unit. Consequently \(s\) is a local frame at every point, and the morphism \(\OO_X\to\mathcal N\) defined by \(1\mapsto s\) is an isomorphism.
4. If \([\mathcal L]\) has order \(r\) in \(\Pic(X)\), then for every \(n>0\),\[ \boxed{\dim\abs{nD}= \begin{cases} 0,&r\mid n,\\ -1,&r\nmid n. \end{cases}} \]The least positive period is \(r\).
3.If \(r\nmid n\), the sheaf \(\mathcal L^n\) is nontrivial and torsion, so step 3 forces its space of global sections to be zero.Subtracting one gives the two dimensions.Divisibility by \(r\) is unchanged on replacing \(n\) by \(n+r\), so the function has period \(r\).If \(p>0\) were any period, its values at \(n=r\) and \(n=r+p\) would both be zero, forcing \(r\mid(r+p)\) and hence \(r\mid p\).This proves the least-period assertion, including \(r=1\).
5. Q.E.D.
Steps <1>1–<1>2 prove (a), and steps <1>3–<1>4 prove (b).