The complete linear system of \(D\) is the set of effective divisors linearly equivalent to it, \[ \abs{D} = \ts{ D' \geq 0 \st D' \sim D } \cong \PP H^0(X, \OO_X(D)) . \] A linear system is a linear subspace of \(\abs{D}\), that is, \(\PP V\) for a subspace \(V \subseteq H^0(X, \OO_X(D))\). A point \(p\) is a base point when \(p \in \supp D'\) for every \(D'\) in the system; the base locus \(\Bs\abs{D}\) is the set of base points, and the system is base-point free when this is empty.
(Sections and degree on a curve)
Let \(D\) be a divisor on a nonsingular projective curve \(X\) over \(k = \bar k\). If \(\abs{D} \neq \emptyset\), that is \(\ell(D) \neq 0\), then \(\deg D \geq 0\). If moreover \(\deg D = 0\), then \(D \sim 0\) and \(\mcl(D) \cong \OO_X\). [@Har10a]
(Degenerate)
A morphism \(X \to \PP^n\), and equivalently the invertible sheaf with sections or the linear system \(\abs{V}\) defining it, is degenerate if the image of \(X\) is contained in a hyperplane.
\(p\) is a base point of \(\PP V\) exactly when \(s_p \in \mfm_p \mcl_p\) for every \(s \in V\). So \(\abs{D}\) is base-point free if and only if \(\OO_X(D)\) is generated by global sections.
The divisors in \(\abs{D}\) are the zero loci of the nonzero sections of \(\OO_X(D)\), two sections giving the same divisor if and only if they differ by a nonzero scalar; so \(\dim\abs{D}=h^0(\OO_X(D)) - 1\), which on a curve Riemann–Roch computes as \(\deg D-g+h^0(\OO_X(K-D))\).
Dependencies
D-DIVODThe sheaf $\OO_X(D)$ and effective divisors as sectionsuses
Backlinks
T-BERTINIBertini's theoremusesT-DIVMAPPNThe morphism to projective space attached to a linear systemuses