Linear system of divisors

Builds on Projective variety Cartier divisor Invertible sheaf

Definition

Let $X$ be a normal projective variety over an algebraically closed field $k$ , and let $D$ be a Cartier divisor on $X$ . Two Cartier divisors are linearly equivalent if their difference is principal. The complete linear system $\abs{D}$ is the set of effective Cartier divisors on $X$ that are linearly equivalent to $D$ . Let $V = H^0(X, \mathcal{O}_X(D))$ , a finite-dimensional vector space, where $\mathcal{O}_X(D)$ is the invertible sheaf of $D$ . Each nonzero $s \in V$ has a divisor of zeros $(s)_0 \in \abs{D}$ , and $s \mapsto (s)_0$ gives a bijection between $(V \setminus \set{0})/k^*$ and $\abs{D}$ . A linear system is a subset of $\abs{D}$ of the form $\set{(s)_0 : s \in W \setminus \set{0}}$ for a linear subspace $W \subseteq V$ . Its dimension is $\dim W - 1$ .

Remarks

The base locus of a linear system is the intersection of the supports of all its divisors; the system is base-point-free if the base locus is empty.

Example: on $X = \mathbb{P}^n_k$ with $D$ a hyperplane, $V$ is the space of linear forms in $n+1$ variables. So $\abs{D}$ is the set of all hyperplanes, a linear system of dimension $n$ , and it is base-point-free.

Non-example: on $\mathbb{P}^1_k$ , the set of divisors $\set{p, \; p + q}$ for two points $p, q$ is not a linear system. The two divisors have degrees $1$ and $2$ , so they are not linearly equivalent and do not lie in one $\abs{D}$ .

A base-point-free linear system of dimension $r$ , with a basis $s_0, \dots, s_r$ of $W$ , gives a morphism $X \to \mathbb{P}^r_k$ , $x \mapsto [s_0(x) : \dots : s_r(x)]$ .

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Tag	`B4Z8C36HEB`
URI	`m2://B4Z8C36HEB`
Kind	definition
Writing	W0 stub
MSC	14C20
Defines	Linear system of divisors
Uses	Projective variety, Cartier divisor, Invertible sheaf

Linear system of divisors

Definition

Remarks

Reading path

Further reading

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