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Let $X$ be a normal projective variety and $D$ a Cartier divisor on it. A point $p\in X$ is called a fixed point of $|D|$ if $p \in \operatorname{supp}(D')$ for any $D'\in |D|$. Here $|D|$ is the linear system of $D$ defined by $|D|=(H^0(X,O(D))\setminus0)/\mathbb{C}^\times\cong \mathbb{P}^{h^0(X,O(D))-1}$.

Could anyone give me a simple example of $X$ and $D$, where the fixed point set of $|D|$ is non-empty?

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up vote 1 down vote accepted

If $X$ is a smooth projective curve (e.g. a Riemann surface if you work over $\textbf C$) and $D$ is the divisor of a point $P\in X$, then $|D|$ is just $P$ (unless $X\cong\textbf P^1$, in which case $|D|=\textbf P^1$).

As another example, consider this pencil ($=$ one-dimensional linear system) on the projective plane: for $t$ ranging $\textbf P^1$, the corresponding divisor is given by the equation $y(x-tz)$: the $x$-axis plus a "vertical line". All these divisors meet at the point at infinity, and there is no other base point (even if the $x$-axis appears in each divisor!)

A useful characterization says that $|D|$ is base-point-free if and only if $\mathscr O_X(D)$ is generated by its global sections. Similarly, for non-complete linear systems: $\textbf PV^\ast\subseteq |D|$ is base-point-free if and only if $\mathscr O_X(D)$ is generated by the global sections in (a basis of) $V$. Translation: a linear system determines a rational map $\psi:X\dashrightarrow \textbf PV^\ast$, and $x\in X$ is a base point if and only if $\psi$ is not defined at $x$ (which means that all the sections generating $V$ vanish at $x$). Just take $V=H^0(X,\mathscr O_X(D))$ if you only care about complete linear systems.

Another example of a linear system with base points: let $k$ be the algebraic closure of $\textbf F_2$ and let $V\subset H^0(\textbf P^2_k,\mathscr O(3))$ be spanned by the cubics

\begin{equation} x^2y+xy^2,\,x^2z+z^2x,\,y^2z+z^2y. \end{equation}

Then the linear system $\textbf PV^\ast$ (which is two-dimensional) has $7$ base points: look at those points in $\textbf P^2_k$ at which the above sections vanish simultaneously, and you will find that this happens exactly when the coordinates all lie in $\textbf F_2$. So the base locus here is the finite projective plane $\textbf P^2_{\textbf F_2}$.

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I totally forgot about the useful criterion that $|D|$ is base-point-free iff $O(D)$ is generated by global sections. Then the ideal sheaf $I_{Z}=O(Z)$ of subvariety $Z\subset X$ is not generated by global sections and thus not base=point-free. I don't quite understand your second example. Aren't points on $y=0$ base-point of this linear system? – M. K. Sep 30 '12 at 22:58
@M. K. You are right: if $(x:y:z)$ are homogeneous coordinates on $\textbf P^2$, then the base locus of that linear system is the copy of $\textbf P^1$ given by $y=0$. What I had in mind (but, I'm sorry, I did not write it correctly!) is an example by Kleiman (, p. 12), involving inhomogeneous coordinates. – Brenin Oct 1 '12 at 18:10
Thank you for clarifying the example. I will check the Keliman's article too. By the way, in my comments $O(Z)$ is typo and should be $O(-Z)$. – M. K. Oct 1 '12 at 18:45
  1. If $D<0$, then every point of $X$ is a fixed point of $|D|=\emptyset$.

  2. If $X$ is a smooth projective curve of genus $\ge 1$, then for any rational point $P$ of $X$, $P$ is a fixed point of $|P|$ because otherwise, there would exits a rational function on $X$ with only a simple pole and this would imply that $X$ is the projective line.

  3. Let $X$ be as a curve as above. Let $P\in X$. Then $|2P|$ has no fixed point if and only if $P$ is a so called Weierstrass point. On a hyperelliptic curve, they are the ramification points of the canonical degree $2$ map $X\to \mathbb P^1$.

  4. Let $f$ be a rational map from $\mathbb P^n$ to $\mathbb P^m$. Let $D$ be the pullback of a hyperplane by $f$. Then $D$ has no fixed points if and only if $f$ is defined everywhere. So take for example f equal to a projection to a linear subspace, then $D$ has fixed points.

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Thank you for providing me with many examples. Just a quick question, is every point of $X$ a fixed point if $|D|=\emptyset$? The second example is a easy good example. – M. K. Sep 30 '12 at 22:36
@M.K.: The first example is kind of cheating. But well logically it is correct because $|D|=\emptyset$ means there is no condition to verifiy for $p\in X$. – user18119 Oct 1 '12 at 7:45

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