I'm following Beauville's book on Complex Algebraic Surfaces.
If $D$ is a divisor on a surface $S$, we write $|D|$ for the set of all effective divisors linear equivalent to $D$ and we call it a complete linear system on a surface $S$. At page 14, Beauville says that $|D|$ can be identified with the projective space associated to the vector space $H^0(S,\mathcal{O}_S(D))$, so if i well understood, this means
$$|D|=\mathbb{P}H^0(S,\mathcal{O}_S(D)).$$
But i think it should be $|D|=\mathbb{P} H^0(S,\mathcal{O}_S(D))^*.$
Could you help me in making this identification more clear?
The linear system $|D|$ is the set of all effective divisors linearly equivalent to $D$. Consider the map $H^0(S,\mathcal{O}_S(D))\to|D|$ where $f\mapsto\mbox{div}(f)+D$ (here I am identifying $H^0(S,\mathcal{O}_S(D))$ with the vector space of all rational functions $f$ such that $\mbox{div}(f)+D\geq0$). This map is surjective, since if $E\in |D|$, then $E=\mbox{div}(f)+D$ for some $f$ ($E$ is linearly equivalent to $D$). If $\mbox{div}(f)+D=\mbox{div}(g)+D$, then $\mbox{div}(f/g)=0$, and so $f=\lambda g$ for some non-zero constant $\lambda$. Therefore the fibers are precisely the one dimensional linear subspaces of $H^0(S,\mathcal{O}_S(D))$, and so $|D|$ can be seen as the projectivization of $H^0(S,\mathcal{O}_S(D))$.