Let $X$ be a (compact) Riemann surface. Let $D$ be a divisor. In Rick Miranda's book on Riemann surfaces, on page 160, there is a bijection between
- Base-point-free linear systems of dimension $n$ on $X$
- holomorphic maps $\phi: X \to \mathbb P^n$ with nondegenerate image, up to linear coordinate changes
Suppose $|D|$ is nonempty and corresponds to a basis $f_0, \dots f_n$ in $L(D)$. Without assuming $|D|$ is base-point-free, I can still get a map $\phi: X \to \mathbb P^n$ by $p \mapsto [f_0(p): \dots : f_n(p)]$. If you're worried about common poles or zeros at $p$, we can essentially divide or multiply by some power of $z$ in local coordinates with $p$ corresponding to $0$ (see Lemma 4.2 on page 154).
So why do we need base-point-free? Don't all nonempty linear systems give meromorphic functions into projective space? Why do we need to bother with linear systems to get maps into projective space? Shouldn't just having $n+1$ meromorphic functions be enough?
To prove the bijection above I do indeed require the linear system to be base-point-free.
Best Answer
Let $D$ be a divisor with $|D|\neq\emptyset$. Then we may replace $D$ with a linearly equivalent effective divisor and assume that $D$ is effective. Then base points of $D$ are contained in the support of $D$ and if $B$ is the divisor of base points, then $|D|=|D-B|$.So, the map given by $|D|$ is the same as the one given by the base point free divisor $D-B$.
So, the correct way of saying what Miranda says is that 1 is equivalent to 2 with an additional condition on 2, namely, $\phi^*(\mathcal{O}_{\mathbb{P}^n}(1))=\mathcal{O}_X(D)$.