Once again I am confused after thinking too much about something I thought I already understood...
Let $\mathcal{L}$ be an invertible sheaf on a smooth projective curve $X$ such that $\deg \mathcal{F} \geq 2g$. Denote by $\Gamma(\mathcal{L})$ the global sections of $\mathcal{L}$. Then it is known that $\Gamma(\mathcal{L})$ has no base-points. That is, there exists no point $P \in X$ such that all elements $f \in \Gamma(\mathcal{L})$ vanish simultaneously at $P$.
The proof is something like this: If $P \in X$, one considers an invertible sheaf $\mathcal{L}(-P)$; it satisfies $\deg \mathcal{L}(-P) = \deg \mathcal{L} - 1$, and the global sections $\Gamma(\mathcal{L}(-P))$ are precisely the global sections of $\mathcal{L}$ which vanish at $P$. Then one can show (for example, by Serre duality) that $\dim \Gamma(\mathcal{L}) = \dim \Gamma(\mathcal{L}(-P)) - 1$.
Now for my question, suppose that $\deg \mathcal{L} \geq 2g+1$. Let $P \in X$. According to the above, $\deg \mathcal{L}(-P) \geq 2g$, so $\Gamma(\mathcal{L}(-P))$ has no base points, but all global sections $\Gamma(\mathcal{L}(-P))$ vanish at $P$!
Obviously I am making some mistake in what I've stated above. Could someone tell me what it is?
Thanks in advance.
This is indeed quite an interesting and subtle question!
The point is that when $\mathcal{L}(-P)\subset \mathcal{L}$, every section $s$ of $\mathcal{L}(-P)$ is a section of $\mathcal{L}$ but its order of vanishing at $P$ seen in $\mathcal{L}(-P)$ is one less than its order of vanishing seen in $\mathcal{L}$.
Proof:
Since the problem is local, we may assume we have a coordinate $z$ on $X$ with $z(P)=0$ and suppose that $\mathcal{L}$ is trivial.
A section $s$ of $\mathcal{L}$ is then locally of the form $z^k u(z)$ with $u$ non-vanishing.
The order of vanishing of $s$ at $P$ is then $k$.
The line bundle $\mathcal{L}(-P)$ on the other hand has as local basis around $P$ the function $z$ and thus the section $s=z^k u=(z^{k-1} u)\cdot z$ has order $k-1$ at $P$ if seen as a section of the bundle $\mathcal{L}(-P) $.$\;$ qed
In particular, if the order of vanishing of $s$ in $\mathcal L$ is 1, its order of vanishing in $\mathcal{L}(-P)$ is 0 and $s$ does not vanish at $P$ in $\mathcal{L}(-P)$ although $s$ vanishes at $P$ if seen in $\mathcal L$.
This solves the paradox.