Base-points and invertible sheaves 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. 
 A: 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.
