If $C$ is a curve with genus $g$ and $k$ a field, I'm stuck in something I'm sure easy, I think I'm forgetting some basic things.
Define $\Omega(D)=\{\omega\in\Omega;div(\omega)\ge D\}$ and $\delta(D)=\dim_k(\Omega(D))$. I'm trying to prove this equivalence:
There is a diferencial $w$ with $\text{ord}_P(w)=n-1\ \Leftrightarrow \delta((n-1)P)=1+\delta(nP)$
In another words:
There is a diferencial $w$ with $\text{ord}_P(w)=n-1\ \Leftrightarrow \dim_k(\Omega((n-1)P))=1+\dim_k(\Omega(nP))$
I'm studying Fulton's Algebraic Curves book and I couldn't prove this equivalence.
Thanks in advance
Let me first make a trivial but useful remark: $\Omega(nP)\subseteq\Omega((n-1)P)$.
Now, suppose that $\delta((n-1)P)=1+\delta(nP)$. Thus, there must exists $w$ such that $w\in\Omega((n-1)P), w \notin\Omega(nP)$. This implies that $div(w)\geq (n-1)P$ and $ord_{P}(w)<n$, hence $ord_{P}(w)=n-1$. We have proved the first implication.
Suppose now that there exists such $w$. This means that $\delta((n-1)P)-\delta(nP)\geq 1$. We want to prove that $\delta((n-1)P)-\delta(nP)\leq 1$. Using Riemann-Roch, we have:
$0\leq l(nP)-l((n-1)P)=\delta(nP)+n+1-g-\delta((n-1)P)-(n-1)-1+g=\delta(nP)-\delta((n-1)P)+1\Rightarrow\delta((n-1)P)-\delta(nP)\leq1$
and we have finished.