$l(rP)\le l((r-1)P)+1$

Question

If $C$ is a curve of genus $g$, I'm trying to prove the dimension of the divisor $rP$ associated to this curve is less or equal than the dimension of the divisor $(r-1)P+1$, where $r\in \mathbb N$.

In another words, I want to prove: $l(rP)\le l((r-1)P)+1$.

Remark: my background is Fulton's Algebraic Curves book.

Thanks in advance

Answer 1

By Riemann-Roch, you have $$l(rP)= r+1-g +l(K-rP) $$ and $$ l((r-1)P) +1 = r + 1 - g +l(K-(r-1)P).$$

So the difference between these two are measured by $l(K-rP)-l(K-(r-1)P)$. Below is a more elaborate explanation:

But clearly $l(K-rP) \leq l(K-(r-1)P)$: if $f \in l(K-rP)$, this means by definition that $K-rP + (f) \geq 0$.

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But $0 \leq K-rP+(f)=K-(r-1)P-P+(f) \leq K-(r-1)P + (f)$, so that $f \in l(K-(r-1)P)$ also. Hence $l(K-rP) \leq l(K-(r-1)P)$.