For an effective divisor $D\ge 0$ on a curve $Y$, define $$\lvert D\rvert =\{ D' \in \mathrm{Div}(Y) \mid D'\ge 0 \;\text{ and }\; D' \sim D \}$$
where $D\sim D$ means $\exists$ a rational function $f$ such that $D' =D+ (f)$.
Prove that $\lvert D\rvert$ is isomorphic to $\Bbb P^{\ell}$ for $\ell =\ell (D)-1$.
I read this proposition (from Griffiths and Phillip's book introduction to algebraic curve) while studying Clifford's theorem.
I want to learn how to prove this proposition. Please show me this proof. Thank you for helping.
$\ell(D)$ by definition is the dimension of the space $$L(D)=\{ f | D+ (f) \geq 0\}$$ Note further that for a constant $a$, $(af)=(f)$. Now there are $f_1, \ldots ,f_{\ell}$ forming a basis of $L(D)$. (Sorry I changed your notation, I use $\ell=\ell(D)$) Thus for any $D^{\prime} \in |D|$ there is $f$ such that $D^{\prime}=D+(f)$ and $f$ is unique up to an multiplicative constant. Since if $(f)=0$ then $f$ is constant. Thus there are $a_1, \ldots , a_{\ell} \in k$ such that $D^{\prime}=D+(a_1 f_1+\cdots a_{\ell}f_{\ell})$ The mapping $$D^{\prime}\mapsto (a_1, \ldots ,a_{\ell} )$$ then defines an isomorphism of $|D|$ with $\mathbb{P}^{\ell -1}$.