I already searched on the site and there is several topics which deal with this question, but actually it doesn't make it clear as crysal to me. For the context, we take $X$ a good variety (let say smooth), and $D$ a divisor on $X$. Then, we denote by : $\mathcal{L}(D) = \{ f \in k(X)^\times \, | \, \text{div}(f)+D \geq 0\} \cup \{0 \}$. We take $(f_0, \dots, f_n)$ a base of $\mathcal{L}(D)$. It induces a rational map $\phi = [f_0 : \dots f_n]$ from $X$ to $\mathbb{P}^n$. We then suppose that $D$ is very ample. By definition, it means that $\phi$ is an embedding.
I want to show that the complete linear system $|D|$ equal to $\phi^*(H)$, where $H$ is an hyperplane of $\mathbb{P}^n$.
Attempt (induced by the book of Shafarevich and the book of Miranda) : We denote by $D'= -\text{hcd}(\text{div}(f_0), \dots, \text{div}(f_n))$. It means that if $\text{div}(f_i) = \sum_j k_{ij}C_j$ for all $i$, then we have $D' = \sum_j -\min_i(k_{ij})C_j$. Let $H$ an hyperplane of $\mathbb{P}^n$ given by the equation : $\sum_{i=0}^n \lambda_ix_i = 0$, where the $\lambda_i$ are not all zero.
Let $x \in X$ and $i$ such that $f_i(x) \neq 0$. Then, $\phi^*(H)$ has the local equation on $x$ : $\frac{1}{f_i} \sum_{j=0}^n \lambda_jf_j$. Moreover, $\text{div}(\sum_{j=0}^n \lambda_jf_j) + D'$ has local equation on $x$ given by : $\sum_{j=0}^n \lambda_jf_j \cdot \frac{1}{h}$ where $uh = f_i$ for some $u \in \mathcal{O}_{x}^\times$. Then, we find that $\text{div}(\sum_{j=0}^n \lambda_jf_j) + D' = \phi^*(H)$.
Then, all we need to prove is that $D=D'=-hcd(\text{div}(f_0), \dots, \text{div}(f_n))$. But I don't succeed to prove that it's true ?
Please don't use line bundle theory on your answer, I'm not familiar with it at the moment.
Thank you !
I think I finally find the answer. Maybe someone could correct me if I'm wrong ?
We know that we have equivalence between linear systems $L$ of dimension $n$ without fixed components and morphisms $\phi : X \rightarrow \mathbb{P}^1$ with image not contained in a hyperplane, up to projective isomorphism.
Now, if the linear system $|\phi| = |D'|$ (notations from Miranda's book) don't have any fixed components, then, as for all $Q \in D'$, $Q$ can be written as $Q = \text{div}(f) + D'$ where $\text{div}(f) \in \mathcal{L}(D)$ and $D' = hcd(\text{div}(f_0), \dots \text{div}(f_n)$, then we have : $Q \geq -D+D'$. Then, it implies that $D'=D$.
If $\phi(X)$ is contained in any hyperplane, then the $(f_i)_i$ are not a basis of $\mathcal{L}(D)$, which leads to a contradiction.