Very ample divisors on $\mathbb{P}^n$

Question

Suppose $D$ is a divisor on a curve $X$, and we can embed it into some projective space, say $\mathbb{P}^n$. Then if $D$ is very ample, why it has to be linearly equivalent to a hyper section H of $\mathbb{P}^n$?

Any help is appreciated!

Answer 1

This is not true. Take $X= \Bbb P^1$ embedded as a line in $\Bbb P^2$. Then for any point $p \in X$, $2p$ is very ample but it is not equivalent to an hyperplane section since a linear section is a point.

It is true if the embedding $\phi : X \to \Bbb P^n$ is induced by a divisor linearly equivalent to $D$ but then it is almost tautological.