This question is based on a line in the proof of corollary IV.3.3 in Hartshorne's Algebraic Geometry.
The first line of the proof goes: "if $D$ is an ample divisor (on a curve $X$), then some multiple is very ample, so $nD\sim H$, where $H$ is a hyperplane section for a projective embedding."
This line suggests that there is some obvious relationship between hyperplane sections and very ample divisors, namely that any very ample divisor is linearly equivalent to a hyperplane section.
Why is this true? Is this done somewhere else in Hartshorne? (I've been looking for a long time, but I can't seem to find it anywhere else!)
Best Answer
Look at Chapter II Theorem 7.1, $\phi:X \longrightarrow P^n$ a very ample line bundle $\phi^*(O(1))$ is the pull back of O(1) bundle on $P^n$. But $O(1)$ is a hyperplance $H$, so in the image of $\phi$, the very ample divisor is the intersection of $X\cap H$