Relationship between very ample divisors and hyperplane sections 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!)
 A: 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 $X\cap H$
A: If $D$ is ample ($\Longleftrightarrow \deg(D)>0$) on a curve $X$, then $nD$ is very ample for a large $n$ and gives enough global sections to define an immersion $i: X \longrightarrow \mathbb{P}^n$ s.t. $\mathcal{O}_X(nD) \cong i^*\mathcal{O}_{\mathbb{P}^n}(1)$.
If your $X$ is proper in the first place then $i$ is a closed immersion.
Effective divisors on $\mathbb{P}^n$ corresponding to the invertible sheaf $\mathcal{O}_{\mathbb{P}^n}(1)$ (or say, "the complete linear system of $\mathcal{O}_{\mathbb{P}^n}(1)$") consists exactly of hyperplanes in $\mathbb{P}^n$.
Suppose $\xi_1,\dots,\xi_r$ are generic points of $X$ and $H$ is a hyperplane in $\mathbb{P}^n$ NOT containing any of them (so that "the pullback of $H$ by $i$ is defined"), then $i^*H = X \cap H$ is an effective divisor on $X$. Such divisors are called "hyperplane sections" and there are plenty of them.
Now $i^*H$ is a global section of $i^*\mathcal{O}_{\mathbb{P}^n}(1) \cong \mathcal{O}_X(nD)$, then of course $i^*H \sim nD$.
