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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!)

Misja
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    The very ampleness stuff is back in Chapter II. The way a very ample sheaf is defined is that we can evaluate the global sections generating the sheaf to give projective coordinates for a closed immersion into projective space. Under that scheme, the hyperplane sections are precisely the elements of the very ample linear system. – John Brevik Mar 19 '15 at 16:42
  • Ok, thanks. I think I got it now! – Misja Mar 19 '15 at 16:55

2 Answers2

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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 $X\cap H$

Shuhang
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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 $i^*H \sim nD$.

QU Binggang
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