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!)

  • 5
    $\begingroup$ 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. $\endgroup$ – John Brevik Mar 19 '15 at 16:42
  • $\begingroup$ Ok, thanks. I think I got it now! $\endgroup$ – Misja Mar 19 '15 at 16:55

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 $X\cap H$


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