I am reading a book about the history of algebraic geometry. In this I came upon an interesting definition of a very ample divisor:

If $X$ is a smooth projective variety, a divisor $D$ on $X$ is called very ample if it is the section of an immersion of $X$ in a projective space $\mathbb{P}^r$ with a hyperplane of $\mathbb{P}^r$ not containing $X$.

I've been using Hartshorne's definition that a very ample line bundle (divisor) is one which induces a closed embedding and such that the pullback of standard twisting sheaf is isomorphic to the line bundle.

How are these two definitions related? I don't see how the hyperplane from the first definition has anything to do with the definition in Hartshorne.


If $X$ is not too bad (I believe normal should be enough) you have an equivalence between Cartier divisor, line bundle and invertible sheaf. A line bundle $L$ gives you a Cartier divisor (up to linear equivalence) as the zero set of a section $s : X \to L$.

(Edit : as Mohan said, one should really put Cartier here : in fact, the sheaf $O_X(D)$ is an invertible sheaf, i.e a line bundle if and only if $D$ is Cartier)

Now, in term of $\mathcal O(1)$ on $\Bbb P^n$, a section $s$ of this bundle is by definition an homogenous polynomial of degree $1$, and its zero set will be an hyperplane $H$. Now, the pullback of $\mathcal O(1)$ to $X$ is $\mathcal O(1)_{|X}$ and of course $Z(s_{|D}) = Z(s) \cap D = D \cap H$. This shows why the definition of Hartshorne also gives you an hyperplane section.

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    $\begingroup$ Do be careful about what one means by a divisor. Cartier divisors give rise to line bundles, but Weil divisors do not. They coincide only for smooth varieties (locally factorial is enough) in general. $\endgroup$
    – Mohan
    Sep 29 '17 at 19:45
  • $\begingroup$ Thanks for your comment, I will edit my answer. $\endgroup$ Sep 29 '17 at 19:51
  • $\begingroup$ If we view $X$ as a subsheme of $\mathbb{P}^n$, are very ample divisors of $X$ always the result of an intersecting $X$ with the affine hyperplanes of $\mathbb{P}^n$? $\endgroup$
    – user7090
    Sep 29 '17 at 21:52
  • $\begingroup$ No, a very ample divisor $D$ correspond to an embedding into $\mathbb P(H^0(X,O_X(D)))$, and so is a priori unrelated to the embedding in $\Bbb P^n$. In fact, if $D$ is an hyperplane section of $X \subset \Bbb P^n$ then more or less by definition, the set of all hyperplane sections of $X$ will be exactly the set of divisors linearly equivalent to $D$ (hence the name "linear equivalence"). If this is not detailed in Hartshorne, I would suggest to looks at Rick Miranda's book about Riemann surfaces, which is a bit less ambitious but definitely give a good intuition about divisors on curves. $\endgroup$ Sep 29 '17 at 22:05
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    $\begingroup$ @Jadwiga : one example, if you take $X = \Bbb P^1$ embedded as a line, you will get $H \cap X = \{pt\}$ always (until $H = X$). On the other hand, a divisor of degree $2$ will correspond to the embedding $X \to \Bbb P^2$ where $X$ is now a conic, a divisor of degree $3$ is the twisted cubic and for any $d \in \Bbb N$ you get an embedding $\Bbb P^1 \to \Bbb P^{N}$ by sending a point to the evoluation at all monomials of degree $d$. It's called the Veronese embedding. $\endgroup$ Sep 30 '17 at 7:44

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