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Edit: Apparently the answer is "no", so what is an example of two curves of genus g, and a divisor of degree d on each, such that one is very ample and the other is not?

Question as originally stated:

Suppose X is a complete nonsingular curve (smooth proper integral scheme of dimension 1 over C) and D ∈ DivX.

I have heard that very ampleness of D is determined entirely by its degree and the genus of X. How can this be done explicitly?

Thanks!

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5 Answers 5

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Let $C_1$ be a hyperelliptic curve of genus $g \geq 3$ (example: $y^7 = x^2 + 1$ for $g = 3$), and $C_2$ be a non-hyperelliptic curve of the same genus $g$ (for example, the Klein quartic with $g = 3$ again: I'll use it in the form $y^7 = x^2(x-1)$).

Then let $K_1$, $K_2$ be the canonical divisors of $C_1$, $C_2$ respectively (In the example above: $K_1 = 4[\infty_1]$ and $K_2 = 4[\infty_2]$, where $\infty_1, \infty_2$ are the "points at infinity" on the curves above: note that the projective curves corresponding to the equations above both have singularities at infinity, but in both cases these singularities can be resolved to give a single point at infinity on each curve.) Then $K_1$ is not very ample but $K_2$ is: this is a standard application of Riemann-Roch (see also Hartshorne, Chapter IV, Proposition 5.2.) The reason that $K_1$ is not very ample is that the map $C_1 \rightarrow P^2$ associated to $K_1$ is a double cover of a rational curve (a conic, in fact) in $P^2$, rather than an embedding.

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Not entirely by its degree; see Hartshorne Chapter 4, section 3 (Prop. 3.1 and corollary 3.2) and section 5.

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    $\begingroup$ Hart. Prop 5.2 says the canonical divisor K on a curve C of genus g > 1 is very ample iff C is not hyperelliptic. $\endgroup$ Commented Oct 19, 2009 at 4:28
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I'm late to the game, but I would like to point out that the answer is systematically no. One class of examples. Suppose g>2 for simplicity. In that case any general line bundle of degree 2g is very ample and special ones are not. This can be seen by using the criteria that a line bundle is very ample iff for any effective divisor $D$ of degree 2, $h^0(L(-D) = h^0(L)-2$. One checks using R-R that this holds iff $L$ is not of the form $L= K_C(D)$ where $D$ is an effective divisor of degree 2. Line bundles of the form $L= K_C(D)$ are a 2 dimensional subset of the g dimensional (Picard) variety. This can be expanded upon.

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The other way to construct examples is just by considering curves in a given projective space. For example, a plane curve $C$ of degree $d$ has genus $g = \frac{(d-1)(d-2)}2$. One can, of course, rephrase this as saying that $\mathcal{O}_C(1)$ is a very ample divisor of degree $d$ on $C$. On the other hand, as $d < g$, a general divisor of degree $d$ on $C$ does not even have a single global section.

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I am maybe misunderstanding, but I think that ypur example is fulfilled by a couple of curves of gwnus bigger than 3, one hyperelliptic and the other not. If you consider the canonical divisor, it is very ample on the non-hyperelliptic whereas it factors through the hyperelliptic pencil on the hyperelliptic curve.

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    $\begingroup$ This seems to be a duplicate of Alison Miller's accepted answer. $\endgroup$
    – user5117
    Commented Feb 13, 2014 at 15:40
  • $\begingroup$ ha! I was answering from my cell phone and I did not notice it... $\endgroup$
    – IMeasy
    Commented Feb 14, 2014 at 7:43

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