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What is the easiest way to prove that a divisor $D$ is very ample if and only if $l(D - P - Q) = l(D) - 2$ for all points $P, Q \in C$. It seems like it might be a consequence of the Riemann-Roch theorem, but I am not sure how to deduce this from the said theorem.

glebovg
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  • I don't see how to make this pop out of Riemann-Roch. One thing RR would help you say is that if $\deg D \geq 2g+1$ then $D$ satisfies this condition. – Hoot Nov 25 '14 at 11:16

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Hint: Show first that it's globally generated. This should be simple since

$$\ell(D-P)\leqslant \ell(D-P-Q)+1=\ell(D)-1$$

so $\ell(D)-\ell(D-P)\ne 0$. Show that this implies that $\mathcal{L}$ is globally generated (Hint: think about what this means in terms of 'having a section which doesn't vanish there').

Then, you want to show that the map $\varphi_D:C\to \mathbb{P}^m$ is a closed embedding. You want to show that it separates tangent vectors and points. Think, onc again, about what this means (similar to the parenthetical hint of the last paragraph) and show that your condition implies this.

Then use the fact in, say Hartshorne, that separating tangent vectors and points is enough to be a closed embedding (morally this is because your morphism is then a proper monomorphism, which is a closed embedding).

Remark: The last step of the above (if you look at Hartshorne's proof) uses the fact that $k=\bar{k}$. This isn't necessary since you can just tensor up to $\bar{k}$, prove it's a closed embedding (since separating tangent vectors and points isn't effected by field extension) and then conclude your original morphism was a closed embedding.

Alex Youcis
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  • I do not understand your proof. It is too convoluted for me. – glebovg Nov 25 '14 at 08:26
  • @glebovg It's not a proof, it's a hint for a proof. The goal was for you to fill in the necessary steps. What are you having difficulties with? – Alex Youcis Nov 25 '14 at 08:30
  • The hints do not help me at all. I do not mean to insult you. Algebra is not my strong suit, so I do not understand what you are trying to communicate. – glebovg Nov 25 '14 at 08:38
  • @glebovg You have to be more specific than that. Do you know what all the words mean (globally generated, very ample, etc.)? What book are you using? Or, what is the language that you are using? – Alex Youcis Nov 25 '14 at 08:40
  • I do not know what 'globally generated' and 'closed embedding' mean. Also, what do you mean by 'a section which does not vanish'? Moreover, what does it mean to 'separate' tangent vectors and points? We are not using Hartshorne's book, so I do not think I can cite results therein. In fact, I think Hartshorne actually proves what I want to prove in his book. We mostly used Shafarevich and Fulton. – glebovg Nov 25 '14 at 08:49
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    @glebovg This is very confusing to me. If you don't know what a closed embedding is, then I don't know how you are defining a very ample divisor? Can you tell me your definition please? – Alex Youcis Nov 25 '14 at 08:54
  • I was about to do that. Let $C$ be a projective nonsingular irreducible curve. Let $D$ be a divisor on $C$. Suppose $l(D) = n > 0$ and $L(D) = \langle f_1, \ldots, f_n \rangle$. Consider the map $\varphi_D : C \to \mathbb{P}^{n - 1}$ defined by $\varphi_D(P) = (f_1(P) : \cdots : f_n(P))$. We say that $D$ is very ample if $\varphi_D$ gives an isomorphism to its image. – glebovg Nov 25 '14 at 09:02
  • @glebovg That's essentially what a closed embedding is, but you require that the image be closed (which it will automatically be since $C$ is projective). I don't know what techniques are available to you in, say, Shafarevich. The separating tangent vectors/points technique is just a criterion for which your sections $f_i\in L(D)$ give a closed embedding $\varphI_D:C\to\mathbb{P}^m$. You should look at the proof of this in Hartshorne (I think you shoul d be able to understand it). Once you do, you can quickly check (as I suggested above) that your condition implies that your $f_i$ separate – Alex Youcis Nov 25 '14 at 09:06
  • tangent vectors and points. Also, globally generated comes up in your definition. Not every line bundle $L$ has a set of sections which gives a map to projective space. Globally generated essentially just means that it has enough global sections to give such a map. – Alex Youcis Nov 25 '14 at 09:07
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    Dear glebovg, take a look at Miranda's "Algebraic Curves and Riemann Surfaces", Proposition 4.20 (and everything leading up to it). This will show you the definitions that Alex is using in a simple way. – rfauffar Nov 25 '14 at 14:01
  • @rfauffar Thanks. – glebovg Nov 26 '14 at 05:06