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Let $V$ be a smooth irreducible projective curve over an algebraically closed field $k$, embedded in some projective space $\mathbb{P}^n$, and let $[H]$ be the induced hyperplane divisor class on $V$.

Question. Suppose $D$ is an effective divisor on $V$ such that $D$ is in $[H]$. Is $D$ necessarily realised as the intersection divisor of some hyperplane?

It is quite straightforward to see that the intersection divisor of any hyperplane in general position is indeed in $[H]$ – this seems to be the whole point behind linear equivalence.

I can prove this for plane curves using Bézout's theorem: after all, if $H$ and $D$ are linearly equivalent, then $\deg H = \deg D$, and so if $H - D$ is the divisor of a rational function $h$, then it must be possible to find linear forms $F$ and $G$ such that $h = F / G$, and then $H = V (F)$ while $D = V (G)$. But what about the case where $n > 2$?

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  • $\begingroup$ I don't understand your proof that $h$ can be represented by a quotient of two linear forms. Did you use some well-known theorem here? $\endgroup$ – Yuchen Liu May 7 '13 at 7:31
  • $\begingroup$ If $F$ and $G$ are homogeneous of degree $m$ and coprime, then $F / G$ will be a rational function with $m \deg H$ zeros and $m \deg H$ poles, counted with multiplicity, by Bézout's theorem. $\endgroup$ – Zhen Lin May 7 '13 at 8:03
  • $\begingroup$ Yes, but why can you choose $F$, $G$ such that their degree are both $1$? $\endgroup$ – Yuchen Liu May 7 '13 at 11:04
  • $\begingroup$ Well, if if $h$ is a rational function, then there are some coprime $F$ and $G$ such that $h = F / G$. If they weren't of degree $1$ then that would contradict the condition on the degrees of the divisors. $\endgroup$ – Zhen Lin May 7 '13 at 11:10
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    $\begingroup$ By Max Noether's AF+BG theorem, $div(G)=H+E$ implies we can find a polynomial $G_1$ s.t. $E=div(G_1)$ and $\deg G_1=\deg G -1$. Again by AF+BG theorem, $div(F)=D+div(G_1)$ implies we can find a polynomial $F_1$ s.t. $D=div(F_1)$ and $\deg F_1=\deg F-\deg G_1=1$. $\endgroup$ – Yuchen Liu May 8 '13 at 5:52
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Definition:
If $\mathcal I=\mathcal I_V\subset \mathcal O_{\mathbb P^n}$ is the ideal sheaf defining the curve $V$, we have the exact sequence of sheaves on $\mathbb P^n$: $$0\to \mathcal I\to \mathcal O_{\mathbb P^n} \to \mathcal O_V \to 0$$ which gives after twisting by $\mathcal O_{\mathbb P^n}(1)$: $$ 0\to \mathcal I(1)\to \mathcal O_{\mathbb P^n} (1)\to \mathcal O_V (1) \to 0 $$ The long associated cohomology sequence has as fragment $$o \to \Gamma (\mathbb P^n,\mathcal I(1))\to \Gamma (\mathbb P^n,\mathcal O_{\mathbb P^n} (1))\to \Gamma (V,\mathcal O_V (1)) \to H^1(\mathbb P^n,\mathcal I(1)) \to H^1(\mathbb P^n, \mathcal O_{\mathbb P^n} (1))=0.$$ Your problem is whether the second morphism is surjective, or equivalently if $ H^1(\mathbb P^n,\mathcal I(1))=0$.
These equivalent properties are called linear normality of $V$.
Beware that linear normality is not an invariant of $V$, contrary to usual normality, but depends on the embedding of $V$ into $\mathbb P^n$.

Useful Criterion:
If $\Gamma (\mathbb P^n,\mathcal I(1))=0$, which just means that $V$ is not included in a hyperplane, then the linear map $0\to \Gamma (\mathbb P^n,\mathcal O_{\mathbb P^n} (1))\to \Gamma (V,\mathcal O_V (1)) $ is injective and linear normality, its surjectivity, is simply equivalent [since $\dim \Gamma (\mathbb P^n,\mathcal O_{\mathbb P^n} (1))=n+1$ ] to the equality $$\dim \Gamma (V,\mathcal O_V (1)) =n+1$$ Complete intersections:
It is known (Hartshorne, Exercise III 5.5.(a), page 231) that complete intersections are linearly normal and this confirms your calculation for curves in $\mathbb P^2$, since they are automatically complete intersections.

Not linearly normal:
On the other hand, the rational quartic curve $C\subset \mathbb P^3$ described parametrically by $(x^4:x^3y:xy^3:y^4)$ is not linearly normal: this is a particular case of the wonderful example given by Yuchen Liu in his answer.
So $C$ is not a complete intersection. This can be confirmed as follows:
A non-plane complete intersection of degree $4$ can only be the intersection of two quadrics (by Bézout's theorem), but then a smooth intersection of two quadrics has genus $1$ and cannot be the rational curve $C$: just look at the formula for the genus in Hartshorne's Exercise 7.2. (d), Chapter I, page 54.
The most elementary and direct way however to see that $C$ is not linearly normal (and thus answers your question in the negative) is to remark, computing dimension with Riemann-Roch, that $\dim \Gamma (C,\mathcal O_C (1)) =5\neq 3+1$, thus violating the equality in the Useful Criterion .

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  • $\begingroup$ About the proof in plane curves, I guess that any smooth curve in $\mathbb{P}^3$ cannot be projected to $\mathbb{P}^2$ being smooth after the projection. My idea to prove this was to prove the secant of the curve in $\mathbb{P}^3$ must be three-dimensional, hence contains the whole space. Then I need a lemma: the image of a smooth curve under projection is still smooth if and only if the center of projection doesn't lie in the secant of the curve. Is this proof correct? How to make it rigorous? $\endgroup$ – Yuchen Liu May 7 '13 at 13:33
  • $\begingroup$ Dear jerry, your proof is almost correct. You must also make sure that the center of projection does not lie on any tangent to the curve. A reference is Hartshorne, Chapter IV, Proposition 3.4, page 309. By the way, are you Chinese? I hope you will excuse my curiosity, but I see you study in Beijing and I'm puzzled by your perfect English. I'm a language freak, I'm afraid... :-) $\endgroup$ – Georges Elencwajg May 8 '13 at 12:06
  • $\begingroup$ Yes, I'm Chinese. Thank you for praising my English:) I guess the reason is that I am used to write in English and think in English when studying math, and also I have many chances to talk to Western people in English, since many foreign professors visited here in recent years. $\endgroup$ – Yuchen Liu May 8 '13 at 12:33
  • $\begingroup$ Dear jerry, at the risk of making you blush I can only reiterate my admiration for the linguistic and mathematical talent you display at your age. $\endgroup$ – Georges Elencwajg May 8 '13 at 12:41
  • $\begingroup$ I have edited my answer to reflect Yuchen Liu's happy decision to use his real name on this site. $\endgroup$ – Georges Elencwajg Nov 20 '14 at 11:00
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Your question is equivalent to the proposition that the linear system of hyperplane divisors of a smooth curve $V$ is complete. This is not true for $n\geq 3$, as the following example will show:

For any $n\geq 3$, consider $\sigma_n:\mathbb{P}^1\rightarrow \mathbb{P}^n$ defined by $[x,y]\mapsto[x^{n+1},x^ny,\cdots,\widehat{x^2y^{n-1}},xy^n,y^{n+1}]$, then $\sigma_n$ is an embedding (you can check this in affine coordinates), and $\sigma_n(\mathbb{P}^1)$ is a projection of degree $(n+1)$ rational normal curve in $\mathbb{P}^{n+1}$ to $\mathbb{P}^n$. Therefore, the linear system of hyperplane divisors of $\sigma_n(\mathbb{P}^1)$ is a codimension $1$ subspace of $|\mathcal{O}(n+1)|$, so it is not complete. To find such a divisor $D$, let $p=\sigma_n([1,0])$, $q=\sigma_n([0,1])$, let $D=(n-1)p+2q$, then $D$ is not a hyperplane divisor of $\sigma_n(\mathbb{P}^1)$.

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  • $\begingroup$ +1 for this brilliant example. Do you know any reference around this kind of result or did you make it up on the spot? $\endgroup$ – Georges Elencwajg May 7 '13 at 13:10
  • $\begingroup$ @GeorgesElencwajg: Here is an idea that I came up with: every smooth curve in projective space of higher dimension can be projected to a 3-dimensional subspace with the image still smooth. Also, I like your answer very much, I haven't heard about linearly normal before! $\endgroup$ – Yuchen Liu May 7 '13 at 13:26

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