Is a divisor in the hyperplane class necessarily a hyperplane divisor?

Question

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 $n > 2$?

Answer 1

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 $\sigma_n(\mathbb{P}^1)$.

Answer 2

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 .