I am trying the following problem from Hartshorne:
Original Problem (IV.3.2.i): Let $X$ be a plane curve of degree 4. Show that the effective canonical divisors on $X$ are exactly the divisors $X \cdot L$, where $L$ is a line in $\mathbb{P}^2$.
Easy part: I know that all divisors of the form $X \cdot L$ are canonical effective divisors, so there is a map $\phi:\{\text{lines in } \mathbb{P}^2 \} \to |K|: L \mapsto X \cdot L$, where $K$ is a canonical divisor of X. I need to show that this map is surjective.
My idea: Let $\mathbb{P}^2_c$ be the classical projective space, i.e., the subespace of closed points in $\mathbb{P}^2$. As $\ell(K) = g = 3$, there are bijections $\{\text{lines in } \mathbb{P}^2 \} \leftrightarrow \mathbb{P}^2_c \leftrightarrow |K|$ and i want to show that $\phi: \mathbb{P}^2_c \to \mathbb{P}^2_c$ is a morphism of algebraic varieties. It is easy to see that $\phi$ is injective (by Bezout's theorem). So it would follow that $\phi(\mathbb{P}^2)$ is a closed subset of $\mathbb{P}^2$ of dimension 2 and then $\phi$ is surjective.
The question: Is there an easy way to see that $\phi$ is a morphism? I tried opening in affine charts of $\mathbb{P}^2_c$ but nothing came out.
Maybe this is not an easy way to do the original problem, but this idea sounds pretty to me.
Best Answer
"Lines in $\Bbb P^2$" is exactly the projectivization of the global sections of $\mathcal{O}_{\Bbb P^2}(1)$. So the map you're looking for is the projectivization of the map $\Gamma(\Bbb P^2,\mathcal{O}_{\Bbb P^2}(1))\to\Gamma(X,\omega_X\cong\mathcal{O}_{\Bbb P^2}(1)|_X)$. But this is an isomorphism: starting from the exact sequence $$0\to\mathcal{I}_X\to \mathcal{O}_{\Bbb P^2}\to\mathcal{O}_X\to 0,$$ twist by $1$, identify $\mathcal{O}_X(1)$ with $\omega_X$, take global sections, and see that $\Gamma(\Bbb P^2,\mathcal{I}_X(1))=0$ as $X$ is not contained in a line to get that the map on global sections is injective. Surjectivity follows either from Riemann-Roch and calculating $\dim\Gamma(X,\omega_X)=\dim\Gamma(\Bbb P^2,\mathcal{O}_{\Bbb P^2}(1)$ or the fact that $\mathcal{I}_X\cong\mathcal{O}_{\Bbb P^2}(-3)$, so its $H^1$ vanishes.
You're right that this is slightly dressed up compared to the simplest way to solve the problem, but that's not always a bad thing, and sometimes this sort of perspective can be useful in other arguments.