Degree of a Projective Curve

Question

Let $f \in k[X,Y,Z]$ be a homogeneous polynomial of degree $d$ and $V_+(f) := C \subset \mathbb{P}^2_k$ interpreted as closed subscheme (therefore $1$-dimensional, proper $k$-scheme) of $\mathbb{P}^2$. Let $i:C \to \mathbb{P}^2$ be the inclusion morphism.

The degree of $C$ is by definition defined via $\deg(C) := \deg(i^*\mathcal{O}_{\mathbb{P}^2}(1))$, where $\mathcal{O}_{\mathbb{P}^2}(1)$ is the tautological line bundle.

Here the $deg$-map from $Pic(C) \to \mathbb{Z}$ for an invertible sheaf $\mathcal{L}$ on $C$ is defined via $deg(\mathcal{L}) := \chi(\mathcal{L})- \chi(\mathcal{O}_C)$, where $\chi$ is the euler characteristics.

How to deduce that $d=\deg(C)$ holds?

Answer 1

Edit: Short answer: If we assume as known that $$\tag{1}\chi(\mathbb{P}_k^n,\mathscr{O}_{\mathbb{P}_k^n}(m))=\binom{n+m}{n},$$ then we can immediately use the short exact sequence

$$0\rightarrow \mathscr{O}_{\mathbb{P}_k^n}(-d)\rightarrow\mathscr{O}_{\mathbb{P}_k^n}\rightarrow i_\ast\mathscr{O}_C\rightarrow 0,$$

(which corresponds, on the level of graded modules, to $0\rightarrow S(-d)\overset{f\cdot}{\rightarrow}S\rightarrow S/(f)\rightarrow 0$), its respective shifts and the additivity of the Euler characteristic to the effect that $$\chi(\mathscr{O}_C(m))=\chi(i_\ast\mathscr{O}_C(m))=\chi(\mathscr{O}_{\mathbb{P}_k^n}(m))-\chi(\mathscr{O}_{\mathbb{P}_k^n}(m-d))$$ for any $m$, in particular, $$\deg(C)=\chi(\mathscr{O}_C(1))-\chi(\mathscr{O}_C)=\binom{2+1}{2}-\binom{2+1-d}{2}-1+\binom{2-d}{2}=d$$

Another edit: I just realized I really didn't like my old answer, so let me replace the rest of it with only a proof of (1), for completeness. Recall the definition of the Hilbert polynomial:

The Hilbert Polynomial $P_{\mathscr{F}}(X)\in\mathbb{Q}[X]$ for some coherent $\mathscr{O}_{\mathbb{P}_K^n}$-module $\mathscr{F}$ on is the unique polynomial satisfying $$P_{\mathscr{F}}(r)=\dim_k(H^0(\mathbb{P}_k^n,\mathscr{F}(r)))$$ for all $r\gg 0$ (i.e. for all $r\geq r_0$ for some fixed $r_0$).

Now, consider $\mathscr{F}:=\mathscr{O}(m)$ (writing $\mathscr{O}:=\mathscr{O}_{\mathbb{P}_k^n}$ as a shorthand) and write $S:=k[x_0,\ldots,x_n]$ for the homogeneous coordinate ring. Then, for $r\geq -m$, we have (counting the number of polynomials of the respective degree in $n+1$ variables)

$$\dim_k(H^0(\mathbb{P}_k^n,\mathscr{O}(m+r)))=\dim_k(S_{m+r})=\binom{n+m+r}{n}, $$

which clearly is a polynomial in $\mathbb{Q}[r]$. On the other hand, the map

$$r\mapsto \chi(\mathscr{O}(m+r))$$

is known to be a polynomial as well (as a sum of binomial coefficients containing integers and $r$), so in order to verify

$$\chi(\mathscr{O}(m+r))=P_{\mathscr{O}(m)}(r)=\binom{n+m+r}{n}\quad\text{for all }r\in\mathbb{Z}, $$

which gives the desired result for $r=0$, it certainly suffices to prove that both sides agree on infinitely many integers $r\in\mathbb{Z}$. But for $m>n$, we certainly know that $H^j(\mathbb{P}_k^n,\mathscr{O}(m))=0$ for all $j>0$, so for any $r>m-n$, we have

$$\chi(\mathscr{O}(m+r))=\sum(-1)^j\dim_k(H^j(\mathbb{P}_k^n,\mathscr{O}(m+r)))=\dim_k(H^0(\mathbb{P}_k^n,\mathscr{O}(m+r)))=P_{\mathscr{O}(m)}(r).$$