The genus of a smooth plane curve in terms of its degree

(Local scheme dimension.) Work over $\bar{k}$. Smoothness of $C_{\bar{k}}$ follows from [F1]. In a standard projective chart containing a closed point $p$, the actual equation is the dehomogenization $f$ of $F$; because $F$ is nonzero, $f$ is nonzero. If ${\mathfrak{m}}_p$ is the maximal ideal of $p$ in $\bar{k}[u,v]$, then $f\in {\mathfrak{m}}_p$ and the local ring $\bar{k}[u,v{]}_{{\mathfrak{m}}_p}\mathrm{/}(f)$ has dimension one by [F3]. The same calculation on all three charts shows every projective component has dimension one. This keeps the dimension argument at the level of affine scheme local rings.

(Square-freeness.) Suppose over $\bar{k}$ that an irreducible homogeneous factor $G$ occurs in $F$ with multiplicity at least two. Choose a linear form $\mathrm{\ell }$ not divisible by $G$. By [F4], $V_{+}(G,\mathrm{\ell })$ is nonempty, so choose a closed point $p$ in that scheme. In a projective chart through $p$, the local equation $f$ of $C_{\bar{k}}$ is divisible by the square of the local equation $g$ of $G$, hence $f\in {\mathfrak{m}}_p^2$. Its two partial derivatives vanish in the residue field at $p$. The local ring has dimension one by [F3], so [F2] says this actual one-equation presentation is not regular. But it is a local ring of the smooth scheme $C_{\bar{k}}$, which is regular by [F1], a contradiction. Thus $F$ is square-free over $\bar{k}$.

(Geometric irreducibility.) If this square-free $F$ were reducible over $\bar{k}$, write it as $F=GH$ with $G,H$ coprime nonconstant homogeneous forms. [F4] gives a closed point $p\in V_{+}(G,H)$. In a chart through $p$, the local equation is $f=gh$ with $g,h\in {\mathfrak{m}}_p$, so again $f\in {\mathfrak{m}}_p^2$ and its Jacobian row has rank zero. The dimension-one local ring is not regular by [F2], contradicting [F1]. Thus $F$ is irreducible over $\bar{k}$. Since a polynomial ring over a field is a UFD, this irreducible form is prime; $(F)$ is prime over $\bar{k}$ and $C_{\bar{k}}$ is integral. Therefore $C$ is geometrically integral. It is proper because it is a closed subscheme of ${\mathbf{P}}_k^2$: by [F10], the closed immersion and projective-space structure morphism are proper, as is their composite.

(The hyperplane degree.) Over $k$, choose a linear form $\mathrm{\ell }$ coprime to $F$: if $d=1$ choose a nonproportional form, and if $d>1$ any nonzero linear form is coprime to the geometrically irreducible $F$. Its restriction $s=\mathrm{\ell }{\mathrm{\mid }}_C$ is a nonzero section of ${\mathcal{O}}_C(1)$, whose zero scheme is the scheme-theoretic intersection $C\cap V_{+}(\mathrm{\ell })$. At each closed point $x$ of this finite intersection, [F5] identifies the local intersection length with the order of $s$ in the DVR ${\mathcal{O}}_{C,x}$: the DVR normal form and quotient-length theorem give $\mathrm{length}({\mathcal{O}}_{C,x}\mathrm{/}(s))=n$ when $s=u{t}^n$ with $t$ a uniformizer. Thus the weighted sum of the local lengths is the degree of the divisor of $s$, namely $\mathrm{deg}({\mathcal{O}}_C(1))$. By [F4] the same weighted sum is $d\cdot 1=d$. Therefore $\mathrm{deg}({\mathcal{O}}_C(1))=d$.

(All twists, with the negative case rational.) By [F6], the twisting identities give ${\mathcal{O}}_C(m)={\mathcal{O}}_C(1{)}^{\otimes m}$ for every integer $m$, using dual powers when $m<0$. Let $D$ be the divisor of the rational section $s=\mathrm{\ell }{\mathrm{\mid }}_C$ from step 4.1. The rational section $s^{\otimes m}$ of ${\mathcal{O}}_C(m)$ has divisor $mD$ for every integer $m$. If $m>0$ it is also a global section; for $m<0$ it is only a rational section and is not asserted to be global. By [F5] and step 4.1, $\mathrm{deg}({\mathcal{O}}_C(m))={\mathrm{deg}}_k(mD)=md$ for every integer $m$.

(Adjunction and the canonical degree.) By [F6], ${\omega }_C\cong {\mathcal{O}}_C(d-3)$, and a canonical divisor is the divisor of a nonzero rational section of this line bundle. Applying [F7] and step 5.1 gives $2g-2={\mathrm{deg}}_k(K_C)=\mathrm{deg}({\mathcal{O}}_C(d-3))=d(d-3)$.

Solving the identity of step 6.1 yields $g=(d^2-3d+2)\mathrm{/}2=(d-1)(d-2)\mathrm{/}2$. Substitution gives genus one for $d=3$ and genus three for $d=4$. The Axiom of Choice [F9] is inherited from the duality, dimension, Jacobian, and Bezout suppliers used above. The proof preserves the original arbitrary-field and all-characteristic scope; the negative twist in adjunction is handled by a rational section.