Genus–degree formula

Remarks

One proof uses the adjunction formula: the canonical bundle of $C$ is $\mathcal{O}_C(d - 3)$, of degree $d(d - 3)$, and this equals $2g - 2$. Another uses the exact sequence $0 \to \mathcal{O}_{\mathbb{P}^2}(-d) \to \mathcal{O}_{\mathbb{P}^2} \to \mathcal{O}_C \to 0$. It gives $\chi(\mathcal{O}_C) = 1 - (d - 1)(d - 2)/2$, so the arithmetic genus of every plane curve of degree $d$, smooth or not, is $(d - 1)(d - 2)/2$. Example: lines and conics have genus $0$, smooth cubics have genus $1$, and smooth quartics have genus $3$. Smoothness is needed: the nodal cubic $y^2 z = x^2(x + z)$ has degree $3$ but geometric genus $0$. For an irreducible plane curve whose singular points $p$ are ordinary with multiplicities $r_p$, the geometric genus is $(d - 1)(d - 2)/2 - \sum_p r_p (r_p - 1)/2$.

Going deeper

For a student

The formula shows that not every genus occurs for smooth plane curves: the values are $0, 1, 3, 6, 10, \ldots$. So a curve of genus $2$ is never a smooth plane curve, although it can be drawn in the plane with singular points. The formula is a special case of the adjunction formula, and its singular version is the basic tool to compute the genus of a plane curve from its equation.