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  "body": "\n## Statement\n\nLet $k$ be an [[algebraically closed field|0PF6TM8TWT]] and let $C \\subset \\mathbb{P}^2_k$ be a smooth [[plane curve|EWS4WFTA7W]] of degree $d \\ge 1$ in the [[projective plane|ZT9A6JK8QF]].\nThen the [[genus|6W8TDV9VVN]] of $C$ is\n$$g(C) = \\frac{(d - 1)(d - 2)}{2}.$$\n\n## Remarks\n\nOne proof uses the adjunction formula: the canonical bundle of $C$ is $\\mathcal{O}_C(d - 3)$, of degree $d(d - 3)$, and this equals $2g - 2$.\nAnother uses the exact sequence $0 \\to \\mathcal{O}_{\\mathbb{P}^2}(-d) \\to \\mathcal{O}_{\\mathbb{P}^2} \\to \\mathcal{O}_C \\to 0$.\nIt gives $\\chi(\\mathcal{O}_C) = 1 - (d - 1)(d - 2)/2$, so the [[arithmetic genus|H03GKRVBJ1]] of every plane curve of degree $d$, smooth or not, is $(d - 1)(d - 2)/2$.\nExample: lines and conics have genus $0$, smooth cubics have genus $1$, and smooth quartics have genus $3$.\nSmoothness is needed: the nodal cubic $y^2 z = x^2(x + z)$ has degree $3$ but geometric genus $0$.\nFor an irreducible plane curve whose singular points $p$ are ordinary with multiplicities $r_p$, the geometric genus is $(d - 1)(d - 2)/2 - \\sum_p r_p (r_p - 1)/2$.\n\n## Explanations\n\n### L1 human\n\nFor a smooth curve in the plane, the degree alone fixes the genus.\nDegree $1$ and $2$ give spheres, degree $3$ gives a torus, and degree $4$ gives a surface with three handles.\nThe genus grows roughly like half the square of the degree.\n\n### L2 human\n\nThe formula shows that not every genus occurs for smooth plane curves: the values are $0, 1, 3, 6, 10, \\ldots$.\nSo a curve of genus $2$ is never a smooth plane curve, although it can be drawn in the plane with singular points.\nThe 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.\n",
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