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The projective completion has identity $O=[0:1:0]$. For this Weierstrass equation of an elliptic curve, negation isFor distinct affine points $P,Q$, draw their chord; for $P=Q$, use the tangent with its intersection multiplicity. If the third intersection is $R$, define $P\oplus Q=-R$. A vertical chord pairs $P$ with $-P$ and gives their sum $O$; also $P\oplus O=P$. A vertical tangent at a point satisfying $2y+1=0$ gives $2P=O$.

For a nonvertical line $y=mx+b$, substitution into the cubic givesThus if $P=(x_1,y_1)$ and $Q=(x_2,y_2)$, the sum has coordinatesHere $m=(y_2-y_1)/(x_2-x_1)$ for a chord, and implicit differentiation givesfor a nonvertical tangent. These formulas include the repeated-root case. Associativity can be seen without a lengthy coordinate calculation: a line cuts the cubic in a divisor equivalent to $3(O)$, and $P\mapsto [(P)-(O)]$ identifies the curve with its degree-zero divisor class group. The chord-and-tangent operation becomes addition of divisor classes, hence is associative.

Let $H$ denote a hyperplane-section divisor of a projective plane curve. The adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

Let the smooth projective curve be the projective completion of $f(x,y)=0$, of degree $d$, and suppose after interchanging $x$ and $y$ if necessary that $f_y=\partial f/\partial y$ is nonzero. A basis of its holomorphic differential forms isEquivalently one may use $-x^i{y}^{j}dy/f_x$. These expressions agree on their common domain because differentiating $f(x,y)=0$ gives $f_{x}dx+f_{y}dy=0$. The allowed monomials are precisely the homogeneous polynomials of degree at most $d-3$, in accordance with the adjunction formula $K_C\cong {\mathcal{O}}_C(d-3)$.

The projective completion ofis the smooth plane quarticThe coordinate relabelling $[A:B:C]=[Y:X:Z]$ turns this equation into the standard Klein quartic equation $A^{3}B+B^{3}C+C^{3}A=0$.