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For a projective variety $X\subseteq {\mathbb{P}}^n$ and a projective hyperplane $H$, the hyperplane section is the scheme-theoretic intersection $X\cap H$. The defining linear equation restricts to a global section of the line bundle ${\mathcal{O}}_X(1)$.

To view its zero scheme as an effective Cartier divisor, the restricted section must be a non-zero-divisor locally. For an integral $X$, this holds when $H$ does not contain $X$. If $X\subseteq H$, the restriction is the zero global section and the intersection is all of $X$, rather than a codimension-one Cartier divisor. More generally one must avoid the associated points that would make the local equation a zero divisor.

For a projective plane curve not contained in a line $H$, intersection multiplicities record tangencies in the section. The hyperplane-section divisor of a projective plane curve has total degree equal to the degree of the curve. The restricted bundle ${\mathcal{O}}_X(1)$ exists whether or not the particular section is regular.

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 $H=V(L)$ for a nonzero homogeneous linear polynomial $L$. Because $H$ does not contain $X$, the restriction $L{|}_X$ is a nonzero section of ${\mathcal{O}}_X(1)$. Its zero divisor is the hyperplane-section divisor of a projective plane curvewhere $I_p(X,H)$ is the local intersection multiplicity; equivalently, it is the order of vanishing of $L{|}_X$ at $p$.

The curves $X$ and $H$ have no common component and have degrees $d$ and $1$. The Bézout theorem therefore gives