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For a smooth degree-$d$ projective plane curve $X$, the Adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. A hyperplane section has degree $d$, so

The arithmetic genus of a degree-$d$ projective plane curve isEquivalently, its structure sheaf has Euler characteristic $\chi ({\mathcal{O}}_X)=1-p_a$.

If a projective line $H$ does not contain the projective plane curve $X$, its hyperplane-section divisor iswhere $I_p(X,H)$ is the intersection multiplicity. The Bézout theorem gives $\mathrm{deg}[X\cap H]=(\mathrm{deg}X)(\mathrm{deg}H)=\mathrm{deg}X$.

If a smooth degree-$d$ projective plane curve had genus two, the genus-degree formula would give $(d-1)(d-2)=4$, equivalently $d^2-3d-2=0$. Its discriminant is $17$, so it has no integral solution for the degree $d$.

An isomorphism of schemes over $\mathbb{R}$ induces a bijection on rational points. A real point of this projective plane curve would be represented by a nonzero real triple with $t_0^2+t_1^2+t_2^2=0$. Each summand is nonnegative, so every coordinate would vanish. Thus $X(\mathbb{R})=\varnothing$, whereas $[1:0]\in {\mathbb{P}}^1(\mathbb{R})$. The two real schemes are not isomorphic. The obstruction is the real conic without real rational points; nonsingularity alone does not make a real conic a projective line over its ground field.

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

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

A plane cubic is a projective plane curve defined by a nonzero homogeneous polynomial of degree three. Its arithmetic genus is one by the genus-degree formula, and its Hilbert polynomial is $3m$. A singular cubic can have normalization of smaller genus; the Hilbert polynomial measures the arithmetic genus.

This smooth projective plane curve has no real rational points: a sum of three real squares is zero only when all three coordinates vanish, which is forbidden in projective space. It is therefore not isomorphic over $\mathbb{R}$ to the projective line, which has rational points. Over $\mathbb{C}$ it becomes a nonsingular conic isomorphic to a projective line. Thus the ground field changes whether a geometric genus-zero curve admits a projective-line parametrization.

A smooth complex degree-two projective plane curve is a projective linear image of the degree-two Veronese map and is diffeomorphic to a sphere. Its class in ${\mathbb{CP}}^2$ is twice the class of a projective line. Its self-intersection number is four, so its oriented normal bundle has Euler class evaluating to four.