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Projection from a point of a smooth plane quartic gives a degree-three map to the projective line. A degree-two map would make the curve hyperelliptic, but its canonical map is its plane embedding, whereas a hyperelliptic canonical map is not an embedding. Its gonality is therefore three.

For a smooth plane quartic $C\subset {\mathbb{P}}^2$, the Adjunction formula gives $K_C\cong {\mathcal{O}}_C(1)$. Hence every line section $C\cap \ell$ is a canonical divisor.

Homogenizing the affine equation gives the projective closureIts partial derivatives areIf all three vanished at a point of $X$, then $X=0$ and either $Y=0$ or $Z=0$. If $Y=0$, the equation $F=0$ forces $Z=0$; if $Z=0$, the equation $F_Z=0$ forces $Y=0$. Either conclusion gives the forbidden zero triple of homogeneous coordinates. The Jacobian criterion therefore shows that $X$ is a smooth projective curve; more specifically, it is the projective closure of y cubed equals x to the fourth plus one, a smooth plane quartic.

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 $(4-1)(4-2)/2=3$.

Choose a point $p\in C$ and project from $p$ to a line not containing $p$. Every line through $p$ meets the quartic in four points counted with intersection multiplicity by Bézout theorem; removing the fixed intersection at $p$ leaves a generic fibre of three points. This projection from a point on a plane curve therefore defines a nonconstant morphismof degree three, so the gonality is at most three.

It cannot be one, because a degree-one morphism between smooth projective curves is an isomorphism of algebraic varieties, whereas $C$ has genus three and the projective line has genus zero. It cannot be two either: that would make $C$ a hyperelliptic curve. The canonical map of a hyperelliptic curve is not a closed embedding, but for a smooth plane quartic the canonical map of a smooth plane curve is exactly its given embedding $C\hookrightarrow {\mathbb{P}}^2$, since $K_C\cong {\mathcal{O}}_C(1)$. This contradiction proves

For a divisor on an algebraic curve $D$ on a smooth projective curve $X$ of genus $g$, the Riemann-Roch theorem stateswhere $K_X$ is a canonical divisor. Taking $D=0$ gives $1-\ell (K_X)=1-g$, so $\ell (K_X)=g$. Taking $D=K_X$ then gives

To obtain a uniform projective embedding, choose a divisor $D$ of degree $2g+1$. Since every divisor appearing below has degree greater than $2g-2$, Riemann--Roch givesThe first two equalities show that the complete linear system of a divisor $|D|$ has no base point. The strict drops in the last two comparisons show respectively that its sections separate distinct points $P,Q$ and tangent directions at $P$. Thus $D$ is a very ample divisor, in accordance with the general fact that a high-degree divisor is very ample on a smooth projective curve, and its sections define a closed embeddingThe ambient dimension therefore depends only on $g$.

The Riemann-Hurwitz formula for a nonconstant morphism $h:C_1\to C_2$ of degree $n$ isChoose a smooth plane quartic $C$, so $g(C)=3$, and form the product of projective varieties $S=C\times C$. This is a smooth projective variety of dimension two. If $B\subset S$ is an irreducible curve, pass to its normalization $\tilde{B}$. At least one coordinate projection $\tilde{B}\to C$ is nonconstant, since otherwise $B$ would be a point. For that projection, Riemann--Hurwitz givesso the geometric genus of $B$ is at least three. Hence $S$ is the required surface; this is the product surface without low-genus curves construction.

Finally let $X\subset {\mathbb{P}}^2$ be a smooth plane curve of degree $d$ and let $p\notin X$. After a projective change of coordinates, take $p=[0:0:1]$. Projection away from $p$ isIts two homogeneous coordinate functions cannot vanish simultaneously on $X$, because their common zero in ${\mathbb{P}}^2$ is $p$. The criterion for a morphism of algebraic varieties therefore shows that the restrictionis a morphism. A fibre is the intersection with a line through $p$, and a general such line meets $X$ in $d$ points counted with multiplicity. Thus the projection of a plane curve from an exterior point has degree $d$.

By the genus of a smooth plane curve, $g(X)=(d-1)(d-2)/2$. Applying Riemann--Hurwitz to $\pi$ and its ramification divisor $R_{\pi }$ givesEvery ramification point contributes at least one to this degree, so

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 $A^{3}B+B^{3}C+C^{3}A=0$.