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For a nonhyperelliptic smooth projective curve of geometric genus four, the canonical map is an embedding of degree six. Dimension counts $10>h^0(2K)=9$ and $20>h^0(3K)=15$ supply a quadric and a cubic not divisible by it. The quadric is irreducible because the curve is nondegenerate. Degree six and unmixedness of a complete intersection then identify the intersection with the curve as a scheme.

First use the canonical map to obtain the embedding. Its canonical divisor has degree six and $h^0(K_C)=4$. Over an algebraically closed ground field, every effective degree-two divisor on an algebraic curve $E$ satisfies $h^0(E)=1$: a second section would give a nonconstant rational function with poles bounded by $E$, hence a map $C\to {\mathbb{P}}^1$ of degree at most two. Degree one would make $C$ rational, and degree two would make it hyperelliptic, both excluded here.

The Riemann-Roch theorem now givesThus the complete linear system of divisors $|K_C|$ separates every pair of points and every tangent direction, including the tests $E=2p$. By the length-two criterion for a very ample linear system, it is a very ample linear system. Consequently the canonical map embeds $C$ in ${\mathbb{P}}^3$, with degree of a projective curve six. Its image is a nondegenerate projective variety, since the four canonical sections are linearly independent.

The space of homogeneous quadrics in four variables has dimension ten, while the Riemann-Roch theorem givesRestriction therefore has a nonzero kernel, giving a quadric $Q$ containing $C$. This quadric is irreducible: a reducible quadric is a union of two planes, or a double plane, and an integral curve lying in it would lie in a plane, contradicting nondegeneracy.

Similarly, homogeneous cubics form a twenty-dimensional space, whereasThere are at least five independent cubics vanishing on $C$. The multiples of $Q$ by linear forms give only four dimensions, so choose a cubic $F$ vanishing on $C$ which is not divisible by $Q$.

The polynomials $Q,F$ are a regular sequence, hence their intersection $D$ is a projective complete intersection of pure dimension one and degree $2\cdot 3=6$. It contains $C$, which already has degree six. The unmixedness of a complete intersection excludes embedded components. Equal degrees force $C$ to be the only irreducible component and force multiplicity one at its generic point. By generic reducedness with no embedded components, this makes $D$ reduced, so $D=C$ as schemes. We have provedThis is a canonical genus-four curve as a quadric-cubic intersection. The quadric may be smooth or a cone; the argument does not require it to be smooth. The same construction descends over the ground field when the curve is geometrically nonhyperelliptic, since the restriction maps and canonical embedding are defined there.

The Riemann-Hurwitz formula for a nonconstant degree-$n$ morphism
$f:C\to C^{\prime }$ of smooth projective connected curves in characteristic zero iswhere $R_f$ is the effective ramification divisor. If $g(C^{\prime })\ge 2$ and
$g(C)<g(C^{\prime })$, the right side is at least $2g(C^{\prime })-2>2g(C)-2$, impossible. If $g(C^{\prime })=1$, then $g(C)=0$ and the left side is $-2$ while the right side is nonnegative. Thus every morphism is constant.

Choose $p\in C_d$ and project from $p$ to the pencil of lines through it. Bézout's theorem says that a general such line meets $C_d$ in $p$ plus $d-1$ further points, so the resolved projectionhas degree $d-1$. Since the genus of a smooth plane curve is
$g=(d-1)(d-2)/2$, Riemann--Hurwitz givesEvery branch point receives at least one ramification point, hence

The adjunction formula givesSince $\mathrm{deg}D=d$ and $\mathrm{deg}{K}_{C_d}=d(d-3)$, linear equivalence
$D\sim K_{C_d}$ would force $d=d(d-3)$, hence $d=4$. It is therefore impossible for $d\ge 5$.

For a smooth plane quartic, projection from a point of the curve gives a degree-three map, so its gonality is at most three. Its genus is three and
$K_C\sim D$, so its plane embedding is the canonical map. A genus-three curve with a degree-two map to ${\mathbb{P}}^1$ would be hyperelliptic, and its canonical map would factor through that double cover and map onto a conic rather than embed the curve. Thus no degree-two map exists, and a degree-one map is excluded by the genus. The quartic's gonality isFor a genus-one curve, Riemann--Roch supplies a degree-two map to ${\mathbb{P}}^1$, while degree one would make it isomorphic to ${\mathbb{P}}^1$. Its gonality is therefore