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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.

If a line bundle on a smooth projective curve has both $h^0(\mathcal{L})>0$ and $h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$, the displayed inequality holds. Write these dimensions as $r,s$. The product dimension bound for sections on a curve gives $r+s-1\le g$, and the Riemann-Roch theorem gives $r-s=\mathrm{deg}\mathcal{L}+1-g$. Adding gives $2r\le \mathrm{deg}\mathcal{L}+2$.

On a smooth projective curve, the exact sequence $0\to \mathcal{O}(D-P)\to \mathcal{O}(D)\to k(P)\to 0$ increases the Euler characteristic by one. Iteration for positive and negative coefficients gives $\chi (\mathcal{O}(D))=\mathrm{deg}D+1-g$. Serre duality identifies $h^1(\mathcal{O}(D))$ with $h^0(\mathcal{O}(K-D))$, proving $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$.

Every rational map from a smooth projective curve to a projective variety extends uniquely to a morphism. At a missing point, the discrete valuation ring of the curve lets one divide homogeneous coordinates by their smallest valuation, leaving regular coordinates of which at least one is a unit.

The Fermat cubic $x^3+y^3+z^3=0$ is a smooth projective curve: its three first partial derivatives vanish simultaneously only at the forbidden zero vector. The genus of a smooth plane curve formula gives genus one.

The embedding line bundle has at least four independent global sections. If it is nonspecial, the Riemann-Roch theorem gives $h^0=6-g$, so $g\le 2$. If it is special, the Clifford inequality for curves gives $h^0\le 3$, a contradiction. The smooth projective curve need not span its full ambient projective space.