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Choose a nonzero meromorphic differential on a Riemann surface $\eta$ and write $K=(\eta )$. Multiplication defines a linear mapIt takes values in the holomorphic differential forms because $(f\eta )=(f)+K\ge 0$. It is injective since $\eta \ne 0$. It is surjective because, for any holomorphic differential form $\omega$, the ratio $\omega /\eta$ is a meromorphic function and has $(\omega /\eta )+K=(\omega )\ge 0$. Thus this map identifies $\mathrm{HD}(C)$ with the canonical Riemann-Roch space.

Furthermore, $\mathcal{L}(0)=\mathbb{C}$: a meromorphic function with no poles is holomorphic on the compact Riemann surface $C$, and hence constant by the maximum modulus principle. Applying the Riemann-Roch theorem with $D=0$ givesTherefore

LetAt a finite ramification point $P_{\alpha }=(\alpha ,0)$, the uniformizer $t=y$ from part b satisfies $x-\alpha =c_{\alpha }{t}^3+O(t^6)$. ConsequentlyAt every other affine point, $x-x(p)$ is a uniformizer and $y$ is a unit, so again ${\nu }_p(\omega )=0$.

At $P_{\infty }$, take the uniformizer $t=u=X/Y$ used above. Since $v=Z/Y=t^4$ times a local unit,It follows that ${\nu }_{P_{\infty }}(dx)=-4$ and ${\nu }_{P_{\infty }}(y^2)=-8$, so the valuation of a rational differential isThuswhose degree four agrees with $2g_X-2$. For the resulting canonical divisor $K_X=4P_{\infty }$, the functions $1,x,y$ have pole orders $0,3,4$ at $P_{\infty }$ and no other poles. They therefore belong to the canonical Riemann-Roch space. Since $\ell (K_X)=g_X=3$, they form a basis:Equivalently, multiplying by $\omega$ gives the basisof holomorphic differential forms, also matching the general description of holomorphic differentials on a smooth plane curve.