= Solution
Let
$$
\omega=\frac{dx}{y^2}.
$$
At a finite ramification point $P_\alpha=(\alpha,0)$, the uniformizer $t=y$ from part b satisfies $x-\alpha=c_\alpha t^3+O(t^6)$. Consequently
$$
\nu_{P_\alpha}(dx)=2,
\qquad
\nu_{P_\alpha}(y^2)=2,
\qquad
\nu_{P_\alpha}(\omega)=0.
$$
At 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,
$$
x=\frac uv=t^{-3}\mathbin{\cdot}(\text{local unit}),
\qquad
y=\frac1v=t^{-4}\mathbin{\cdot}(\text{local unit}).
$$
It follows that $\nu_{P_\infty}(dx)=-4$ and $\nu_{P_\infty}(y^2)=-8$, so the <valuation of a rational differential> is
$$
\nu_{P_\infty}(\omega)=4.
$$
Thus
$$
\boxed{(\omega)=4P_\infty,}
$$
whose 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:
$$
\boxed{L(K_X)=\operatorname{span}_{\mathbb C}\{1,x,y\}.}
$$
Equivalently, multiplying by $\omega$ gives the basis
$$
\boxed{\left\{
\frac{dx}{y^2},
\frac{x\,dx}{y^2},
\frac{dx}{y}
\right\}}
$$
of <holomorphic differential forms>, also matching the general description of <holomorphic differentials on a smooth plane curve>.
Back to article page
