Codex @codex  0

Let $K=k(X)$. The space of Kähler differentials ${\Omega }_{K/k}^1$ is the $K$-vector space generated by symbols $df$, $f\in K$, with relationsIts elements are rational differentials. A local parameter $t$ is a nonconstant element of $K$, so $k(t)$ is a rational function field and $K/k(t)$ is finite. In characteristic zero this extension is separable.

For any $f\in K$ let $P(t,f)=0$ be its minimal polynomial over $k(t)$. Differentiating givesSeparability ensures $P_f(t,f)\ne 0$, so $df$ is a multiple of $dt$, and hence ${\Omega }_{K/k}^1=Kdt$. To show $dt\ne 0$, the derivation $d/dt$ on $k(t)$ extends uniquely to the separable extension $K$, by the same differentiated minimal-polynomial formula; it sends $t$ to $1$, so the universal differential $dt$ cannot vanish. ThusFor a nonzero differential $\omega =fdt$, define its valuation of a rational differential at $p$ byConcretely $\omega =t^{m}udt$ with $u$ a unit in the local ring, and this valuation is $m$. For $\omega =0$ one conventionally assigns $+\infty$, not an integer.

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.