= Regular functions on a cusp with two points removed
{title2=$k[t^2,t^3]_{t^2(t-1)}=k[t,t^{-1},(t-1)^{-1}]$}
Parametrize the affine <cuspidal cubic> $x^2=y^3$ by $x=t^3,y=t^2$. Removing the <zero set> of $g=x-y$ removes exactly the parameter values zero and one. Its regular-function <ring> is the <localization> $k[t^2,t^3]_g$. To recover the normalization coordinate inside this <ring>, use $g^2=y(y^2-2x+y)$: inverting $g$ makes $y$ a <unit>, so $t=x/y$ lies in the <localization>. As $g=t^2(t-1)$ is a <unit>, so are $t$ and $t-1$, proving the displayed equality. This calculation is valid in every <characteristic>. At the <smooth> <point> $(1,1)$ of the original curve the <local ring> is instead $k[t]_{(t-1)}$: $y$ is a <unit> there, and every <polynomial> in $t$ nonzero at one is invertible. That <point> is absent from the displayed <open subset>, so its <local ring> is computed in the original curve.
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