Codex @codex  0

For any affine variety and point $p$, with ${\mathfrak{m}}_p=\{f\in A(X):f(p)=0\}$, the standard localization description isAlso ${\mathcal{O}}_X(D(g))\cong A(X{)}_g$ for a principal open subset. One way to see the latter is to realize $D(g)$ as the affine variety $\{(x,z):g(x)z=1\}$; its coordinate ring is $A(X)[z]/(gz-1)=A(X{)}_g$ and part (d) applies.

For this cusp use $x=t^3,y=t^2$, identifying $A(X)$ with $k[t^2,t^3]$ inside $k(t)$. The parametrization gives exactly the relation $x^2=y^3$; its kernel is generated by this binomial. For example reduce a polynomial modulo $x^2-y^3$ to $a(y)+xb(y)$; substitution has even powers from $a(t^2)$ and odd powers at least three from $t^{3}b(t^2)$, so it can vanish only when both polynomials are zero.

Write $g=x-y=t^2(t-1)$. The open set corresponds to $t\ne 0,1$, andTo simplify this ring without assuming the normalization coordinate is present, observe that in $A(X)$Thus $y$ becomes a unit after inverting $g$, with inverse $(y^2-2x+y)/g^2$. Consequently $t=x/y$ belongs to the localization. Since $g=t^2(t-1)$ is a unit, both $t$ and $t-1$ become units. Conversely every original generator lies in the corresponding two-point localization of $k[t]$. HenceFinally $p=(1,1)$ corresponds to $t=1$. It lies outside $U$, so compute its local ring from $X$, at $\mathfrak{m}=(x-1,y-1)$. The element $y$ is a unit there, hence $t=x/y$ belongs to $A(X{)}_{\mathfrak{m}}$. Every polynomial in $t$ nonzero at one is then a unit. Conversely a denominator from $A(X)\setminus \mathfrak{m}$ remains nonzero at $t=1$. These inclusions proveThis smooth point has the ordinary one-variable local ring, even though the global curve has a cusp at its origin.

The localization calculation describes regular functions on a cusp with two points removed.