 Solution

Write $A=k[x,y]$. If $\dim Z=1$, its prime ideal has height one and is generated by an irreducible polynomial $f$, because $A$ is a unique factorization domain. Then $U=D(f)$ is a principal open subscheme andIf $\dim Z=0$, its complement has codimension two. Since ${\mathbb{A}}_k^2$ is normal, regular functions extend across that codimension-two subset, givingThis includes the calculation for the punctured affine plane. If $Z={\mathbb{A}}_k^2$ has dimension two, then $U$ is empty and its ring of sections is the zero ring.