What is Proj $\mathbb{C}[x,y][z]/\langle xz-yz\rangle$?

Question

Assuming that $x,y$ have weight $0$ and $z$ has weight $1$,
$$ R= \mathbb{C}[x,y][z]/\langle xz-yz\rangle = \mathbb{C}[x,y]\oplus ( \oplus_{i\geq 1}\mathbb{C}[x]z^i), $$ what closed subvariety is $\text{Proj } R$? How do you see that? Could this be the empty set?

Answer 1

Non, it is not empty. Denote by $A=\mathbb C[x,y]$, $f=x-y\in A$. Then $R=A[z]/fz$. So $$\mathrm{Proj}(R)=V_+(f)\cup V_+(z)=V_+(f)=\mathrm{Proj}((A/fA)[z])=\mathrm{Spec}(A/fA)\simeq \mathbb A^1_{\mathbb C}.$$ The middle equality is because $V_+(z)=\emptyset$ ($z$ generates the irrelevant ideal).