TITLE: Is it true that $(\mathbb{C}[x,y]/(xy))_{(x,y-k)}=\mathbb{C}[t]_{(t)}$?
QUESTION [6 upvotes]: I want to show that $(\mathbb{C}[x,y]/(xy))_{(x,y-k)}=\mathbb{C}[t]_{(t)}$.
But I think that the leftside has 2 variables, but rightside has only 1. Is it possible that the two are isomorphic?

REPLY [5 votes]: When $k=0$, as indicated by Ted, the answer is no. The ring $\mathbf{C}[t]_{(t)}$, which is the local ring of the affine line at the origin, is a discrete valuation ring, and in particular, is an integral domain. However, the localization of $\mathbf{C}[x,y]/(xy)$ at $(x,y)$, which is the local ring of the union of the coordinate axes in $\mathbf{A}_\mathbf{C}^2$ at the origin, is not a domain. Geometrically speaking, the two irreducible components of $V(xy)$ (the $x$ and $y$-axes) both pass through the origin, and this implies that the local ring at this point cannot be a domain (because minimal primes of the local ring are in bijection with irreducible components passing through the origin). Concretely, $x,y$ are two elements of the ring which are not zero, but whose product $xy$ is zero. In the case where $k\neq 0$, as explained by Ted, this argument does not apply because $y$ is a unit and therefore $x$ actually is zero.