Is it true that $(\mathbb{C}[x,y]/(xy))_{(x,y-k)}=\mathbb{C}[t]_{(t)}$? 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?
 A: 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. 
A: Assuming $k \ne 0$, yes.  
Intuitively: $\mathbb{C}[x,y]/(xy)$ corresponds to the union of the 2 axes, and localizing at the ideal $(x,y-k)$ corresponds to looking in a "neighborhood" of the point $(0,k)$.  When we localize at $(0,k)$, the $x$-axis does not matter anymore because $(0,k)$ is not on the $x$-axis.  So we are localizing just the $y$-axis ($\mathbb{C}[y]$) at the point $k$ (the ideal $y-k$), which is of course isomorphic to $\mathbb{C}[t]$ localized at $t$ (both are localizing a line at a point on it).
Somewhat more rigorously, if you localize $\mathbb{C}[x,y]/(xy)$ at $P = (x, y-k)$, the element $y$ (which is not in $P$) becomes invertible.  But $xy=0$ in $\mathbb{C}[x,y]/(xy)$, and since $y$ is invertible in the localization, we have $x=0$ in the localization.  So the localization is isomorphic to $\mathbb{C}[0, y]$ localized at $(0, y-k)$, which is just the same as $\mathbb{C}[y]$ localized at $(y-k)$.
The precise isomorphism sends $t \mapsto y-k$ is one direction, and $x \mapsto 0, y \mapsto t+k$ in the other direction.
