Localizing C[x,y]/(xy) to get a direct product of Laurent polynomials. I am trying to show $$\left(\mathbb C[x,y]/(xy)\right)_{x+y}\cong \mathbb C[x^{\pm 1}] \times \mathbb C[y^{\pm 1}].$$ In class, my professor did the example $$\left(\mathbb C[x,y]/(xy)\right)_{x} \cong \mathbb C[x^{\pm 1}]$$ by using the fact that $A_f \cong A[t]/(tf - 1)$ for a commutative ring $A$. I am trying to mimic this approach. I get that $$\left(\mathbb C[x,y]/(xy)\right)_{x+y} \cong \left(\mathbb C[x,y]/(xy)\right)[t]/(t(x+y)-1)$$ $$=\mathbb C[x,y,t]/(xy, t(x+y) - 1).$$ I am not really sure how to simplify this quotient to get what I need. I would really appreciate any help! Thanks.
 A: Since $\Bbb C[x^{\pm 1}]\times \Bbb C[y^{\pm 1}]$ has two orthogonal idempotents, $(1,0)$ and $(0,1)$, we need to find two orthogonal idempotents in $\Bbb C[x,y,t]/(xy,t(x+y)-1)$. I claim that $xt$ and $yt$ suffice: it's not hard to see that $xt+yt=t(x+y)=1$, while seeing that $(xt)^2=xt(1-ty)=xt$ (and similarly for $(yt)^2=yt$) takes a touch more insight. This means that $R=\Bbb C[x,y,t]/(xy,t(x+y)-1)$ is isomorphic to $Rxt\times Ryt$.
Now we can check that $Rxt\cong \Bbb C[x^{\pm 1}]$: writing a general element in $R$ as $p(x,y,t)$, we can use the relation $xy=0$ to see that $p(x,y,t)xt=p(x,0,t)xt$, and then  our calcluation that $xt$ is idempotent means that we can write this uniquely as $xt(c+\sum c_ix^i + \sum d_jt^j)$ for $c,c_i,d_j\in \Bbb C$. Since $x^2t\cdot xt^2=x^3t^3=xt$, we can define a surjective map sending $x^2t\mapsto x$ and $xt^2\mapsto x^{-1}$, which is injective by the claim that every element of $Rxt$ can be uniquely written as $xt(c+\sum c_ix^i + \sum d_jt^j)$.
Intuitively, what's going on here is that you're taking the variety $V(xy)$, the union of two coordinate axes, and throwing away all the points lying on $V(x+y)$, the antidiagonal. This gives you two disjoint copies of a punctured line, which both have coordinate algebra $\Bbb C[x^{\pm 1}]$. By disjointness, the coordinate algebra of the pair is just the product.
