$k[x,y]/(xy-1)$ isomorphic to $k[x,\frac{1}{x}]$ I mean clearly one uses the isomorphism $\phi$ that sends to $x$ to $x$ and $y$ to $\frac{1}{x}$. And also clearly is $(xy-1)\subseteq\ker(\phi)$. I just struggle to prove the other inclusion.
Can you guys maybe help me?
 A: So we have a polynomial $$p(x,y)=\sum_{j,k\ge0}a_{j,k}x^jy^k$$such that $$p(x,1/x)=0,$$and we want to show that there exists a polynomial $r(x,y)$ with $$p(x,y)=(1-xy)r(x,y).$$
In general $$p(x,1/x)=\sum_{n\in\Bbb Z}b_nx^n,$$where $$b_n=\sum_{j-k=n}a_{j,k}.$$Since $p(x,1/x)=0$ we have $$\sum_{j-k=n}a_{j,k}=0$$for every $n\in\Bbb Z$. Let $$p_n(x,y)=\sum_{j-k=n}a_{j,k}x^jy^k.$$
Assume first that $n\ge0$. Then we have $$p_n(x,y)=x^n\sum_{j-k=n}a_{j,k}(xy)^k=x^nq_n(xy),$$where $$q_n(t)=\sum_{j-k=n}a_{j,k}t^k.$$Now $q_n(1)=0$, so there exists a polynomial $r_n(t)$with $$q_n(t)=(1-t)r_n(t).$$Hence $$p_n(x,y)=(1-xy)x^nr_n(xy).$$
We've shown that $1-xy$ divides $p_n(x,y)$ for $n\ge0$. The proof for $n<0$ is similar, except factoring out $y^{-n}$ from $p_n$ instead of $x^n$.
A: For any commutative ring $k$, the ring $k[x; 1/x]$ is characterized by the following "mapping property": for any other ring $R$ containing $k$ and a unit $u \in R$, there is a homomorphism $\psi: k[x; 1/x] \to R$ such that $\psi(a) = a$ for $a \in k$ and $\psi(x) = u$. 
We can apply this to $k[x,y]/(xy-1)$: since $k \cap (xy-1)$ is the trivial ideal, $k$ is isomorphic to its image in the quotient ring $k[x,y]/(xy-1)$.
Further, $x$ is a unit there, with inverse $y$.
So construct a homomorphism $\psi: k[x;1/x] \to k[x,y]/(xy-1)$ by mapping $\psi(a)=a$ for $a \in k$ and $\psi(x) = x$. This will be an inverse to your homomorphism.
