Show that $k[x,y]/(xy-1)$ is not isomorphic to a polynomial ring in one variable. 
Let $R=k[x,y]$ be a polynomial ring ($k$, of course, is a field). Show that
  $R/(xy-1)$ is not isomorphic to a polynomial ring in one variable.

 A: Suppose $R\simeq K[T]$, where $K$ is a commutative ring. Then $K$ is an integral domain and since $\dim K[T]=1$ we get $\dim K=0$ (why?). It follows that $K$ is a field. So $k[X,X^{-1}]\simeq K[T]$. Now use this answer.
A: In the lines of the accepted answer, one can show that the global dimension of $k[x,x^{-1}]$ is $1$, so that if it is isomorphic to a polynomial ring, it must be isomorphic to one of the form $D[y]$ with $D$ a semisimple commutative domain, which is thus a field. 
A: We have $k[x,\frac{1}{x}]$ as a ring. Form a polynomial ring $k[x,\frac{1}{x}][y]$ with y as the variable. Construct an evaluation map $\phi$ $$\phi:k[x,\frac{1}{x}][y] \to k[x,\frac{1}{x}], \text{with }x\to x, \frac{1}{x} \to \frac{1}{x}, y \to \frac{1}{x}$$ then $\phi$ is obviously a ring morphism. Now by factor theorem, if $f\in k[x,\frac{1}{x}][y], f\in \ker\phi$, then $f(1/x) = 0$, which gives $(y-1/x)|f$. Note that 1/x is a taken as a constant here.
Now if $f\in k[x][y] \subset k[x,\frac{1}{x}][y] $, then $f = g(y-1/x), g \in k[x,\frac{1}{x}][y]$. Now you only need to show that g is in $k[x][y]$ and then we've shown $k[x][y]/(xy-1) \cong k[x,1/x]$, now check for the accepted answer by user26857.
A: Let $k$ be a field and $R:=k[x,y]/\langle xy-1\rangle$. Easy to check that $R$ is a localization of $k[x]$ by the set $\{x^n:n\in \mathbb{Z}_{\geq 0}\}$, i.e. any element of $R$ may be written as $$\sum_{i=m}^n \lambda_ix^i, \ \ \lambda_i \in k, m,n\in \mathbb{Z}.$$ This ring has Krull dimension of $1$, hence it may be isomorfic to $F[t]$ only if $F$ is a field.
Suppose that the necessary isomorphism is possible. We know that any ring homomorphism $f\colon F[t]\to R$ may be determined as a ring homomorphism $F\to R$ and a mapping $\{t\}\to R$.
We know that $(F[t])^\times= F^\times \ (i)$. Let's check what is possible $f(t)$. It is neither $0$, because otherwise $\ker f\ni t$ nor $\lambda \in k^\times$ because of $(i)$. Elements of form $\lambda x^n$ are also invertible in $R$, hence they cannot be images of $t$.
Now only one possibility is left. It is that $f(t)$ is a «polynomial», i.e. $$f(t)=\sum_{i=m}^n \lambda_ix^i, \ \ m<n.$$ Then the image of $$P:=t-f^{-1}(\sum_{i=m+1}^n \lambda_ix^i)$$ is a unit but $P$ is non-invertible (it's easy to see if you would find $R^\times$), a contradiction.
