How can I prove a coordinate ring is not isomorphic to a polynomial ring Let $Z$ be the plane curve $xy=1$. I would like to prove that $A(Z)$ is not isomorphic to a polynomial ring in one variable over $k$.
I'm already prove that the coordinate ring is $A(Z)=k[x,y]/(xy-1)$, but I couldn't finish the question.
I have a guess that $A(Z)\cong k[x,1/x]$, but I can't prove it formally yet and even I would I don't know what to say with this information. 
I really need help.
Thanks a lot.
 A: 1) The isomorphism $f:k[x,y]/(xy-1)\to k\left[x,\frac1x \right]$ is obtained from the morphism $F:k[x,y]\to k\left[x,\frac1x \right]:u(x,y) \mapsto u(x,\frac 1x)$ (which is clearly surjective) by passing to the quotient at the source by $ker(F)=(xy-1)$ (which last equality you have to check, of course).  
2) In $k[x,1/x]$ the  non-constant element $x\in k[x,1/x]\setminus k$ is invertible, whereas in $k[x]$ only the elements in $k^*$, the nonzero constants, are invertible.
Hence  the $k$-algebras $k[x,1/x]$and $k[x]$ are not isomorphic. 
A: Your guess is right. $k[x,y]/(xy-1)\cong k[x,\frac1x]$ and an isomorphism is given by $$\varphi:k[x,y]/(xy-1)\to k\left[x,\frac1x \right]$$ by $\varphi(x)=x$ and $\varphi(y)=\frac1x$. You can show that $\varphi$ is indeed a ring homomorphism (precisely because you have the relation $xy=1$). It's also not hard to show that $\varphi$ is an isomorphism: it's obviously surjective and $\ker\varphi=\{0\}$.
Edit (Thanks to Georges for pointing this out):
So $k[x,y]/(xy-1)$ is not a field because $(xy-1)$ is not a maximal ideal. $(xy-1)\subset (x-1,y-1)$ which is a maximal ideal. 
If you want to think about this in an algebraic geometry setting, think of the affine 2-space over $k$, which is just $\textrm{Spec}\ k[x,y]$. Assuming that $k$ is algebraically closed, the maximal ideals correspond to closed points and the curve $xy=1$ is obviously not a point. 
