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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.

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    $\begingroup$ Just because $x+y$ is in $X=R/(xy-1)$ (which is not even true, only $x+y+(xy-1)$ is in $X$, that does not mean that $X$ is not isomorphic to some polynomial ring in one variable. For example, the ring generated by $x+y$ is isomorphic to a polynomial ring in one variable. $\endgroup$ – 5xum Aug 31 '15 at 6:53
  • $\begingroup$ Well, among other things, if it was $k'[z]$, for some field $k'$, you'd know it'd be a euclidean domain (And thus a principal ideal domain). Can you find an ideal in your new ring that is not principal? That'd do it. (Note: I don't know if this is true or not, but it's a good place to start) $\endgroup$ – Alan Aug 31 '15 at 6:54
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    $\begingroup$ Try thinking about units. $\endgroup$ – moonlight Aug 31 '15 at 6:54
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    $\begingroup$ @Learner: Perhaps, to be more specific, think about whether or not units generate each of the two rings $A[z]$ and $R/(xy-1)$ and whether that is a property invariant under isomorphism. $\endgroup$ – moonlight Aug 31 '15 at 7:18
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    $\begingroup$ @JyrkiLahtonen That's OK by me. These things can go either way sometimes. $\endgroup$ – rschwieb Sep 2 '15 at 0:32
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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.

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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.

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