Show that $\Bbb C[x,y]/I$ is not isomorphic to $\Bbb C[z]$ Let $I$ be the ideal in the ring $\Bbb C[x,y]$ defined by
$$I=\{f\in\Bbb C[x,y]:f(x,y)=0\text{ for all $x,y\in\mathbb C$ such that $xy=1$}\}.$$

Question: Show that $\Bbb C[x,y]/I$ is not isomorphic to $\Bbb C[z]$.

My algebra is a bit rusty, so I can't find a way to prove this.

First of all, can we say that $f\in I$ if and only if $f(x,y)=(xy-1)g(x,y)$ for $g\in\Bbb C[x,y]$? So $I=\langle xy-1\rangle$?

I am trying to start by assuming there is a homomorphism
$$\varphi:\Bbb C[x,y]\to\Bbb C[z]$$
with kernel $I$, but I get nowhere with this.
 A: Here is an attempt to answer my own question. Please feel free to comment (either negatively or positively):

Suppose
  $$\varphi:\Bbb C[x,y]/(xy-1)\to \Bbb C[z]$$
  is an isomorphism. Then,
  $$\varphi(x)\varphi(y)=\varphi(xy)=\varphi(1)=1,$$
  so $\varphi(x),\varphi(y)\in\Bbb C[z]$ are units. But $\Bbb C[z]^\times=\Bbb C^\times$ so $\varphi(x)$ and $\varphi(y)$ are constant polynomials. But then $\varphi$ sends every polynomial to a constant one, and hence is not surjective.

A: 
We have to prove $$\{f\in\Bbb C[x,y]:f(a,b)=0\text{ for all $a,b\in\mathbb C$ such that $ab=1$}\}=(xy-1),$$ that is, $$I(V(xy-1))=(xy-1).$$

Work in $\mathbb C[x,x^{-1},y]=\mathbb C[x,x^{-1}][y]$ instead of $\mathbb C[x,y]$, and write $$f(x,y)=(xy-1)g(x,x^{-1},y)+r(x,x^{-1}).$$ (One can perform a long division since the coefficient of the polynomial $xy-1$ is invertible in $\mathbb C[x,x^{-1},y]$.)
Now multiply this equation by $x^n$ for some $n\ge0 $ with the following property: $x^ng(x,x^{-1},y)\in\mathbb C[x,y]$ and $x^nr(x,x^{-1})\in\mathbb C[x]$. Then $x^nf(x,y)=(xy-1)g_1(x,y)+r_1(x)$ in $\mathbb C[x,y]$, and from $f(a,a^{-1})=0$ for all $a\in\mathbb C^\times$ we get $r_1(x)=0$, so $x^nf(x,y)\in(xy-1)$. But $x^n$ and $xy-1$ are coprime, and thus $f(x,y)\in(xy-1)$.
