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.

  • $\begingroup$ Do you mean isomorphic as rings or as $\mathbb {C}$-algebras? $\endgroup$ – Dune Oct 4 '15 at 9:18
  • $\begingroup$ @Dune As rings. $\endgroup$ – 123 Oct 4 '15 at 9:23
  • $\begingroup$ What could possibly be $\varphi(x)$ ? Note that $x$ is invertible, so must $\varphi(x)$. Also, note that $x-a$ is not invertible for $a\neq 0$. $\endgroup$ – Roland Oct 4 '15 at 9:40
  • $\begingroup$ @Roland Why wouldn't $x-a$ be invertible? $\endgroup$ – 123 Oct 4 '15 at 10:19
  • $\begingroup$ The first part of the question is also here: math.stackexchange.com/questions/1050291/…. $\endgroup$ – user26857 Jan 22 '17 at 22:45

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.

  • $\begingroup$ Your argument is basically right, but not complete. (1) You don't have to assume $I = \langle xy -1 \rangle$ for this (besides it is true). (2) Why exactly does $\varphi$ take all elements to constant polynomials? Note that it need not to be $\mathbb{C}$-linear. $\endgroup$ – Dune Oct 4 '15 at 13:44
  • $\begingroup$ For any $a\in\mathbb C^\times$ we also get $\varphi(a)\in\mathbb C^\times$, hence $\operatorname{Im}\varphi\subseteq\mathbb C$, a contradiction. (For the answer to be true we have to assume that $\varphi$ is unitary.) $\endgroup$ – user26857 Jan 22 '17 at 23:06

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 $f(x,y)\in(xy-1)$.


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