I've been given a homework problem that requires me to show that the rings $\mathbb{C}[x,y]/(y - x^2)$ and $\mathbb{C}[x,y]/(xy-1)$ are not isomorphic.

This is my attempt at a solution:

For $\mathbb{C}[x,y]/(y - x^2)$, we can parametrize in the following way: $x = t$ and $y = t^2$. Then this ring is isomorphic to $\mathbb{C}[t]$.

For $\mathbb{C}[x,y]/(xy-1)$, we can parametrize $x = t$ and $ y = 1/t$. Then this is isomorphic to $\mathbb{C}[t, 1/t]$.

But $\mathbb{C}[t, 1/t]$ is not isomorphic to $\mathbb{C}[t]$.

Am I on the right track? If not, any helpful hints?

  • 3
    $\begingroup$ @user26857 Okay, let me try harder. If there existed an isomorphism $\Phi$ from $\mathbb{C}[t,1/t]$ to $\mathbb{C}[t]$, then it would have to satisfy $\Phi(t) \cdot \Phi(1/t) = 1$. So then $\Phi(t)$ and $\Phi(1/t)$ must be units in $\mathbb{C}[t]$; but that can only happen if they are constants. It follows that $\Phi$ maps only onto constants, a contradiction. Is this correct? $\endgroup$ – Andrea Dec 3 '14 at 19:14
  • $\begingroup$ @Andreas Your attempted proof works under the hypothesis that $\Phi$ is unitary, that is, $\Phi(1)=1$. One has to remark that $\Phi(a)\Phi(1/a)=1$ for any $a\in\mathbb C^\times$, so the non-zero constants are also sent to non-zero constants. Now can conclude that $\operatorname{Im}\Phi\subseteq\mathbb C$, a contradiction. $\endgroup$ – user26857 Jan 22 '17 at 23:10

Here is my favorite way to tell $\Bbb C[t]$ and $\Bbb C[t,t^{-1}]$ apart.

Do you know that the units of $\Bbb C[t]$ are just the nonzero constant functions? That means that the subset $\Bbb C$ is the set of units along with the zero element, which is additively closed.

But what about $\Bbb C[t,t^{-1}]$? Is the sum of two units again a unit or zero? Or can you find two units such that their sum is neither a unit nor zero? (It's easy!)

(Added later)

I just wanted to highlight how general this solution is. Using the same reasoning, you can show that for any field $F$, $F[t,t^{-1}]$ is not isomorphic to a polynomial ring over any field whatsoever. It doesn't depend on the field being $\Bbb C$ nor does it stipulate that the fields be shared between the two.

Given any fields $F_1, F_2$, the units of $F_1[t]$ with zero are additively closed, while the units of $F_2[t,t^{-1}]$ do not share this property.


If two rings are isomorphic, then their unit groups are isomorphic.

In this case $\mathbb{C}[t]^{\times}\simeq \mathbb{C}^{\times}$, while $\mathbb{C}[t,t^{-1}]^{\times}\simeq\mathbb C^{\times}\times\mathbb Z$. But the groups $\mathbb C^{\times}$ and $\mathbb C^{\times}\times\mathbb Z$ are not isomorphic for an obvious reason: the first one is divisible, while the second is not.


To show that $\mathbb C[t,\frac 1t] \not \simeq \mathbb C[t]$, you can find their automorphism groups.

Any automorphism of the left ring must send $t$ to something invertible. The only invertible element of $\mathbb C[t,\frac 1t]$ are the constants and $t^n$ for $n \in \mathbb Z \backslash \{ 0\}$. Thus the automorphisms are given by $t \mapsto ct^n$ for $n \neq 0$.

In particular, the automorphism group is abelian!

Now, what about the automorphism of $\mathbb C[t]$? Hint: It contains translations and multiplications by scalar, and it is not abelian.

  • $\begingroup$ As mere rings (not $\mathbb{C}$-algebras), you're missing automorphisms; e.g. if $\sigma$ is any automorphism of $\mathbb{C}$, then it is also an automorphism of $\mathbb{C}[t]$ when applied to the coefficients of a polynomial. $\endgroup$ – user14972 Jan 23 '17 at 8:29

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