Show that quotient rings are not isomorphic

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?

Solution 1:

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.

Solution 2:

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.