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?
 A: 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.
A: 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.
A: 
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.
