On isomorphism of two quotient of polynomial rings

Question

$\mathbf{The \ Problem \ is}:$ Show that the ring $R_1=\mathbb{C}[x,y,z]/(xy-z^2)$ is not isomorphic to $R_2 = \mathbb{C}[x,y,z]/(xy-z)$ .

$\mathbf {My \ approach}:$ There was a hint stating to show that $R_2$ is a UFD and $x$ is irreducible in $R_1$.

Now, $x$ is not prime in $R_1$ as $x\mid xy=z^2$ but then primality of $x$ would mean $x\mid z$, then $R_1$ will not be a UFD.

I have only read up to free modules and Gauss' lemma, a proof involving those things only will be very helpful for me.

A small hint is warmly appreciated.

Answer 1

Hint: Consider the surjective map $x \mapsto x, y \mapsto y, z \mapsto xy$ from $\mathbb C[x,y,z]$ to $\mathbb{C}[x,y]$. Can you show that its kernel is $(xy-z)$? This would give an iso $R_2 \simeq \mathbb C[x,y]$.

This shows that $R_2$ is a UFD, since $A$ being a UFD implies $A[T]$ is an UFD.

Hint': Now we're left with showing that $R_1$ is not a UFD. By its very definition, the equality $xy = z^2$ holds in this ring (I'm abusing notation and dropping the classes). Do you see why this equation is troubling, factorization-wise?

Indeed, we have $z^2 = z \cdot z = xy$. These are two factorizations for the same element, prove this by showing that $x,y,z$ remain irreducible in the quotient and are not associates