# On isomorphism of two quotient of polynomial rings

$$\mathbf{The \ Problem \ is}:$$ 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.

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