Prove that $k[x,y,z,w]/(xy-zw)$, the coordinate ring of $V(xy-zw) \subset \mathbb{A}^4$, is not a unique factorization domain 
I want to show that $k[x,y,z,w]/(xy-zw)$, the coordinate ring of $V(xy-zw)\subset\mathbb{A}^4$, is not a unique factorization domain. 

Morally, all we need to do is find some nonzero element that can be written in two distinct ways as a product of irreducible elements, but nothing is coming to mind immediately. Thoughts? 
 A: 
Let $S$ be an integral domain. Then $R=S[X,Y,Z,W]/(XY-ZW)$ is not an UFD. 

First of all, it isn't hard to prove that $R$ is also an integral domain. (See here.)
In the following we denote by $x,y,z,w$ the residue classes of $X,Y,Z,W$ modulo the ideal $(XY-ZW)$. 


*

*$x$ is irreducible. 


Consider $R$ as a graded ring (with the grading inherited from $S[X,Y,Z,W]$). Assume that $x$ has a factorization into two non-zero non-unit elements. Since $x$ has degree one, the two factors must be a degree one element $ax+by+cz+dw$ with $a,b,c,d\in S$, and a degree zero element $s\in S$. (If $S$ is a field, one can stop here.) This gives $$(1-sa)X+sbY+scZ+sdW\in(XY-ZW),$$ so in $S[X,Y,Z,W]$ a degree one element belongs to the ideal $(XY-ZW)$ whose the non-zero elements are of degree at least two. Consequently, $(1-sa)X+sbY+scZ+sdW=0$ and this implies $sa=1$, so $s$ is a unit, a contradiction. 


*

*$x$ is not prime. 


We have $R/(x)\simeq (S[Z,W]/(ZW))[Y]$ and this is not an integral domain.
Remark. When $S$ is a field one can prove that the divisor class group of $R$ is (isomorphic to) $\mathbb Z$.
