How to show that $\Bbb Z[x,y,z,w]/(xw-zy)$ is not a UFD I am trying to show that $R=\Bbb Z[x,y,z,w]/(xw-zy)$ is not a UFD. 


Let $I=(xw-zy)$.

Let $X=x+I$, $Y=y+I$, $Z=z+I$, and $W=w+I$. My guess is that $X$ is irreducible and therefore $(X)$ is a prime ideal. Now, $XW=ZY$, so that $ZY\in(X)$. I want to show that $Z,Y\notin(X)$ and conclude that $R$ is not a UFD.

However, I cannot confirm my guess because I don't even know how $R$ looks like and how to find an argument to show $X$ is irreducible. I think it suffices to prove that $X$ is irreducible. If it is so, the argument can be applied to $Y$, $Z$, and $W$. Therefore, $Z,Y\notin(X)$.

Any help?
Since $xw = zy$ in $R$, it suffices to show that $x, w, z, y$ are all irreducible. Note that $(xy - zw)$ is a graded prime ideal in $\mathbb{Z}[x,y,z,w]$ (apply Eisenstein at the prime $w$ in the ring $\mathbb{Z}[w,y,z][x]$), so $R$ is a positively graded domain, with degree $0$ subring $\mathbb{Z}$, and if we write $\deg(a)$ for the degree of the highest nonzero component in $a$, then $\deg(ab) = \deg a + \deg b$ for any $a, b \in R$.

Now if $x = rs$ for some $r, s \in R$, then $1 = \deg x = \deg r + \deg s$, but this implies one of $\deg r, \deg s = 0$, say $\deg r = 0$. Then $r \in \mathbb{Z}$, but the only elements of $\mathbb{Z}$ dividing $x$ in $R$ are $\pm 1$ (verify this!). Thus $r$ is a unit, so $x$ is irreducible, and by symmetry, $y, z, w$ are all irreducible. Thus $xw = zy$ are two distinct factorizations into irreducibles, so $R$ is not a UFD.

By the way, $(x)$ is not a prime ideal in $R$: 

$$R/(x) \cong \mathbb{Z}[x,y,z,w]/(x, xw - zy) = \mathbb{Z}[x,y,z,w]/(x, zy) \cong \mathbb{Z}[y,z,w]/(yz)$$ 

is not a domain. This gives another way to see that $R$ is not a UFD, as $x$ is an irreducible element that is not prime.
Consider the element $a = xw$. Then
$$ xw = xw - zy + zy \overset{mod \ I}{\equiv} 0 + zy = zy $$
so in $R$ we have $xw = a = zy$. Hence, $R$ is not a UFD.