I was reading about divisor class groups, and I was wondering the following.
Let $R=\mathbb{C}[X,Y,Z,W]/(XZ-YW)$, and let $x,y,z,w$ be the images of $X,Y,Z,W$ in $R$, respectively. Is there a way to determine all the non-principal height one primes?
I know that four of them are $(x,y)$, $(x,w)$, $(z,y)$ and $(z,w)$. Are there any others?
There are in fact many more non-principal height one primes in $R$: e.g. $(xw-y^2, yz-w^2)$ (recall that $(XW-Y^2, YZ-W^2, XZ-YW)$ is the ideal of the twisted cubic, and is a height $2$ prime in $k[X,Y,Z,W]$). One can see this example via algebraic geometry:
$R$ is the coordinate ring of the affine cone over the quadric surface $Q = V(xz-yw) \subseteq \mathbb{P}^3$. Now $Q$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$ (under the Segre embedding), so $\text{Cl}(Q) \cong \mathbb{Z} \oplus \mathbb{Z}$. Thus any divisor $C$ on $Q$ has a bidegree $(a,b) \in \mathbb{Z} \oplus \mathbb{Z}$. Furthermore, $C$ has genus $(a-1)(b-1)$ and degree $a+b$, and for any $a, b > 0$ (if $k$ is algebraically closed), there exist irreducible curves $C$ of type $(a,b)$ on $Q$.
An irreducible curve $C$ on $Q$ gives rise to a height $1$ prime in $R$, which is principal iff $C$ is a complete intersection of $Q$ with another hypersurface $D \subseteq \mathbb{P}^3$, say of degree $d$. But a complete intersection of $Q$ with $D$ has genus $\frac{1}{2}(2)(d)(2+d-4) + 1 = (d - 1)^2$ and degree $2d$, so most choices of $(a,b)$ will not give a complete intersection (in fact, this occurs iff $a = b$). The twisted cubic above corresponds to $(a,b) = (1,2)$.