Is the coordinate ring of SL2 a UFD?

Is the ring $K[a,b,c,d]/(ad-bc-1)$ a unique factorization domain?

I think this is a regular ring, so all of its localizations are UFDs by the Auslander–Buchsbaum theorem. However, I know there are Dedekind domains (which are regular; every local ring is a PID, so definitely UFD) that are not UFDs, so being a regular ring need not imply the ring is a UFD.

With the non-UFD Dedekind domains (at least the number rings), I can usually spot a non-unique factorization, but I don't see any here in this higher dimensional example.

Solution 1:

If $K$ is an algebraically closed field of characteristic $\neq2$, then the ring $K[a,b,c,d]/(ad-bc-1)$ is a UFD.
This results (non trivially) from the Klein-Nagata theorem stating that if $n\geq 5$, the ring $K[x_1,...,x_n]/(q(x_1,...,x_n))$ is factorial for any field $K$ of characteristic $\neq2$ and any non degenerate quadratic form $ q(x_1,...,x_n)$.

Edit
In the comments @Alex Youcis explains why the result is still true for non algebraically closed fields.
I am very grateful for his valuable addition.

Solution 2:

Let $R=K[X,Y,Z,T]/(XY+ZT-1)$. It's easily seen that $R$ is an integral domain.

In the following we denote by $x,y,z,t$ the residue classes of $X,Y,Z,T$ modulo the ideal $(XY+ZT-1)$.

First note that $x$ is prime: $R/xR\simeq K[Z,Z^{-1}][Y]$. Then observe that $R[x^{-1}]=K[x,z,t][x^{-1}]$ and $x$, $z$, $t$ are algebraically independent over $K$. This shows that $R[x^{-1}]$ is a UFD and from Nagata's criterion we get that $R$ is a UFD.