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.

Solutions Collecting From Web of "Is the coordinate ring of SL2 a UFD?"

The answer (probably) depends on $K$.
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)$.

I don’t know what happens to your question over non-algebraically closed fields.

CW version of Justin Campbell and Pete Clark’s answer:

More generally, the coordinate ring of any simply connected, semisimple, linear algebraic group is a UFD. This is proved as the Corollary on page 296 (p. 303 in translation) of Popov (1974). The proof of the corollary from the proposition is explained in §11.2 of Pete Clark’s Factorization notes for those of us for whom the proof was not obvious. This requires knowing the coordinate ring of a linear algebraic group is regular.

Georges Elencwajg’s answer appears very related to §9.4 of Pete’s notes, where indeed the behavior of very similar rings requires characteristic not 2 and algebraic closure to apply.

For some reason, this particular ring is always a UFD, regardless of field.

I am still interested in a solution I can actually understand (so why would the Picard group of SL2 vanish?). The general proof is available in Popov (1974) to those who can read it:

  • Popov, V. L.
    “Picard groups of homogeneous spaces of linear algebraic groups and one-dimensional homogeneous vector fiberings.”
    Izv. Akad. Nauk SSSR Ser. Mat. 38 (1974), 294–322.
    MR357399
    URL:http://mi.mathnet.ru/eng/izv/v38/i2/p294 (original)
    DOI:10.1070/IM1974v008n02ABEH002107 (translation)

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.