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.

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Did you consider applying Nagata's theorem? – Bill Dubuque Jul 19 '12 at 23:09
Note that this is definitely not a Dedekind domain, since it is three-dimensional. – Justin Campbell Jul 19 '12 at 23:15
After a little bit of digging, I'm convinced that the answer is yes over any field. A semisimple algebraic group is factorial if and only if it is simply connected, and $\text{SL}_n$ is simply connected. It is certainly possible that there is a more elementary proof, but I can't find one. – Justin Campbell Jul 19 '12 at 23:37
@Justin: the result you mention is very satisfying. Could you give a reference for it? – Pete L. Clark Jul 20 '12 at 10:45
@Jack: Ah, that helps me out a lot. Because we know the domain is regular, the Picard group is isomorphic to the divisor class group, so we have an integrally closed domain with trivial divisor class group, and such a thing must be a UFD. See $\S 11.2$ of math.uga.edu/~pete/factorization2010.pdf for a discussion of these points. I could turn this comment into an answer if you like... – Pete L. Clark Jul 20 '12 at 15:12

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.

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If it is easy to explain the relation between the quadratic form and SL2, I'd love to see it. I suspect $k[a,b,c,d,e,f,g,h]/(ah+bg-cf-de)$ [a symplectic quadratic form, I hope] is also a UFD, due to its large number of variables? This could also suffice in my application (which should work for all fields, but suffices for algebraically closed fields of characteristic equal to 2). – Jack Schmidt Jul 20 '12 at 20:51
Dear Jack, yes the ring you mention is a UFD since indeed the large number of variables makes the quadratic form non-degenerate. (I hope your mentioning that the characteristic is equal to $2$ is a typo!) – Georges Elencwajg Jul 20 '12 at 22:34

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:

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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.

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This is an excellent demonstration of Nagata's criterion at work. @JackSchmidt: if you are still looking for a solution you can actually understand, you won't find a better one than this – zcn Aug 26 '14 at 19:33