Is the coordinate ring of SL2 a UFD?

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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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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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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.