# 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.

• Did you consider applying Nagata's theorem? Jul 19 '12 at 23:09
• 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. Jul 19 '12 at 23:37
• @Justin: the result you mention is very satisfying. Could you give a reference for it? Jul 20 '12 at 10:45
• @PeteL.Clark: Popov 1974 showed simply connected algebraic groups have vanishing picard groups. Someone on math overflow claims having a vanishing picard group means the coordinate ring is a UFD. A few authors use the word "factorial" for a variety whose picard group vanishes. I don't understand Popov's paper, nor do I know anything about Picard groups beyond Dedekind domains. Jul 20 '12 at 15:05
• @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... Jul 20 '12 at 15:12

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.

• Dear @GeorgesElencwajg, how can one uses Klein-Nagata theorem which deals with at least five variables in this case when we have only four? (I have a guess: to replace that $1$ by an $e^2$, where $e$ is a fifth variable, but I'd be glad if you could give me some more details or a reference. Swan does something similar in Theorem 5 from his paper Vector bundles and projective modules.) Aug 25 '16 at 9:27
• Dear @user26857: yes, as I mentioned in the answer, the result does not follow immediately from Nagata. I am afraid I don't remember what I had in mind exactly but your idea is excellent : adding a variable $e$ and changing the quadratic form to $a^+b^2+c^2+d^2-e^2$ gives you the graded ring of a smooth quadric $Q\subset \mathbb P^4_K$. That quadric has cyclic class group, generated by the intersection $H$ of $Q$ with the hyperplane $e=0$. (to be followed) Aug 25 '16 at 20:14
• Thus the original affine quadric $Q\setminus H$ has zero class group, thanks to the exact sequence in Hartshorne page 133. Hence the ring $R$ of that normal (actually even smooth) quadric is a UFD. Aug 25 '16 at 20:15
• Hey Georges, I found this question/answer by accident while googling. Anyways, we're really asking what $\text{Pic}(\text{SL}_2)$ is. One can compute this geometrically (it follows from the fact that $\text{SL}_2$ is simply connected) or you can, as you suggest, use Nagata's theorem. That said, it does extend to non-algebraically closed by the following argument. From the five-term-sequence for the Hocschild-Serre spectral sequence for $X_{\overline{k}}\to X$ (let's assume that $k$ is perfect) Nov 7 '16 at 10:51
• you get the sequence $0\to H^1(G_k,\mathcal{O}(X_{\overline{k}})^\times)\to \text{Pic}(X)\to \text{Pic}(X_{\overline{k}})^{G_k}$. Now, since $\text{SL}_2$ is semisimple one can show that $\mathcal{O}_{\text{SL}_2}(\text{SL}_2)=k^\times$ (valid over any $k$, including $\overline{k}$) and thus this first cohomology term vanishes by Hilbert's Theorem 90. Since $\text{Pic}(\text{SL}_{2,\overline{k}})$ vanishes, so then must $\text{Pic}(\text{SL}_2)$. Nov 7 '16 at 10:51

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.

• 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

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: