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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.

user26857
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Jack Schmidt
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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
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    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
  • One advantage of this approach: it works for $n > 2$. – Justin Campbell Jul 19 '12 at 23:43
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    @Justin: the result you mention is very satisfying. Could you give a reference for it? – Pete L. Clark Jul 20 '12 at 10:45
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    @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. – Jack Schmidt Jul 20 '12 at 15:05
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    @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 http://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
  • ...integrally closed Noetherian domain (hence a Krull domain), I should have said. – Pete L. Clark Jul 20 '12 at 15:19
  • @Pete: I like the first sentence of section 11. 11.2 has cleared up the Picard group. If you can say anything about Popov's proof, that'd make a great answer. – Jack Schmidt Jul 20 '12 at 15:27
  • @Pete: I wrote up a very short answer, if only to make the link to your notes and Popov's paper (thanks Justin) prominent. If you (or anyone) can describe Popov's proof, I'd appreciate it. I have a feeling it works by translating height 1 prime -> generator of group of invertible ideals -> projective rank 1 -> "line bundle"? -> something about a homogeneous space. Your notes explain the first transition; the second transition is known to me through Lam's textbook, but the remaining transitions are a mystery. – Jack Schmidt Jul 20 '12 at 20:55
  • @Jack: thanks for doing this, especially posting the link to Popov's paper. I'm afraid I will not have the chance to look carefully at it for at least a little while. In the meantime, you might find parts of my commutative algebra notes http://math.uga.edu/~pete/integral.pdf, especially $\S 6$ and $\S 19$. (Note that the latter is unfinished and in fact doesn't yet contain everything that is written up in my factorization notes. Writing these things up takes time...for instance, the time spent properly learning the material!) – Pete L. Clark Jul 20 '12 at 22:15
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    @JustinCampbell - I'm having trouble believing the "only if" in the statement "factorial if and only if simply connected". Maybe I'm missing something obvious, but I think of $GL(1,k)$ as factorial (coordinate ring is $k[x,x^{-1}]$) but not simply connected (e.g., squaring is a nontrivial isogeny). What am I missing? – Ben Blum-Smith Oct 05 '20 at 12:22

3 Answers3

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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.

Georges Elencwajg
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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
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    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.) – user26857 Aug 25 '16 at 09:27
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    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) – Georges Elencwajg Aug 25 '16 at 20:14
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    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. – Georges Elencwajg Aug 25 '16 at 20:15
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    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) – Alex Youcis Nov 07 '16 at 10:51
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    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 $\text{Pic}(\text{SL}_2)$. – Alex Youcis Nov 07 '16 at 10:51
  • Impressive! Thanks a lot, @Alex. – Georges Elencwajg Nov 07 '16 at 13:00
  • @GeorgesElencwajg Thanks :) I continually benefit from your answers--it's about time I return the favor. – Alex Youcis Nov 07 '16 at 13:10
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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.

user26857
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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
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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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Jack Schmidt
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