Is the ring of regular functions of a simple complex connected linear algebraic group factorial?

Question

The question is basically the subject. I have found references in the case of simply-connected algebraic groups (the field is arbitrary, the group is not necessarily affine):

Popov, Picard groups of homogeneous spaces of linear algebraic groups and one-dimensional homogeneous vector bundles.

The fact that $\mathcal{O}(G)$ is factorial in the case of simply-connected groups follows from the fact that $\text{Pic}(G) = 0$. What if the group is not simply-connected?

Any references and thoughts would be appreciated.

Answer 1

The answer is no: a semi-simple algebraic group $G$ over a field $k$ has the property that $A_G:=\mathcal{O}_G(G)$ is a UFD if and only if $G$ is simply connected.

First observe that as $G$ is smooth over $k$, it is a domain (as it is connected) it is regular and thus locally factorial and normal. So, by [Hartshorne, II.6.16] we have that $\mathrm{Pic}(G)=\mathrm{Cl}(G)$, an thus by [Hartshorne, II.6.2] we have that $A_G$ is a UFD if and only if $\mathrm{Pic}(G)=0$. The claim then follows from the following fact.

Fact ([Milne, Corollary 18.26]): Let $G$ be a semi-simple algebraic group over a (perfect) field $k$. Then, there is a canonical isomorphism of functors on extensions of $k$, $\mathbf{Pic}(G)\cong {\pi}_1(G)^\vee$.

Let me explain this notation:

• $\mathbf{Pic}(G)$ is the functor which sends an (separable) extension $k'/k$ to $\mathrm{Pic}(G_{k'})$,
• let me denote by $G^\mathrm{sc}\to G$ is the universal cover of $G$ (see [Milne, Definition 18.7] -- and note that by the universal property, one can show that $G^\mathrm{sc}_{k'}=(G_{k'})^\mathrm{sc}$) and then $\pi_1(G)=\ker(G^\mathrm{sc}\to G)$ (a finite etale commutative group scheme),
• for a finite etale commutative group scheme $A$ over $k$, let me denote by $A^\vee$ the Cartier dual which represents $\mathscr{Hom}(A,\mathbb{G}_{m,k})$ (the sheaf Hom from $A$ to $\mathbb{G}_{m,k}$ -- the Hopf algebra $\mathcal{O}(A^\vee)$ is just the $k$-dual Hopf algebra to the commutative and cocommutative Hopf $k$-algebra $\mathcal{O}(A)$).

Remark 1:

If you don't care about varying fields at all (e.g. you only care about the case when $k$ is algebraically closed) then this just reduces to the statement that $\mathrm{Pic}(G)=\pi_1(G)^\vee$ where the left-hand side is the usual thing, $\pi_1(G)$ is the usual thing, and now $(-)^\vee$ can be taken to mean Pontryagin dual.

The intuition for this fact is actually quite simple if you take the following (non-obvious) fact as given: $\mathrm{Pic}(G)$ is torsion (in fact it's finite -- for instance, the sketch below even shows this once you know that $\pi_1(G)$ is finite). In this case, it then comes in the case when $k$ is algebraically closed of chraracteristic $0$ (and for general characteristic $0$ fields by Galois descent) from combining the following facts:

• by the Kummer sequence $$1\to \mu_{n,k}\to \mathbb{G}_{m,k}\to \mathbb{G}_{m,k}\to 1$$ one has an exact sequence $$1\to A_G^\times/(A_G^\times)^n\to H^1(G,\mu_{n,k})\to \mathrm{Pic}(G)[n]\to 1$$,
• $A_G^\times=k^\times$ by this post so the first term vanishes,
• $\mu_{n,k}\cong \mathbb{Z}/n\mathbb{Z}$ and so $$H^1(G,\mu_{n,k})=H^1(G,\mathbb{Z}/n\mathbb{Z})=\mathrm{Hom}(\pi_1(G),\mathbb{Z}/n\mathbb{Z}).$$

Indeed, from these three facts we see that we get the equality

$$\mathrm{Pic}(G)[n]=\mathrm{Hom}(\pi_1(G),\mathbb{Z}/n\mathbb{Z}),$$

and passing to the limit over all $n$ gives the result.

Remark 2:

For the above sketch (although not actually for the proof given in Popov/Milne) it's important to know that $\pi_1(G)$ is (the group scheme associated to the) the $\mathrm{Gal}(\overline{k}/k)$-module associated to the usual (etale) fundamental group of $G_{\overline{k}}$.

References:

[Harsthorne] Hartshorne, Robin. Algebraic geometry. Vol. 52. Springer Science & Business Media, 2013.

[Milne] Milne, James S. Algebraic groups: the theory of group schemes of finite type over a field. Vol. 170. Cambridge University Press, 2017.