# Picard group of connected linear algebraic group

Here's a statement:

Suppose $$G$$ is a connected linear algebraic group over a field $$k$$, then $$Pic(G)$$ is a finite group.

I know this is true when $$k=\mathbb{C}$$. My question is does this true for abitrary field $$k$$? If not, how about furthermore when $$G$$ is smooth or even reductive? Is there any reference?

Thanks for any help.

• I think there is a proof for this fact for every spilt reductive group in chapter 18 of the Milne book algebraic groups – ali Nov 29 '20 at 10:57

$$\DeclareMathOperator\Pic{Pic}$$The statement is false over most imperfect fields, even for smooth affine group schemes. In particular, it is false over any separably closed imperfect field $$k$$. I will give an example over imperfect fields of characteristic at least $$3$$, but it is not difficult to adapt it to work in char. 2 as well.

The statement is, however, correct over fields of characteristic $$0$$ or if $$G$$ is reductive. This is proven in Prop. 4.5 of Knop, Kraft, Luna, Vust - Local properties of algebraic group actions (DOI). The proof is written under the assumption that $$\operatorname{char} k=0$$ but it can be adapted to char. $$p$$ in the case of reductive $$G$$.

Example: Let $$k$$ be a separably closed imperfect field of characteristic $$p>2$$, and $$U=\operatorname{Spec} k[x,y]/(y^p-x-ax^p)$$ for some $$a\in k \setminus k^p$$.

Remark: $$U$$ is a naturally a subgroup of $$\mathbf{G}_a^2$$. This is a so-called $$k$$-wound form of $$\mathbf{G}_a$$. This, in particular, means that $$U$$ is isomorphic to $$\mathbf{G}_a$$ over the algebraic closure of $$k$$, i.e. $$U_{\bar{k}} \cong \mathbf{G}_{a, \bar{k}}$$.

Claim: $$\Pic(U)$$ is infinite.

One easily checks that its Zariski closure $$C\mathrel{:=}V(Y^p-XZ^{p-1}-aX^p)$$ inside $$\mathbf{P}^2_k$$ is a regular curve of genus $$\frac{(p-1)(p-2)}{2}>0$$ such that $$C\setminus U$$ is a point $$P$$ with residue field $$k(a^{1/p})$$. So we have an exact sequence $$\mathbf{Z}P \to \Pic(C) \to \Pic(U) \to 0$$ that induces an inclusion $$\Pic^0(C) \to \Pic(U)$$. So it suffices to show that $$\Pic^0(C)$$ is infinite. Now the Picard functor $$\Pic^0_{C/k}$$ is representable by a $$k$$-smooth group scheme of dimension $$\frac{(p-1)(p-2)}{2}$$. Therefore, $$\Pic^0(C)=\mathbf{Pic}^0_{C/k}(k)$$ is infinite as $$\dim \mathbf{Pic}^0_{C/k}>0$$ and $$k$$ is separably closed.

• Nice! Maybe you should make clear that you view $U$ as a subgroup of $\Bbb{G}_a^2$. – abx Nov 29 '20 at 6:36
• A nice answer! Thanks a lot! – jasonlzy Nov 29 '20 at 6:52
• Is there any reference for proof when char $k>0$ and $G$ is reductive? – jasonlzy Nov 29 '20 at 6:55
• @jasonlzy I am not aware of a reference, where the result is stated over any field. The essential idea of the argument should be to reduce to the case of $G$ that is semisimple and simply connected. Then show that $\mathrm{Pic}(G)=0$ for a semisimple, simply connected $k$-group. The second step is explained here mathoverflow.net/questions/273762/…, I can provide more details on the first step later if you are interested in it. – gdb Nov 29 '20 at 22:22
• @gdb I hope there're more details since I failed to work out by myself. Thanks! – jasonlzy Nov 30 '20 at 1:57