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.

  • $\begingroup$ I think there is a proof for this fact for every spilt reductive group in chapter 18 of the Milne book algebraic groups $\endgroup$ – 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.

  • $\begingroup$ Nice! Maybe you should make clear that you view $U$ as a subgroup of $\Bbb{G}_a^2$. $\endgroup$ – abx Nov 29 '20 at 6:36
  • $\begingroup$ A nice answer! Thanks a lot! $\endgroup$ – jasonlzy Nov 29 '20 at 6:52
  • $\begingroup$ Is there any reference for proof when char $k>0$ and $G$ is reductive? $\endgroup$ – jasonlzy Nov 29 '20 at 6:55
  • $\begingroup$ @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. $\endgroup$ – gdb Nov 29 '20 at 22:22
  • $\begingroup$ @gdb I hope there're more details since I failed to work out by myself. Thanks! $\endgroup$ – jasonlzy Nov 30 '20 at 1:57

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