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.



Picard group of connected linear algebraic group
$\DeclareMathOperator\Pic{Pic}$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](https://www.researchgate.net/publication/240095738_Local_Properties_of_Algebraic_Group_Actions) ([DOI](http://dx.doi.org/10.1007%2F978-3-0348-7662-9_4)). 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.


