I do not understand fully the proof of Corollary 1.6 from Mumford's Geometric Invariant Theory (§3: Linearization of an invertible sheaf, p 35):
Corollary 1.6
$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\Pic{Pic}$Let $G$ be a connected linear algebraic group $G$ acting on an
algebraic variety over $k$, which is proper over $k$.
(Mumford's variety is a scheme $X/k$ such that
$\overline{X}= X \times \Spec(\overline{k})$ is irreducible
and closed; cf. Chap 0.) Let $\mathcal{L}$ be an invertible
sheaf on $X$.
Then if $ \overline{X} $ is a normal variety, some tensor
power $\mathcal{L}^n$ of $\mathcal{L}$ is always linearizable.
Proof. According to a result of Chevalley [Ch 10], in this case all components of $\Pic(X/k) $ are proper over $k$, hence all reduced components are Abelian varieties. Therefore the connected linear group $G$, being birational to projective space as a variety (?), must act trivially on $\Pic(X/k) $ (Why that's true? comp. 1st question?) […and finally claim follows from previous Prop 1.5 which states that some power $\mathcal{L}^n$ of $\mathcal{L}$ is $G$-linearizable iff some multiple $n \lambda $ of $\lambda \in \Pic(X/k)(k)$ is fixed by $G$, where the $k$-point $\lambda$ of $\Pic(X/k)$ is defined by $\mathcal{L}$.]
Questions:
Why is $G$ as connected linear group birational to
projective space?
And why the above leads to the consequence that $G$ acts trivially on $\Pic(X/k) $? We can reduce it to the question why the action of every connected linear group on an Abelian variety is trivial.
A Remark: Mumford nowhere describes this action of $G$ on $ \Pic(X/k)$ in explicit terms, so I assume he means the most natural one, i.e. for $S \to \Spec(k)$ the action of a geometric point $\alpha \in G$ on $X$ induces an action on $X \times_k S$ via $\alpha \times 1_S$ and this one acts on
$$ \Pic(X/k)(S) = \{ \mathcal{M} \text{ invertible sheaf on } X \times_k S \} / \{ \text{ inv. sheaves of the form } p^*_S(\mathcal{K}) \text{ for } \mathcal{K} \text{ invertible on } S \} $$
via taking pullpacks $\mathcal{M} \mapsto (\alpha \times 1_S)^*\mathcal{M}$. Is it correct?
[Ch10] Chevalley, C: Sur la théorie de la variété de Picard.