Let $G$ be an algebraic group, i.e., an affine reduced, separated $k$-scheme of finite type with structure of a group.
In Armand Borel’s Linear Algebraic Groups Theorem III.10.6(4) says
Theorem 10.6 (3): Let $G$ be a connected, solvable algebraic group. Then $G$ is nilpotent if and only if $G_s$ is a subgroup of $G$. In this case, $G_s$ is a closed subgroup defined over $k$, and $G$ is the direct product $G_s \times G_u$.
The subgroups $G_s, G_u \subset G$ are the semisimple resp. the unipotent parts coming from the Jordan decomposition.
The proof’s strategy is to show that $G = G_s \cdot G_u$ decomposes as a direct product of abstract groups and one observes that $L(G_s) \cap L(G_u) = 0$, where $L(G_s)$ and $ L(G_u)$ are the Lie algebras of $G_s$ resp. $G_u$. Then the proof claims that this already implies that the product structure $G = G_s \cdot G_u$ holds as an algebraic group, because the canonical map $G_s \times G_u \to G$ is bijective and separable.
A similar strategy is also used in the proof of part (4) of the same theorem to show that $G$ decomposes as a semi-direct product $G=T \cdot G_u$ of algebraic groups, where $T$ is a maximal torus.
Question: Why does separability of the bijective product map $G_s \times G_u \to G$ imply that it’s an ismorphisms of algebraic groups?
The book uses two notations of separability: a morphism $f: V \to W$ of algebraic varieties is separable if the field extension $K(W) \subset K(V)$ is separable. There is also a notion for separability for homogeneous spaces: the canonical projection $\pi: G \to G/H$ is separable if the induced map $(d \pi)_{1_G}: L(G) \to L(G/H)$ on Lie algebras at the neutral element $1_G$ is surjective. (Cor. II.6.7)
Since the map $G_s \times G_u \to G$ in my question is the canonical product map, I don’t see how one of these two seemingly unrelated definitions of separability can be applid here. Maybe it’s possible to interpret $G$ somehow as an orbit space/homogeneous space of $G_s \times G_u$. At all, if we want to see $G$ as an orbit space, then we should have a transitive $(G_s \times G_u)$-action on $G$ which coincides with the product structure $(u,v) \to u \cdot v:= m(u,v)$. But with respect to which action?
The most natural choices to define a $(G_s \times G_u)$-action on $G$ might be given via $((u,v), g) \mapsto u \cdot v \cdot g$ or $u \cdot g \cdot v$.
Problem: an action $\Phi \colon H \times S \to S$ by a group $H$ on set $S$ must satisfy the “compatibility relation” with group multiplication $\Phi(h_1 \cdot h_2, s)= \Phi(h_1, \Phi(h_2, s))$ for all $h_1,h_2 \in H, s \in S$. But unfortunately the two natural guesses do not satisfy this rule. So either the argument in the book works differently or there is a less usual $(G_s \times G_u)$-action on $G$ applied. If the second case holds, I wonder why the book doesn’t write this action explicitly down.
And if there is some $(G_s \times G_u)$-action on $G$ established, how this would imply that $G_s \times G_u \to G$ is an isomorphism of algebraic groups? It seems that the central issue in the argument has to do with the separability, but I don’t see why the claim follows from it.
This is not a complete answer, but to large to fit into a comment. My guess is that Borel means separability in the field sense.
Can you show that the morphism $f: G_s \times G_u \to G$ is finite (as a morphism of schemes)? Then it will be an isomorphism of schemes:
As $f$ is finite, it induces a finite, separable extension of function fields $k(G) \hookrightarrow k(G_s \times G_u)$. Because $f$ is bijective, and the extension is separable, it has degree $1$, so the function fields are actually the same.
Now if $\operatorname{Spec} A \subset G$ is open affine, its preimage $ \operatorname{Spec} B = f^{-1}(\operatorname{Spec} A)$ is affine (because $f$ is finite). So you get a finite extension $A \subset B \subset Q(B) = k(G) = Q(A)$. But since $G$ is smooth, $A$ is normal, i.e. integrally closed in $Q(A)$. Thus $A = B$.