Let $G$ be an algebraic group, i.e., an affine reduced, separated $k$-scheme of finite type with structure of a group. In 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 firstly that $G = G_s \cdot G_u$ decomposes as direct product of abstract groups and make the observation 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 product map $m:G_s \times G_u \to G$ 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 in this proof?

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)

The proof argues with Lie algebras, more concretely the proof implies that the induced map $m_L: L(G_s \times G_u) \to L(G)$ on Lie algebras is an isomorphism because $\text{Ker}(m_L)= L(G_s) \cap L(G_u) = 0$ applying a simple dimension count. So it suggest that the proof uses the second notion of separability, namely that one for homogeneous spaces.

But then, how 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. Which one? 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? Does it boils down to something like that realizing the product map $G_s \times G_u \to G$ as homogeneous map the it is an isomorphism iff it is bijective (as set map) + separable?

Remark: This question is identical to this one I asked several weeks ago. Although there is an answer, I doubt that this is what Borel used in his argumentation, since it is expected that the information that $L(G_s) \cap L(G_u) = 0$ should be directly involved in the argument.

  • 1
    $\begingroup$ The argument is straightforward: (1) $f:X \to Y$ is a map of irreducible smooth varieties of the same dimension and the map induced by $f$ on the tangent space at a point $x \in X(k)$ is an isomorphism then $f$ is separable. The statement about Lie algebras in Borel implies that this condition holds for $x$ the identity element. (2) Any separable map of smooth irreducible smooth varieties which is a bijection is an isomorphism. $\endgroup$
    – naf
    Mar 6, 2023 at 4:06
  • $\begingroup$ Proposition AG.18.2 in Borel implies: If $\alpha: V \rightarrow W$ is a dominant injective morphism of irreducible varieties, then $K(V)$ is purely inseparable over $K(W)$. Then note that if an extension is both purely inseparable and separable, it is trivial. $\endgroup$
    – spin
    Mar 7, 2023 at 3:06

1 Answer 1


I don't always know how to make sense of Borel's somewhat old-fashioned approach, but I think the argument is simpler than it might seem: the subgroups $G_s$ and $G_u$ are mutually normalising and intersect trivially, hence commute, so they admit a direct product inside $G$; but then this direct product contains $G(\overline k)$, by the Jordan decomposition, so, since $G$ is smooth, must be all of $G$.

(I just noticed that you say only that $G$ is affine reduced over $k$, but I think you mean to assume either that $G$ is actually smooth, or that $k$ is algebraically closed (in which case reducedness implies smoothness). Borel only considers smooth groups.)


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