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Unipotent radicals of smooth connected trigonalizable groups over perfect fields have normal G_a series

Statement

Assume the Axiom of Choice inherited from the cited smoothness, quotient and reduction suppliers (The Axiom of Choice).

Let $k$ be a perfect field and let $G$ be a smooth connected trigonalizable affine algebraic group over $k$ (Smooth morphism of schemes, Affine schemes and their coordinate rings, Group schemes of finite type over a field). Then the series $G \supseteq G_{0} = G_{u} \supseteq \dots \supseteq G_{r} = 1$ of Trigonalizable groups have a normal series with a multiplicative quotient and additive subgroup quotients can be chosen with every indexed term $G_{i}$ smooth and connected, normal in $G$ , and every quotient $G_{i} / G_{i + 1}$ isomorphic to $G_{a}$ . The separate quotient $G / G_{u}$ remains of multiplicative type.

Facts & Assumptions

Given: AC, a perfect field $k$ and a smooth connected trigonalizable affine $k$ -group $G$ .

[F1]

$G$ has a normal series $G \supseteq G_{0} \supseteq G_{1} \supseteq \dots \supseteq G_{r} = 1$ with $G_{0} = G_{u}$ the largest normal unipotent subgroup of $G$ , $G / G_{u}$ of multiplicative type, and each $G_{i} / G_{i + 1}$ embedded $G / G_{u}$ -equivariantly into $G_{a}$ . (Trigonalizable groups have a normal series with a multiplicative quotient and additive subgroup quotients)

[F2]

Assume AC. For a closed subgroup scheme $H$ of a smooth finite-type group over a perfect field, the identity component of the reduction $(H_{r e d})^{0}$ is a smooth connected closed subgroup with $\dim (H_{r e d})^{0} = \dim H$ ; it is normal when $H$ is normal. (Reduced identity components over perfect fields)

[F3]

Assume AC. In an exact sequence $1 \to N \to H \to Q \to 1$ of finite-type group schemes over a field, if $H$ is smooth and connected then the image $Q$ is smooth and connected; and a smooth connected group over an algebraically closed field with a proper smooth connected normal subgroup of codimension one has quotient of dimension one. (Affine smooth and connected properties in exact sequences of algebraic groups, Connected finite-type groups are geometrically connected)

[F4]

A smooth connected affine unipotent group of dimension one over a perfect field is $G_{a}$ : the one-dimensional classification gives a form split by a finite purely inseparable extension, and a perfect field has no nontrivial such extension. (One-dimensional smooth connected affine groups over perfect fields are additive groups or tori)

[F5]

Over a perfect field, a commutative affine algebraic group has a unique product decomposition into its largest unipotent subgroup and largest subgroup of multiplicative type; both factors are smooth and connected when the group is. This is Milne Theorem 16.13(b) with its proof, and Corollary 16.15, printed pp. 328-329. The derived subgroup of a smooth connected group is smooth connected; a subgroup that is both unipotent and of multiplicative type is trivial. (Properties of the derived subgroup of an algebraic group, A subgroup that is both unipotent and diagonalizable is trivial, Groups of multiplicative type and tori)

[A1]

The Axiom of Choice is inherited through the cited suppliers and is the axiom of The Axiom of Choice.

Proof

Given: AC, a perfect field $k$ and a smooth connected trigonalizable affine $k$ -group $G$ .

1.1F2F3F5F1

First prove that $G_{u}$ is smooth and connected. Put $U_{0} = (G_{u, r e d})^{0}$ , smooth connected and normal in $G$ by [F2], and form $H = G / U_{0}$ . By [F3], $H$ is smooth connected, its kernel $U^{'} = G_{u} / U_{0}$ over $D = G / G_{u}$ is finite unipotent, and $D$ is of multiplicative type. Since $D$ is commutative, $D H \subseteq U^{'}$ . But $D H$ is smooth connected by [F5], so, being finite, it is trivial; thus $H$ is commutative. Decompose $H = H_{u} \times H_{s}$ by [F5]. Its smooth connected unipotent factor $H_{u}$ maps trivially into $D$ and therefore lies in finite $U^{'}$ , so it is trivial. Hence $H$ is of multiplicative type, and its unipotent subgroup $U^{'}$ is trivial by [F5]. Consequently $G_{u} = U_{0}$ is smooth and connected.

2.1F1F2step 1.1

Take the series $G \supseteq G_{0} = G_{u} \supseteq \dots \supseteq G_{r} = 1$ of [F1], and put $H_{i} = (G_{i, r e d})^{0}$ for $0 \leq i \leq r$ . These subgroups are nested, smooth connected and normal in $G$ , with $\dim H_{i} = \dim G_{i}$ , by [F2]. Step 1.1 gives $H_{0} = G_{u}$ , and $H_{r} = 1$ . Since $G_{i} / G_{i + 1}$ embeds in $G_{a}$ , its dimension is at most one, so $\dim H_{i} - \dim H_{i + 1} \leq 1$ .

3.1F3F4step 2.1

Each quotient $H_{i} / H_{i + 1}$ is affine, smooth and connected by [F3], and unipotent as a quotient of a unipotent group. Its dimension is the difference of the dimensions of its source and kernel, hence at most one by step 2.1. A smooth geometrically connected zero-dimensional group is trivial: its geometric fibre is one reduced point and its identity section descends that identification. In dimension one it is $G_{a}$ by the one-dimensional classification.

4.1A1F1step 2.1step 3.1∎

Delete repetitions in $H_{0} \supseteq \dots \supseteq H_{r}$ ; step 3.1 shows that every remaining successive quotient is $G_{a}$ . Prefixing $G \supseteq H_{0} = G_{u}$ retains the multiplicative-type quotient $G / G_{u}$ from the original theorem; the additive successive quotients are precisely those inside $G_{u}$ . Every indexed term is smooth connected and normal in $G$ , as required.

Depends on

Used by

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Sources

J. S. Milne, Algebraic Groups (corrected 2022 printing, Cambridge University Press) (standard reference, not scraped)
Florian Herzig, Linear Algebraic Groups (University of Toronto lecture notes, 2013) (standard reference, not scraped)