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Power maps with exponent prime to the characteristic are bijective on unipotent groups

Statement

Assume the Axiom of Choice for the field-point functor. Let $k$ be a field, let $U$ be an affine unipotent algebraic group over $k$ , and let $e \geq 1$ be an integer with $e \cdot 1_{k} \neq 0$ (every positive integer in characteristic zero, and precisely those prime to $p$ in characteristic $p > 0$ ). Then the map $x \mapsto x^{e}$ is a bijection from $U (k^{a})$ to itself.

Facts & Assumptions

Given: The Axiom of Choice, a field $k$ , an affine unipotent algebraic group $U$ over $k$ , and an integer $e \geq 1$ with $e \cdot 1_{k} \neq 0$ .

[F1]

$U$ has a central series $U = U_{0} \supseteq U_{1} \supseteq \dots \supseteq U_{r} = 1$ of closed subgroup schemes with successive quotients isomorphic to closed subgroup schemes of $G_{a}$ . (Unipotent groups have central series with quotients embedded in G_a)

[F2]

For a closed subgroup scheme $N \subseteq G_{a}$ over an algebraically closed field $K$ , multiplication by $e$ is bijective on $N (K)$ when $e$ is prime to the characteristic. In positive characteristic $p$ , choose integers $d, m$ with $d e = 1 + m p$ ; multiplication by $d$ preserves every additive subgroup and is its inverse. In characteristic zero, a proper closed subgroup has finitely many points, and the additive group has no nontrivial finite subgroup, so $N (K)$ is either $0$ or all of $K$ , where division by $e$ is valid. (Field-valued points and local-ring points)

[F3]

An fppf quotient of finite-type groups has nonempty finite-type fibres, and over an algebraically closed field each such fibre has a rational point. Thus its sequence on rational points is exact. This allows nonsmooth groups and infinitesimal kernels. (Normal subgroup quotients of finite-type group schemes exist as fppf scheme quotients, Over an algebraically closed field, every maximal ideal is an evaluation ideal)

Proof

Given: The Axiom of Choice, a field $k$ , an affine unipotent $U$ , and $e \geq 1$ with $e \cdot 1_{k} \neq 0$ .

1.1F1F2F3induction

Put $K = k^{a}$ and induct on the length of the central series in [F1], deleting repetitions. The group $1$ has a unique $e$ -th root of its only point. Otherwise let $N$ be the last nontrivial term of the series, so $N$ is central in $U$ and embeds in $G_{a}$ ; its power map on $K$ -points is bijective by [F2]. The quotient $Q = U / N$ inherits a shorter central series, and [F3] gives the exact sequence $1 \to N (K) \to U (K) \to Q (K) \to 1$ . By induction the power map on $Q (K)$ is bijective.

2.1F2F3step 1.1discharge-induction∎

For $x \in U (K)$ , take the unique $e$ -th root $\overset{ˉ}{y}$ of its image in $Q (K)$ and lift it to $y \in U (K)$ using [F3]. Then $a = x y^{- e} \in N (K)$ . Choose the unique $n \in N (K)$ with $n^{e} = a$ . Centrality gives $(y n)^{e} = y^{e} n^{e} = x$ , proving existence. If $z^{e} = y^{e}$ for two points of $U (K)$ , quotient uniqueness gives $z = y n$ for some $n \in N (K)$ ; centrality then gives $z^{e} = y^{e} n^{e}$ , so $n^{e} = 1$ and kernel uniqueness forces $n = 1$ . Thus the power map on $U (K)$ is injective as well as surjective, completing the induction. No assertion that a power map is a homomorphism on a noncommutative group is used.

Depends on

Used by

Smooth diagonalizable groups over algebraically closed fields have only principal cocycles into smooth commutative unipotent groups Proposition

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Sources

J. S. Milne, Algebraic Groups (corrected 2022 printing, Cambridge University Press) (standard reference, not scraped)