ISSN 1088-6850(online) ISSN 0002-9947(print)

Rational points of commutator subgroups of solvable algebraic groups

Author: Amassa Fauntleroy
Journal: Trans. Amer. Math. Soc. 194 (1974), 249-275
MSC: Primary 20G15; Secondary 14L15
DOI: https://doi.org/10.1090/S0002-9947-1974-0349860-2
MathSciNet review: 0349860
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Abstract: Let G be a connected algebraic group defined over a field k. Denote by $G(k)$ the group of k-rational points of G. Suppose that A and B are closed subgroups of G defined over k. Then $[A,B](k)$ is not equal to $[A(k),B(k)]$ in general. Here [A,B] denotes the group generated by commutators $ab{a^{ - 1}}{b^{ - 1}},a \in A,b \in B$. We say that a field of k of characteristic p is p-closed if given any additive polynomial $f(x)$ in $k[x]$ and any element c in k, there exists an element $\alpha$ in k such that $f(\alpha ) = c$. Theorem 1. Let G be a connected solvable algebraic group defined over the p-closed field k. Let A and B be closed connected subgroups of G, which are also defined over k, and suppose A normalizes B. $Then\;[A,B]\;(k) = [A(K),B(K)]$. 2. If G, A and B are as above and k is only assumed to be perfect then there exists a finite extension ${k_0}$ of k such that if K is the maximal p-extension of ${k_0}$, then $[A,B](K) = [A(K),B(K)]$.

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