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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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Additional Information

DOI: https://doi.org/10.1090/S0002-9947-1974-0349860-2
Keywords: Algebraic groups, algebra, algebraic geometry
Article copyright: © Copyright 1974 American Mathematical Society

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