Rational points of commutator subgroups of solvable algebraic groups

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 $[A,B](K) = [A(K),B(K)]$.