Rational points of commutator subgroups of solvable algebraic groups

Let G be a connected algebraic group defined over a field k. Denote by G ( k ) 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 ) [A,B](k) is not equal to [ A ( k ) , B ( k ) ] [A(k),B(k)] in general. Here [A,B] denotes the group generated by commutators a b a − 1 b − 1 , a ∈ A , b ∈ B 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 ) f(x) in k [ x ] k[x] and any element c in k, there exists an element α \alpha in k such that f ( α ) = c 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. T h e n [ A , B ] ( k ) = [ A ( K ) , B ( K ) ] 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 {k_0} of k such that if K is the maximal p-extension of k 0 {k_0} , then [ A , B ] ( K ) = [ A ( K ) , B ( K ) ] [A,B](K) = [A(K),B(K)] .