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 | References | Similar Articles | Additional Information

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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Keywords:
Algebraic groups,
algebra,
algebraic geometry

Article copyright:
© Copyright 1974
American Mathematical Society