Powering map by field characteristic is same in algebra and algebra group

From Groupprops
Jump to: navigation, search


Suppose F is a field of characteristic equal to a prime number p and N is an associative algebra over F that admits an algebra group G = \{ 1 + x \mid x \in N \}. (This happens, for instance, if N is nilpotent). Then, the p-power map in N corresponds to the p-power map in G, i.e.:

(1 + x)^p = 1 + x^p \ \forall \ x \in N


The proof simply involves expanding using the binomial theorem and then using characteristic p. Note that we crucially use the fact that 1 commutes with x to apply the binomial theorem.