# Answer on Algebra Question for sanches

Question #16873

Let a be an element in a ring such that ma = 0 = a^(2^r) , where m ≥ 1 and r ≥ 0 are given integers. Show that (1 + a)^(m^r) = 1.

Expert's answer

The proof is by induction on

*r*.The case*r*= 0 being clear, we assume*r >*0. Since*ma*=0, the binomial theorem gives (1 +*a*)*= 1+*^{m}*a*^{2}*b*where*b*is a polynomial in*a*with integer coefficients. Since*m*(*a*^{2}*b*) = 0 and (*a*^{2}*b*)^2^{r−}^{1}=*a^*2*2*^{r*}b^^{r−}^{1}= 0*,*theinductive hypothesis (applied to the element*a*^{2}*b*)implies that 1 = (1+*a*^{2}*b*)^*m*^{r−}^{1}= [(1 +*a*)*]^*^{m}*m*^{r−}^{1}= (1+*a*)^*m*as desired.^{r}Need a fast expert's response?

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