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*)^{m} = 1+*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^{r*}b^2^{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*^{r}as desired.

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