Question

Let $\mathbf{R}$ be a commutative ring with identity of prime characteristic P if $\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{R}$ and $\mathbf{n}\mathbf{\ge }\mathbf{1}$ prove that ${\left(a-b\right)}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{=}{\mathit{a}}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{-}{\mathit{b}}^{{\mathbf{p}}^{\mathbf{n}}}$.

Short answer

${\left(a-b\right)}^{p^n}=a^{p^n}-b^{p^n}$

Step-by-step solution

Definition of bionomial.

The bionomial theorem or bionomial expression describes the algebraic expansion of power of a bionomial

Showing that  (a−b)pn=apn−bpn.

Let $\mathit{R}$ be a commutative ring with identity of prime characteristic $\mathit{p}$ and $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}\mathbf{,}\mathit{n}\mathbf{\ge }\mathbf{1}$.

It will be shown using mathematical induction. So let $\mathit{n}\mathbf{=}\mathbf{1}$. Then

$\begin{array}{c}{\left(a-b\right)}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{=}{\left(a-b\right)}^{{\mathbf{p}}^{\mathbf{1}}}\\ \mathbf{=}{\left(a-b\right)}^{\mathbf{p}}\\ \mathbf{=}{\mathbf{a}}^{\mathbf{p}}\mathbf{+}\left(\begin{array}{c}p\\ 1\end{array}\right){\mathbf{a}}^{\mathbf{p}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{.}\mathbf{........}\left(\begin{array}{c}p\\ r\end{array}\right){\mathbf{a}}^{\mathbf{p}\mathbf{-}\mathbf{r}}{\left(-b\right)}^{\mathbf{r}}\end{array}$

Since each of the middle coefficient $\left(\begin{array}{c}p\\ r\end{array}\right)=\frac{p!}{r!\left(p-r\right)!}$ is an integer. As every term in the denominator is strictly less than the prime P and the factor of P in the numerator does not cancel therefore $\left(\begin{array}{c}p\\ r\end{array}\right)$ is divisible by P say $\left(\begin{array}{c}p\\ r\end{array}\right)=mp$

Since is a commutative ring with identity of prime characteristic . Therefore

$\begin{array}{c}\left(\begin{array}{c}p\\ r\end{array}\right){\mathbf{a}}^{\mathbf{p}\mathbf{-}\mathbf{r}}{\left(-b\right)}^{\mathbf{r}}\mathbf{=}\mathbf{m}\mathbf{p}{\mathbf{1}}_{\mathbf{R}}{\mathbf{a}}^{\mathbf{p}\mathbf{-}\mathbf{r}}{\left(-b\right)}^{\mathbf{r}}\\ \mathbf{=}\mathbf{m}\left(p{1}_R\right){\mathbf{a}}^{\mathbf{p}\mathbf{-}\mathbf{r}}{\left(-b\right)}^{\mathbf{r}}\\ \mathbf{=}\mathbf{m}\left(0_R\right){\mathbf{a}}^{\mathbf{p}\mathbf{-}\mathbf{r}}{\left(-b\right)}^{\mathbf{r}}\\ \mathbf{=}{\mathbf{0}}_{\mathbf{R}}\end{array}$

Thus all the middle terms are zero therefore

${\left(a-b\right)}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{=}{\mathit{a}}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{-}{\mathit{b}}^{{\mathbf{p}}^{\mathbf{n}}}$

Hence the claim follow for $n=1$. Now let the claim is true for $n=k$ then to show claim follow for $n=k+1$

$\begin{array}{c}{\left(a-b\right)}^{{\mathbf{p}}^{\mathbf{n}}}\mathbf{=}{\left({\left(a-b\right)}^{p^n}\right)}^{\mathbf{p}}\\ \mathbf{=}\left(a^{p^n}-b^{p^n}\right)\\ \mathbf{=}{\left(a^{p^k}\right)}^{\mathbf{n}}\mathbf{-}{\left(b^{p^k}\right)}^{\mathbf{n}}\\ \mathbf{=}{\mathbf{a}}^{{\mathbf{p}}^{\mathbf{k}\mathbf{+}\mathbf{n}}}\mathbf{-}{\mathbf{b}}^{{\mathbf{p}}^{\mathbf{k}\mathbf{+}\mathbf{n}}}\end{array}$

Thus the claim follows for $n=k+1$ and hence by mathematical induction

${\left(a-b\right)}^{p^n}=a^{p^n}-b^{p^n}$

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