Exercise 12.1.15

Proof.

(a)

By definition, $${\alpha }_j=\underset{l=1}{\overset{p}{\sum }}{\zeta }_p^{j(l-1)}{x}_l,1\le j\le p.$$

Therefore

$$\begin{array}{llll}\hfill \underset{j=1}{\overset{p}{\sum }}{\zeta }_p^{-j(i-1)}{\alpha }_j& =\underset{j=1}{\overset{p}{\sum }}{\zeta }_p^{-j(i-1)}\underset{l=1}{\overset{p}{\sum }}{\zeta }_p^{j(l-1)}{x}_l& \hfill & \\ \hfill & =\underset{l=1}{\overset{p}{\sum }}\left[\underset{j=1}{\overset{p}{\sum }}{({\zeta }_p^{l-i})}^j\right]x_l& \hfill & \end{array}$$

(b)

$\text{∙}$

If $m\equiv 0\mathit{mod}p$, then ${\zeta }_p^m=1$, so $\underset{j=1}{\overset{p}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{({\zeta }_p^m)}^j=p$.

$\text{∙}$

If $m\not\equiv 0\mathit{mod}p$, then ${\zeta }_p^m\ne 1$, so $$\underset{j=1}{\overset{p}{\sum }}{({\zeta }_p^m)}^j={\zeta }_p^m(1+{\zeta }_p^m+{\zeta }_p^{2m}+\cdots +{\zeta }_p^{(p-1)m})={\zeta }_p^m\frac{1-{({\zeta }_p^m)}^p}{1-{\zeta }_p^m}=0.$$

Thus,

$$\underset{j=1}{\overset{p}{\sum }}{({\zeta }_p^m)}^j=\{\begin{array}{cc}p,\hfill & \text{if }m\equiv 0modp,\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

(c)

With $m=l-i$, part (b) gives $$\underset{j=1}{\overset{p}{\sum }}{({\zeta }_p^{l-i})}^j=\{\begin{array}{cc}p,\hfill & \text{if }l\equiv imodp,\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

Therefore, by part (a),

$$\begin{array}{llll}\hfill \underset{j=1}{\overset{p}{\sum }}{\zeta }_p^{-j(i-1)}{\alpha }_j& =\underset{l=1}{\overset{p}{\sum }}\left[\underset{j=1}{\overset{p}{\sum }}{({\zeta }_p^{l-i})}^j\right]x_l& \hfill & \\ \hfill & =p{x}_i.& \hfill & \end{array}$$

For all $i=1,2,\dots ,p$,

$$\begin{array}{llll}\hfill x_i& =\frac{1}{p}\underset{j=1}{\overset{p}{\sum }}{\zeta }_p^{-j(i-1)}{\alpha }_j& \hfill & \\ \hfill & =\frac{1}{p}\left({\alpha }_p+\underset{j=1}{\overset{p-1}{\sum }}{\zeta }_p^{-j(i-1)}{\alpha }_j\right)& \hfill & \\ \hfill & & \hfill \end{array}$$

Since ${\alpha }_p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{l=1}^p{\zeta }_p^{p(l-1)}{x}_l=x_1+\cdots +x_p={\sigma }_1$, we obtain

$$x_i=\frac{1}{p}\left({\sigma }_1+\underset{j=1}{\overset{p-1}{\sum }}{\zeta }_p^{-j(i-1)}{\alpha }_j\right).$$