Question

Prove (2) in Lemma G.3.

Short answer

Lemma G.3 is proved.

Step-by-step solution

Introduction of Lemma G.3

Consider P be the ring with the coefficient in the ring R and x be the polynomial$\mathbf{(}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}{\mathbf{1}}_{\mathbf{R}}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{.}\mathbf{..}\mathbf{)}$ then for each element$\mathit{a}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{...}\mathbf{)}$ of${\mathit{R}}^{\mathbf{*}}$ also each integer$\mathit{n}\mathbf{\ge }\mathbf{1}$ .

role="math" localid="1658913503266" ${\mathit{x}}^{\mathbf{n}}\mathbf{=}\mathbf{(}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}\mathbf{...}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}{\mathbf{1}}_{\mathbf{R}}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}\mathbf{...}\mathbf{)}$, whererole="math" localid="1658913509721" ${\mathbf{1}}_{\mathbf{R}}$is in position n.

${\mathbf{ax}}^{\mathbf{n}}\mathbf{=}\mathbf{(}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}\mathbf{...}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}\mathbf{a}\mathbf{,}{\mathbf{0}}_{\mathbf{R}}\mathbf{,}\mathbf{...}\mathbf{)}$, where a is in position n.

Prove (2) in Lemma G.3

Since the polynomial $x$can be written as $x=(e_0,e_1,e_2,e_3,\text{\hspace{0.17em}}{e}_4,\mathrm{...})$where $e_i=0_R$for all $i\ne 1$and for $i=1$, have $e_i=1_R$.

By the property of mathematical induction, there is$x^1=x$ thus it is true for $n=1$.

$x^1=(e_0,e_1,e_2,e_3,\text{\hspace{0.17em}}{e}_4,\mathrm{...})$

Let it is true for$n=k$ then,

$x^k=(d_0,d_1,d_2,d_3,d_4,\mathrm{...})$

Where,$d_i=0_R$ for all$i\ne k$ and for $i=k$, have$d_k=1_R$ also,

$a{x}^k=(d_0,d_1,d_2,d_3,d_4,\mathrm{...})$

Where, for $i=k$, have $d_k=a$.

And for $n=k+1$,

$\begin{array}{c}a{x}^{k+1}=(d_0,d_1,d_2,d_3,d_4,\mathrm{...})\\ a{x}^k\cdot x=(d_0,d_1,d_2,d_3,d_4,\mathrm{...})\cdot (e_0,e_1,e_2,e_3,\text{\hspace{0.17em}}{e}_4,\mathrm{...})\\ =d_0{e}_{k+1}+d_1{e}_k+d_2{e}_{k-1}+\mathrm{...}+d_k{e}_1+d_{k+1}{e}_0\end{array}$

This can be written as,

$\begin{array}{c}a{x}^{k+1}=0+0+0+\mathrm{...}+a\cdot 1_R+0\\ =a\end{array}$

So, it is true for $n=k+1$and therefore true for all$n$ by the property of induction.

Hence, $a{x}^n=(0_R,0_R,\mathrm{...},0_R,a,0_R,\mathrm{...})$, where$a$ is in position $n$.