Question

Question: Let R be a commutative ring with identity and $\mathit{a}\mathbf{\in }\mathit{R}$. If ${\mathbf{1}}_{\mathbf{R}}\mathbf{}\mathbf{+}\mathbf{}\mathit{a}\mathit{x}$is a unit in $\mathit{R}\left[x\right]$, show that ${\mathit{a}}^{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{0}}_{\mathbf{R}}$for some integer $\mathit{n}\mathbf{>}\mathbf{0}$. [Hint: Suppose that the inverse of ${\mathbf{1}}_{\mathbf{R}}\mathbf{+}\mathit{a}\mathit{x}$is ${\mathit{b}}_{\mathbf{0}}\mathbf{+}{\mathit{b}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{b}}_{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{b}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}$. Since their product is ${\mathbf{1}}_{\mathbf{R}}\mathbf{,}\mathbf{}{\mathit{b}}_{\mathbf{0}}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$(Why?) and the other coefficients are all ${\mathbf{0}}_{\mathbf{R}}$.]

Short answer

It is proved that,$a^n=0_R$ for some R.

Step-by-step solution

Given part

It is given that R is a ring with identity, where $a\in R$.

And$1_R+ax$ is a unit in $R\left[x\right]$.

Proof part

Assume that $\sum _{i=0}^k{b}_i{x}^i$is the inverse of 1 +ax, so that

$1=(1+ax)\left(\sum _{i=0}^k{b}_i{x}^i\right)$

role="math" localid="1648583410018" $=b_0+(\sum _{i=0}^k(b_i+a{b}_{i-1})x^i)+a{b}_k{x}^{k+1}$

Then, by uniqueness of representation, it can be concluded that $b_0=1,b_i+a{b}_{i-1}=0$for all $1\le i\le k$and $a{b}_k=0$.

When $i=1$, then $b_i+a{b}_{i-1}=0$becomes:

role="math" localid="1648584449619" $$\begin{gathered} b_1+a{b}_{1-1}=0 \\ {b}_1+a{b}_0=0 \\ {b}_1+a=0\{\sin ce,b_0=1\} \\ {b}_1=-a \end{gathered}$$

Next, let $b_i={(-1)}^i{a}^i$for some $1\le i\le k$. Then,

$$\begin{gathered} b_{i+1}+a{b}_i=0 \\ \Rightarrow b_{i+1}={(-1)}^{i+1}{a}^{i+1} \end{gathered}$$

Then by induction, we have,

$b_k={(-1)}^k{a}^k$

Again, from role="math" localid="1648584762326" $a{b}_k=0$, we have

role="math" localid="1648584890172" $$\begin{gathered} {(-1)}^k{a}^{k+1}=0 \\ \Rightarrow a^{k+1}=0 \\ Or{a}^n=0_{R}forsomeR \end{gathered}$$

Hence, the given statement is proved