Question

Let R be a commutative ring with identity and $a\in R$. If$1_R+ax$ is a unit in $R\left[x\right]$, show that$a^n=0_R$ for some integer $n>0$.

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[x].

Proof part

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

role="math" localid="1648451858913" $\begin{array}{rcl}1& =& \left(1+ax\right)\left(\sum _{i=0}^k{b}_i{x}^i\right)\\ & =& b_0+\left(\sum _{i=0}^k\left(b_i+a{b}_{i-1}\right)x^i\right)+a{b}_k{x}^{k+1}\end{array}$

Proof part

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 ab_k=0.

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

$\begin{array}{rcl}{b}_i+a{b}_{1-1}& =& 0\\ b_i+a{b}_0& =& 0\\ b_i+a& =& 0\left\{\because b_0=1\right\}\\ b_i& =& -a\end{array}$

Next, let for some . Then,

$\begin{array}{rcl}{b}_{i+1}+a{b}_i& =& 0\\ & \Rightarrow & b_{i+1}={\left(-1\right)}^{i+1}{a}^{i+1}\\ & & \end{array}$

Then by induction, we have,

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