Question

Let R be a commutative ring with identity and a$\in$R.

If $a^3=0_R$. Show that $1_R+ax$ is a unit in R[x].

If $a^4=0_R$, Show that $1_R+ax$ is a unit in R[x].

Short answer

1. It is proved that 1+ax is a unit.
2. It is proved that 1+ax is a unit.

Step-by-step solution

Polynomial Arithmetic: 

If any given function R[x] is a ring, then the commutative, associative, and distributive laws hold such that the function f(x)+g(x) exists

Polynomial Operations:

(a)

Let us assume, 1-ax+$a^2{x}^2$ , then we have:

role="math" localid="1648450626144" $\begin{array}{rcl}\left(1+ax\right)\left(1-ax+a^2{x}^2\right)& =& 1-ax+a^2{x}^2+ax-a^2{x}^2+a^3{x}^3\\ & =& 1+a^3{x}^3\\ & =& 1+0\\ & =& 1\end{array}$

Clearly, 1+ax is a unit.

Hence proved, 1+ax is a unit.

Polynomial Operations: 

(b)

Let us assume, $1-ax+a^2{x}^2-a^3{x}^3$, then we have:

Clearly, 1+ax is a unit.

Hence proved, 1+ax is a unit.