Question

Let R be a ring with identity and $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$. Assume that a is not a zero divisor. Prove that $\mathit{a}\mathit{b}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$and if and only if $\mathit{b}\mathit{a}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$. [Hint: Note that both $\mathit{a}\mathit{b}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$ and $\mathit{b}\mathit{a}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$ imply$\mathit{a}\mathit{b}\mathit{a}\mathbf{=}\mathit{a}$ (why?); use Exercise 21.]

Short answer

It is proved that $ab=1_{\mathrm{R}}$if and only if $ba=1_{\mathrm{R}}$

Step-by-step solution

Statement of Exercise 21

Exercise 21 considers R as a ring and a as a non-zero element of R which is not a zero divisor. The cancellation law is true for a , that is as follows:

(a) When $ab=ac$ in R, then $b=c$.

(b) When $ba=ca$ in R, then $b=c$.

Show that ab=1R  and if and only if ba=1R 

Assume that $ab=1_R$.

Multiply both sides by a on the right-hand side of the above equation as follows:

$\left(ab\right)a=a$

It is obtained that $a\left(ba\right)=a$ from associativity and $ba=1_R$ from exercise 21.

Assume that $ba=1_R$.

Let us multiply both sides of the above equation by a:

$a\left(ba\right)=a$

It is obtained that $\left(ab\right)a=a$from associativity and $ab=1_R$from exercise 21.

Hence, it is proved that $ab=1_R$ if and only if $ba=1_R$.