Question

A division ring is a (not necessarily commutative) ring $\mathit{R}$with identity

${\mathbf{1}}_{\mathbf{R}}\mathbf{\ne }{\mathbf{0}}_{\mathbf{R}}$that satisfies Axioms 11 and 12 (pages 48 and 49). Thus a field is a commutative division ring.

See Exercise 43 for a non-commutative example.

Suppose$\mathit{R}$is a division ring and$\mathit{a}\mathbf{,}\mathit{b}$are nonzero elements of, $\mathit{R}$.

1. If $\mathit{b}\mathit{b}\mathbf{=}\mathit{b}$, prove that $\mathit{b}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$. [Hint: Let be the solution of $\mathit{b}\mathit{x}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$and note that $\mathit{b}\mathit{\nu }\mathbf{=}{\mathit{b}}^{\mathbf{2}}\mathit{\nu }$.]

Short answer

It is proved that, $b=1_R$.

Step-by-step solution

Axiom: 11:

An integral domain is a commutative ring$R$ with identity$1_R\ne 0_R$that satisfies this axiom:

Whenever$a,b\in R$ and, $ab=0_R$, then,$a=0_R$ or, $b=0_R$.

Axiom: 12:

A field is a commutative$R$ ring with identity,$1_R\ne 0_R$, that satisfies this axiom:

For each,$a\ne 0_R$ in$R$ the equation$ax=1_R$ has a solution in $R$.

Conclusion

Let us consider a non-zero element$b$in the division ring $R$, which satisfies$bb=b$.

Also, let$\nu \in R$satisfying $b\nu =1_R$.

As we know, by the definition of division ring, $R$has an identity element so, $1_R\in R$such that, $b{1}_R=b,\forall b\in R$

Now, using the above conditions we have,

$$\begin{gathered} b{1}_R=b \\ \Rightarrow b{1}_R=bb \\ \Rightarrow \left(b{1}_R\right)\nu =\left(bb\right)\nu \\ \Rightarrow b\left(1_R\nu \right)=b\left(b\nu \right) \\ \Rightarrow b\nu =b{1}_R=b \end{gathered}$$

Hence, by combining,

$$\begin{gathered} b\nu =b; \\ b\nu =1_R \end{gathered}$$

We get,$b=1_R$