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$\mathit{\upsilon }$is the solution of the equation,$\mathit{a}\mathit{x}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$, prove that$\mathit{\upsilon }$is also a solution of the equation$\mathit{x}\mathit{a}\mathbf{=}{\mathbf{1}}_{\mathbf{R}}$. (Remember that$\mathit{R}$may not be commutative.)[Hint: Use part (a) with,$\mathit{b}\mathbf{=}\mathit{\upsilon }\mathit{a}$.]

Short answer

It is proved that, if$\upsilon$ is the solution of the equation, $ax=1_R$, then$\upsilon$ is also a solution of the equation $xa=1_R$.

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,role="math" localid="1653636206022" $b=0_R$.

Axiom 12

A field is a commutative ring$R$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, $a$be a non-zero element in the division ring $R$.

Since, $R$is a division ring, then there exists,$\upsilon \in R$, that satisfies,$a\upsilon =1_R$.

Now, using the associative multiplication, we get,

$$\begin{gathered} \left(\upsilon a\right)\left(\upsilon a\right)=\upsilon \left(a\upsilon \right)a \\ =\upsilon 1_{R}a \\ =\left(\upsilon 1_R\right)a \\ =\upsilon a \end{gathered}$$

As we know that, If $bb=b$, then$b=1_R$ by the part (a) of this question.

Which implies, $\upsilon a=1_R$. Therefore, $\upsilon$is a solution of $xa=1_R$