Question

(a) Show that the ring $\mathit{M}\mathbf{\left(}\mathbf{ℝ}\mathbf{\right)}$ is not a division ring by exhibiting a matrix that has no multiplicative inverse. (Division rings are defined in Exercise 42 of Section 3.1.)

Short answer

The ring $M\left(ℝ\right)$ is not a division ring.

Step-by-step solution

Division ring

A division ring is a ring $R$with identity $1_R\ne 0_R$that satisfies the axioms,

1. whenever role="math" localid="1654243948842" $a,b\in R$ and $ab=0_R$ then role="math" localid="1654243944848" $a=0_R$ or role="math" localid="1654243941168" $b=0_R$.

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

M(ℝ) is not a division ring

Consider a matrix $A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\in M\left(ℝ\right)$and take another matrix $B=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]$in $M\left(ℝ\right)$ then the find the multiplicative inverse as follows:

$$\begin{gathered} AB=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right] \\ =\left[\begin{array}{cc}{b}_{11}& b_{12}\\ 0& 0\end{array}\right] \\ \ne \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Thus, there is no multiplicative inverse of the matrix A in $M\left(ℝ\right)$.

Therefore, $M\left(ℝ\right)$ is not a division ring.