Question

Prove that the field R of real numbers is isomorphic to the ring of all $\mathbf{2}\mathbf{\times }\mathbf{2}$ matrices of the form $\left(\begin{array}{cc}0& 0\\ 0& a\end{array}\right)$, with $\mathit{a}\mathbf{\in }\mathit{R}$. [Hint: Consider the function $\mathit{f}$given by $\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{=}\left(\begin{array}{cc}0& 0\\ 0& a\end{array}\right)$ .]

Short answer

Hence proved that $f$is an isomorphism.

Step-by-step solution

Property of Rings

If any ring R has elements such that, $a,b\in R$, then, addition and multiplication of the function is respectively given by:

$f\left(a+b\right)=f\left(a\right)+f\left(b\right)$

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

Isomorphism

Let us consider a function $f$such that:

$f\left(a\right)=\left(\begin{array}{cc}0& 0\\ 0& a\end{array}\right)$

If this function is isomorphic, the given field and its matrix will be isomorphic.

So, for $a,b\in R$, we have:

$$\begin{gathered} \left(\begin{array}{cc}0& 0\\ 0& a\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& b\end{array}\right) \\ f\left(a\right)=f\left(b\right) \\ f\to injective \end{gathered}$$

As $a=b$ , the function will also be surjective.

Hence proved, $f$ is an isomorphism.