Question

Prove that $\mathit{R}$is isomorphic to the ring $\mathit{S}$of all $\mathbf{2}\mathbf{\times }\mathbf{2}$matrices of the form $\left(\begin{array}{cc}0& 0\\ 0& a\end{array}\right)$, with $\mathit{a}\mathbf{\in }\mathit{R}$.

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 of its elements is respectively given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Isomorphism

Let us consider a ring $S$denoted as:

$S\to \left\{\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)|a\in R\right\}$

And a function $f:R\to S$ such that:

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

This function is surjective.

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

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

Now, checking for addition and multiplication property, we have:

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

role="math" localid="1647888215223" $$\begin{gathered} f\left(ab\right)=\left(\begin{array}{cc}ab& 0\\ 0& ab\end{array}\right) \\ =\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)\left(\begin{array}{cc}b& 0\\ 0& b\end{array}\right) \\ =f\left(a\right)f\left(b\right) \end{gathered}$$

Hence proved$f$is an isomorphism.