Question

Let $\mathit{R}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{S}$ be the ring and let $\overline{\mathbf{R}}$be the subring of $\mathit{R}\mathbf{\times }\mathit{S}$ consisting of all elements of the form $\left(a,0_S\right)$. Show that the function $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{R}$given by $\mathit{f}\left(a\right)\mathbf{=}\left(a,0_S\right)$ is an isomorphism

Short answer

Hence proved that the function$f$ 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

For the given rings $RandS$, the subring we have$\overline{R}$ is expressed as:

$\overline{R}=\left\{\left(a,0_S\right)|a\in R\right\}\subseteq R\times S$.

Also, the function $f:R\to \overline{R}$is defined such that:

$f\left(a\right)=\left(a,0_S\right),\forall a\in R$

From here, we see the function is surjective.

Also, we get:

role="math" localid="1647886760916" $$\begin{gathered} f\left(a\right)=f\left(b\right) \\ \left(a,0_S\right)=\left(b,0_S\right).......\forall a,b\in R \\ a=b \\ f\to injective \end{gathered}$$

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

$$\begin{gathered} f\left(a+b\right)=\left(a+b,0_S\right) \\ =\left(a,0_S\right)+\left(b,0_S\right) \\ =f\left(a\right)+f\left(b\right) \end{gathered}$$

$$\begin{gathered} f\left(ab\right)=\left(ab,0_S\right) \\ =\left(a,0_S\right)\left(b,0_S\right) \\ =f\left(a\right)f\left(b\right) \end{gathered}$$

Hence proved, $f$ is an isomorphism.