Question

Let$R$ be a ring and consider the subset $R*$of $R\times R$defined by$R*$ .

1. If$R={\mathrm{\mathbb{Z}}}_6$ ,list the elements of$R*$ .
2. For any ring$R$ , show that$R*$ is a subring of$R\times R$.

Short answer

1. The required elements are: $\left(0,0\right),\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right)$ .
2. It is proved that $R*$ is a subring of $R\times R$.

Step-by-step solution

Rings

A ringcan be defined as a non-empty set $\mathrm{ℝ}$whose elements deal with basically two operations: addition and multiplication, where elements belong to the real number$\mathrm{ℝ}$ .

Finding sets:(a)

The given set is:

$R*={\mathrm{\mathbb{Z}}}_6=\left\{\overline{)\left(r,r\right)}r\in {\mathrm{\mathbb{Z}}}_6\right\}$

Thus, the elements are given by:

$\left(0,0\right),\left(1,1\right),\left(2,2,\right),\left(3,3\right),\left(4,4\right),\left(5,5\right)$

Proving Axioms:(b) 

Now, if $R$ is a ring, then, for the elements of$r,s\in R*$ we have:

1. $\left(r,r\right)+\left(s+s\right)=\left(r+sr+s\right)\in R*$
   
2. $\left(r,r\right)\left(s,s\right)=\left(rs,rs\right)\in R*$
3. $\left(0_R,0_R\right)\in R*$
4. Also, for the additive inverse:

role="math" localid="1646299372982" $$\begin{gathered} \left(-r,-r\right)\in R* \\ \Rightarrow \left(-r,-r\right)+\left(r,r\right)=\left(0_R,0_R\right)\in R* \end{gathered}$$

Hence it is proved that $R*$ is a subring of $R\times R$.