Question

Let R be a ring without identity. Let T be the set $R\times \mathrm{\mathbb{Z}}$. Define addition and multiplication in T by the given rules:

$\left(r.m\right)+\left(s,n\right)=\left(r+s,m+n\right)$

$\left(r,m\right)\left(s,n\right)=\left(rs+ms+nr,mn\right)$

1. Prove that T is a ring with identity.
2. Let $\overline{R}$consist of all elements of the form (r, 0) in T. Prove that $\overline{R}$ is a subring of T.

Short answer

1. It is proved that $T$ is a ring with identity.
2. It is proved that $\overline{R}$ is a subring of $T$.

Step-by-step solution

Definition of a ring with identity

A ring with identity would be a ring,R which contains an element $1_R$ to satisfy the following axiom:

$ab=ba$ for every$a\in R$ . [Multiplicative identity]

Theorem 3.5 states that with every element $a$ and$b$ of a ring R,

1. $a·0_R=0_R=0_R·a$. Specifically, $0_R·0_R=0_R$.
2. $a\left(-b\right)=-ab$and$\left(-a\right)b=-ab$ .
3. .$\left(-a\right)a=a$
4. .$-\left(a+b\right)=\left(-a\right)+\left(-b\right)$
5. .$-\left(a-b\right)=-a+b$
6. .$\left(-a\right)\left(-b\right)=ab$

When R has an identity, then,

7 .$\left(-1_R\right)a=-a$

Show that T  is the ring with identity

Exercise 23 considers $R$ is a ring and $a,b\in R$. Consider $m$and $n$ as non-negative integers and prove that:

1. $\left(m+n\right)a=ma+na$
2. $m\left(a+b\right)=ma+mb$
3. $\left(ab\right)m=\left(ma\right)b=a\left(mb\right)$
4. $\left(ma\right)\left(nb\right)=mn\left(ab\right)$

a)

$T=R\times \mathrm{\mathbb{Z}}$with $R$is a ring that has no identity.

Consider that$\left(r,m\right),\left(s,n\right),\left(t,p\right)$ as the arbitrary elements in T.

The fact that and are rings will be mentioned several times.

Using addition and multiplication:

1. $\left(r,m\right)+\left(s,n\right)=\left(r+s,m+n\right)\in T\left(r+s\in R,m+n\in \mathrm{\mathbb{Z}}\right)$

2.

$\begin{array}{rcl}\left(r,m\right)\left(\left(s,n\right)+\left(t,p\right)\right)& =& \left(r,m\right)+\left(s+t,n+p\right)\\ & =& \left(r+s+t,m+n+p\right)\\ \left(\left(r,m\right)+\left(s,n\right)+\left(t,p\right)\right)& =& \left(r+s,m+n\right)+\left(t,p\right)\\ & =& \left(r+s+t,m+n+p\right)\end{array}$

Therefore, $\left(r,m\right)+\left(\left(s,n\right)+\left(t,p\right)\right)=\left(\left(r,m\right)+\left(s,n\right)\right)+\left(t,p\right)$.

3.

$\begin{array}{rcl}\left(r,m\right)+\left(s,n\right)& =& \left(r+s,m+n\right)\\ & =& \left(s+r,n+m\right)\\ & =& \left(s,n\right)+\left(r,m\right)\end{array}$

4.

Consider that $0_T=\left(0_R,0\right)$. Then,

$\begin{array}{rcl}\left(r,m\right)+\left(0_R,0\right)& =& \left(r+0,m+0\right)\\ & =& \left(r,m\right)\\ & =& \left(0_r+r,0+m\right)\\ & =& \left(0,0\right)+\left(r,m\right)\end{array}$

5.

$\begin{array}{rcl}\left(-r,-m\right)+\left(s,n\right)& =& \left(r,m\right)+\left(-r,-m\right)\\ & =& \left(r-r,m-m\right)\\ & =& \left(0_R,0\right)\\ & =& 0_T\end{array}$

6.

$$\begin{gathered} \left(r,m\right)\left(s,n\right)=\left(rs+ms+nr,mn\right)\in T \\ \left(ms,nr\in R\Rightarrow rs+ms+nr\in Randmn\in \mathrm{\mathbb{Z}}\right) \end{gathered}$$

7. $\begin{array}{rcl}\left(r,m\right)\left(\left(s,n\right)\left(t,p\right)\right)& =& \left(r,m\right)\left(st+nt+ps+np\right)\\ & =& \left(r\left(st+nt+ps\right)+m\left(st+nt+ps\right)+\left(np\right)r,mnp\right)\end{array}$

Use exercise 23 to obtain:

$\begin{array}{rcl}\left(r,m\right)\left(\left(s,n\right)\left(t,p\right)\right)& =& \left(r,m\right)\left(st+nt+ps,np\right)\\ & =& \left(r\left(st+nt+ps\right)+m\left(st+nt+ps\right)+\left(np\right)r,mnp\right)\end{array}$

Obtain the$\left(\left(r,m\right)\left(s,n\right)\right)\left(t,p\right)$ as follows:

$\begin{array}{rcl}\left(\left(r,m\right)\left(s,n\right)\right)\left(t,p\right)& =& \left(rs+ms+nr,mn\right)\left(t,p\right)\\ & =& \left(\left(rs+ms+nr\right)t+p\left(rs+ms+nr+\left(mn\right)t,mnp\right)\right)\end{array}$

Use exercise 23 to obtain:

$\begin{array}{rcl}\left(\left(r,m\right)\left(s,n\right)\right)\left(t,p\right)& =& \left(rst+m\left(st\right)+n\left(rt\right)+p\left(rs\right)+\left(pm\right)s+\left(pn\right)r+\left(mn\right)t,mnp\right)\\ & =& \left(rst+n\left(rt\right)+p\left(rs\right)+\left(mn\right)t+\left(mp\right)s+\left(np\right)r,mnp\right)\\ & =& \left(r,m\right)\left(\left(s,n\right)\left(t,p\right)\right)\end{array}$

8.

$\begin{array}{rcl}\left(r,m\right)\left(\left(s,n\right)+\left(t,p\right)\right)& =& \left(r,m\right)\left(s+t,n+p\right)\\ & =& \left(r\left(s+t\right)+\left(n+p\right)r+m\left(s+t\right),m\left(n+p\right)\right)\\ & =& \left(rs+rt+nr+pr+ms+mt,mn+mp\right)\\ & =& \left(rs+ms+nr+rt+mt+pr,mn+mp\right)\\ & =& \left(rs+ms+nr,mn\right)+\left(rt+mt+pr,mp\right)\\ & =& \left(r,m\right)\left(s,n\right)+\left(r,m\right)\left(t,p\right)\end{array}$

$\begin{array}{rcl}\left(\left(r,m\right)+\left(s,n\right)\right)\left(t,p\right)& =& \left(r+s,m+n\right)\left(t,p\right)\\ & =& \left(\left(r+s\right)t+p\left(r+s\right)+\left(m+n\right)t,\left(m+n\right)p\right)\\ & =& \left(rt+st+pr+ps+nt,mp+np\right)\\ & =& \left(rt+pr+mt,mp\right)+\left(st+ps+nt,mp+np\right)\\ & =& \left(r,m\right)\left(t,p\right)+\left(s,n\right)\left(t,p\right)\end{array}$

Consider that $1_T$ represents $\left(0_R,1\right)$. The identity in $T$ would be$1_T$ as claimed. Then,

$\left(r,m\right)\left(0_R,1\right)=\left(r{0}_R+1r+m{0}_R,m·1\right)$

Use Theorem 3.5 to obtain:

$\begin{array}{rcl}\left(r,m\right)\left(0_R,1\right)& =& \left(0_R+r+0_R,m\right)\\ & =& \left(r,m\right)\end{array}$

$\left(0_R,1\right)\left(r,m\right)=\left(0_{R}r+m{0}_R+1r,1·m\right)$

Use Theorem 3.5 to obtain:

$\begin{array}{rcl}\left(0_R,1\right)\left(r,m\right)& =& \left(0_R+0_R+r,m\right)\\ & =& \left(r,m\right)\end{array}$

As a result, $T$ is a ring with identity (from Axioms of the ring).

Thus, it is proved that$T$ is a ring with identity.

Show that R  is a subring of T

b)

Theorem 3.6 considers $S$ as the non-empty subset of a ring $R$ in which,

1. $S$ would be closed under subtraction (When $a,b\in S$, then $a-b\in S$ );
2. $S$ would be closed under multiplication (When $a,b\in S$, then$ab\in S$ ).

Consider that$\left(r,0\right),\left(s,0\right)$ as an arbitrary element in $\overline{R}=\left\{\overline{)\left(r,0\right)}r\in R\right\}\underline{\subset }T$.

To show that $\overline{R}$ is closed under subtraction and multiplication as follows:

$\begin{array}{rcl}\left(r,0\right)-\left(s,0\right)& =& \left(r-s,0-0\right)\\ & =& \left(r-s,0\right)\in \overline{R}\left(r-s\in R\right)\\ \left(r,0\right),\left(s,0\right)& =& \left(rs,0\right)\left(rs\in R\right)\end{array}$

As a result, $\overline{R}$ is a subring of $T$ according to theorem 3.6.

Thus, it is proved that $\overline{R}$ is a subring of $T$.