Question

Let R be a ring with identity and let $\mathit{S}\mathbf{=}\left\{n{1}_R\overline{)n\in \mathbb{Z}}\right\}$. Prove that S is a subring of R. [The definition of na with $\mathit{n}\mathbf{\in }\mathbf{\mathbb{Z}}\mathbf{,}\mathit{a}\mathbf{\in }\mathit{R}$ is on page 62. Also see Exercise 27.]

Short answer

It is proved that S is a subring of R.

Step-by-step solution

Theorem

According to theorem 3.6, a non-empty subset S of a ring R is closed under subtraction and multiplication. It can be said that, if $a,b\in S$, then role="math" localid="1648126555180" $a-b\in S$, and $ab\in S$, then S is also a sub-ring of R.

Check for closed under subtraction

The given set is S, let’s prove it is closed under subtraction. Let $m{1}_R,n{1}_R\in S$.

$m{1}_R-n{1}_R=(m-n)1_R$

Now, consider when $m,n\ge 0$. Then,

$\begin{array}{rcl}m{1}_R-n{1}_R& =& \left(\sum _{i=1}^m{1}_R\right)-\left(\sum _{j=1}^n{1}_R\right)\\ & =& \left\{\begin{array}{l}\sum _{i=1}^m{1}_R\mathrm{if}m\ge n;\\ \sum _{i=1}^{m-n}{1}_R\mathrm{if}n>m\end{array}\right.\\ & =& (m-n)1_R\in S\end{array}$

If $m<0,n\ge 0$, then

$\begin{array}{rcl}m{1}_R-n{1}_R& =& \left(\sum _{i=1}^{\left|m\right|}-1_R\right)-\left(\sum _{j=1}^n{1}_R\right)\\ & =& \sum _{j=1}^{\left|m\right|+n}-1_R\\ & =& (m-n)1_R\in S\end{array}$

Next, consider the case$m\ge 0$ and $n<0$.

role="math" localid="1648127629605" $\begin{array}{rcl}m{1}_R-n{1}_R& =& \left(\sum _{i=1}^n-1_R\right)-\left(\sum _{i=1}^{\left|m\right|}{1}_R\right)\\ & =& \sum _{i=1}^{m+\left|n\right|}{1}_R\\ & =& (m-n)1_R\in S\end{array}$

If $m,n<0$.

$\begin{array}{rcl}m{1}_R-n{1}_R& =& \left(\sum _{i=1}^{\left|m\right|}-1_R\right)-\left(\sum _{i=1}^n{1}_R\right)\\ & =& \left\{\begin{array}{l}\sum _{i=1}^{\left|m\right|+\left|n\right|}{1}_R,\mathrm{if}\left|m\right|\ge \left|n\right|\\ \sum _{i=1}^{\left|n\right|-\left|m\right|}{1}_R,\mathrm{if}\left|n\right|\ge \left|m\right|\end{array}\right.\\ & =& (m-n)1_R\in S\end{array}$

Check for closed under multiplication

The given set is S, let’s prove it is closed under multiplication. Let $m{1}_R,n{1}_R\in S$.

$\left(m{1}_R\right)\left(n{1}_R\right)=\left(mn\right)1_R$

Now, consider when $m,n\ge 0$. Then,

role="math" localid="1648128466648" $\begin{array}{rcl}\left(m{1}_R\right)\left(n{1}_R\right)& =& \left(\sum _{i=1}^m{1}_R\right)\left(\sum _{j=1}^n{1}_R\right)\\ & =& \sum _{(i,j)\in \left\{1,...,m\right\}\times \left\{1,...,n\right\}}{1}_R\\ & =& \left(mn\right)1_R\in S\end{array}$

If $m\ge 0,n<0$, then

role="math" localid="1648128532311" $\begin{array}{rcl}\left(m{1}_R\right)\left(n{1}_R\right)& =& \left(\sum _{i=1}^m{1}_R\right)\left(\sum _{j=1}^{\left|n\right|}-1_R\right)\\ & =& \sum _{(i,j)\in \left\{1,...,m\right\}\times \left\{1,...,\left|n\right|\right\}}-1_R\\ & =& \left(mn\right)1_R\in S\end{array}$

Next, consider the case role="math" localid="1648128744755" $m<0$and role="math" localid="1648128764896" $n\ge 0$.

role="math" localid="1648128864751" $\begin{array}{rcl}\left(m{1}_R\right)\left(n{1}_R\right)& =& \left(\sum _{i=1}^{\left|m\right|}-1_R\right)\left(\sum _{j=1}^n{1}_R\right)\\ & =& \sum _{(i,j)\in \left\{1,...,\left|m\right|\right\}\times \left\{1,...,n\right\}}-1_R\\ & =& \left(mn\right)1_R\in S\end{array}$

If $m,n<0$.

$\begin{array}{rcl}\left(m{1}_R\right)\left(n{1}_R\right)& =& \left(\sum _{i=1}^{\left|m\right|}-1_R\right)\left(\underset{j=1}{\overset{\left|n\right|}{\sum -}}{1}_R\right)\\ & =& \sum _{(i,j)\in \left\{1,...,\left|m\right|\right\}\times \left\{1,...,\left|n\right|\right\}}-1_R\\ & =& \left(mn\right)1_R\in S\end{array}$

Hence, S is a subringof R.