Question

Let R be a ring and m a fixed integer. Let $\mathit{S}\mathbf{=}\left\{r\in R|mr=0_R\right\}$· Prove that S is a subring of R.

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, $\mathit{S}$of a ring, $\mathit{R}$is closed under subtraction and multiplication. It can be said that if role="math" localid="1648196546946" $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{S}$, then $\mathit{a}\mathbf{-}\mathit{b}\mathbf{\in }\mathit{S}$and$\mathit{a}\mathit{b}\mathbf{\in }\mathit{S}$, then $\mathit{S}$is also a subring of $\mathit{R}$.

Check for closed under subtraction

Assume $r,s\in S$, such that$mr=0_R$ and $ms=0_R$.

Now, check for the closed property under subtraction.

role="math" localid="1648198120528" $\begin{array}{rcl}m(r-s)& =& \left\{\begin{array}{l}\sum _{i=1}^{m}r-s,m>0\\ 0_R,m=0\\ \sum _{i=1}^{\left|m\right|}-r+s,m<0\end{array}\right.\\ & =& mr-ms\\ & =& 0_R-0_R\\ & =& 0_R\end{array}$

So, the closed property of subtraction is satisfied.

Check for closed under multiplication

Similarly, it is closed under multiplication as follows:

$\begin{array}{rcl}m\left(rs\right)& =& \left\{\begin{array}{l}\sum _{i=1}^{m}rs,m>0\\ 0_R,m=0\\ \sum _{i=1}^{\left|m\right|}-rs,m<0\end{array}\right.\\ & =& \left(mr\right)s\\ & =& 0_{R}s\\ & =& 0_R\end{array}$

Hence, S is a subring of R.