Question

Let r and s be positive integers such that r divides ks + 1 for some k with $\mathbf{1}\mathbf{\le }\mathit{k}\mathbf{\le }\mathit{r}$. Prove that the subset role="math" localid="1659372117979" $\{0,r,2r,...\left(s-1\right)r\}$of role="math" localid="1659372076224" ${\mathbf{\mathbb{Z}}}_{\mathbf{r}\mathbf{s}}$is a ring with identity ks = 1 under the usual addition and multiplication in ${\mathbf{\mathbb{Z}}}_{\mathbf{rs}}$. Exercise 21 is a special case of this result.

Short answer

It is proved that subset $\{0,\mathrm{r},2\mathrm{r},...,(\mathrm{s}-1)\mathrm{r}\}$of ${\mathrm{\mathbb{Z}}}_{rs}$ is a ring with identity ks + 1under the usual addition and multiplication in ${\mathrm{\mathbb{Z}}}_{rs}$ .

Step-by-step solution

Prove S={0,r,2r,...(s-1)r} is a subring of ℤrs 

1. Let kr and lr be some arbitrary elements of $S\left(k,l\in \{0,1,...,\mathrm{s}-1\}\right)$
   $\begin{array}{rcl}kr+lr& =& \left(k+1\right)r=\left(ts+r_1\right)r\\ & =& t\left(rs\right)+\left(r,r\right)\left(**\right)\end{array}$
   Where t is some integer and $r_1\in \left(0,1,...,s-1\right)$
   $\Rightarrow r_{1}r\in \left\{0,r,2r,...,\left(s-1\right)r\right\}=S$
   (By theorem 1.1, it is implied that when k + 1 is divided by s, there exist integers t and r₁ that satisfy mentioned conditions )
   Also, by theorem 1.1 and (**) we can conclude that r₁r is the reminder of kr = lr when divided by rs, which implies
   $kr+lr=r_{1}r\Rightarrow kr+lr\in S$
   
2. By following the same principles as in (i), we get:
   $kr=lr=\left(klr\right)r=\left(ts+r_1\right)r=t\left(rs\right)+r_{1}r$
   Where T is some integer and $r_1\in \left\{0,1,...,s-1\right\}$
   $$\begin{gathered} \Rightarrow r_{1}r\in \left\{r,2r,...,\left(s-1\right)r\right\}=Sandkrlr=r_{1}r \\ \Rightarrow krlr\in S \end{gathered}$$
   
3. $0\in S$
   
4. Let kr be some arbitrary element of$S\left(k\in \left\{0,1,...,s-1\right\}\right)$.
   Let l be an integer such that l = s-k
   We see that $l\in \left\{0,1,...,s-1\right\}$, i.e. $lr\in S$
   Also, $kr+lr=\left(k+l\right)r=sr=0$
   Hence for every element kr in S, the equation kr + x=0 has a solution in S
   Hence by $(i,(ii),(iii),(iv)S$is a subring of ${\mathrm{\mathbb{Z}}}_{rs}$
   Ie S is a ring.

Prove that ks + 1 is an identity in S 

First, it is obvious that k + 1, such that $1\le k<r$and $r/ks+1$ is an element of a ring S

Let for some $lr\in S$

$$\begin{gathered} \left(ks+1\right)lr-lr=\left(ks+1-1\right)lr=kslr \\ \Rightarrow rs/(ks+1)lr-lr \\ \Rightarrow (ks+1)lr=lr \end{gathered}$$

Since the multiplication is commutative, $lr\left(ks+1\right)=lr$

Therefore, ks + 1 is an identity in S

Hence it is proved.