I'm solving a problem and think about it a lot but it doesn't help. How can I solve this problem? The problem is :
Let $r$ and $s$ be positive integers such that $r|ks+1$ for some $k$ with $1 \leq k <r$. Prove that $S:=\{[0],[r],[2r],\cdots ,[(s-1)r]\}$ is a subring of $\mathbb{Z}_{rs}$ and $[ks+1]$ is the multiplicative identity in $S$
To prove that $S$ is a subring of $\mathbb{Z}_{rs}$, I choose $[ur],[vr] \in S$ where $0 \leq u,v \leq (s-1)$. And I'm going to use a theorem that S is a subring of $\mathbb{Z}_{rs}$ if S is closed under subtraction and multiplication. Since $0 \leq |u-v| \leq (s-1)$, in my intuition, $[ur]-[vr]=[(u-v)r] \in S$. And it is clear when $u-v \geq 0$. But I have to prove it holds when $u-v < 0$ And if I prove this I think it would help me to prove $[ks+1]$ is the multiplicative identity in $S$.
Please help me!
First let's show that $S$ is closed under - and $\cdot$. Choose $[ur], [vr] \in S$ where $0 \leq u,v \leq s-1$. So $0 \leq |u-v| \leq s-1$. By division algorithms there are unique $q,R \in \mathbb{Z}$ s.t $(u-v)r=rsq+R (0 \leq R < rs)$. If $u-v \geq 0$ then $(u-v)r <rs$. So $q=0$ so that $R=(u-v)r$. Thus $[(u-v)r] \in S$. If $u-v <0$ then $q=-1$, since $-(s-1) \leq u-v <0$. So $R = (u-v+s)r$, since $-(s-1) + s \leq u-v+s < s$. Thus $[(u-v)r] \in S$. Therefore $S$ is closed under subtraction. Moreover by division algorithm there are unique $Q', \rho$ s.t $urvr=rsQ+\rho$ where $0 \leq \rho < rs$. By applying division algorithm again there are unique $Q', \rho'$ s.t $urv = SQ' +\rho'$ where $0 \leq \rho' <s$. Thus $urvr =(SQ' + \rho')r = sQ'r + \rho' r$. Note that $0 \leq \rho' <s$ then $0 \leq \rho' r <rs$. Thus $\rho' r$ is the remainder of $uurvr$ divided by $rs$. Therefore $[urvr] = [\rho' r] \in S$. So $S$ is closed under multiplication. Hence $S$ is the subring of $\mathbb{Z}_{rs}$. Furthermore, $[ur][ks+1]=[ur(ks+1)]=[urks+ur]=[urks]+[ur]=[0]+[ur]=[ur]$. Thus $[ks+1]$ is the multiplicative identity of S. $\square$