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Given that $R$ is a subring of a commutative ring $S,$ and the additive group $S/R$ is of finite order $n$. If $(m,n)=1,$ I wanted to show that $R/mR$ and $S/mS$ are isomorphic rings.

I know that to show $R/mR≅S/mS,$ we need to write down a well-defined map between the two rings and show it is an isomorphism. My problem is how to define the map! Can anyone guide me? Thanks.

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  • $\begingroup$ What about the composed map $R\subset S\to S/mS$ whose kernel has to be $mR$? (Surjectivity has also to be proved! It reduces to show that any abelian group of order $n$ is $m$-divisible, for $(m,n)=1$.) $\endgroup$ – user26857 Sep 23 '16 at 22:22
  • $\begingroup$ I'd be happy if you could explain further. Thanks $\endgroup$ – Aysha A. Oct 1 '16 at 8:40
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Consider the composition of the inclusion $R\to S$ and the projection $S\to S/mS$: $$f:R\to S\to S/mS,\quad r\mapsto r+mS$$ We have $\ker f=R\cap mS=mR$. We need to prove that $f(R)=S/mS$, so $R/mR\simeq S/mS$ by the first isomorphism theorem.

From $|S/R|=n$ we conclude that $nS\subseteq R$ because the order of any element from $S/R$ divides $n$.

The condition $(m,n)=1$ says that exists $x,y\in \mathbb Z$ such that $mx+ny=1$, so for $s\in S$ we have $$s=(mx+ny)s=m(xs)+n(ys)\in mS+R.$$ Obviously $R+mS\subseteq S$, so $R+mS=S$, and $f(R)=S/mS$.

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