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I am trying to prove the following statement:

Let R be a commutative ring with a unit element, and S be a subring of R of finite index n. Then $\frac{S}{mS}$ is isomorphic to $\frac{R}{mR}$, where m is coprime to n.

I have no idea how I should use the information that m and s are coprime. Any ideas?

Thanks very much.

awr
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2 Answers2

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Hint: Note that $mS=S\cap mR$. Consider $$S\to R\to\frac{R}{mR},$$ that is, the composition of the inclusion $S\to R$ and the projection $R\to\frac{R}{mR}$. What is the kernel? Is it a surjective homomorphism?

Cameron Buie
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We have two facts: first $mS=S\cap mR$; second

$$ \frac{S}{mS}=\frac{S}{S\cap mR}\cong\frac{S+mR}{mR} $$

so all you need to know is that $S+mR=R$.

You have the additional information that $[R:S]=n$ (index as abelian groups), so in particular $nr\in S$ for all $r\in R$. Now we can write

$$1=nx+my$$

for suitable integers $x$ and $y$; thus, if $r\in R$,

$$r = 1r = (nx+my)r = n(xr)+m(yr),$$

which shows that $r\in S+ mR$.

Why is $nR\subseteq S$? You have the abelian group $(R,+)$ and its subgroup $S$ that has index $n$, meaning that $$ \left|\frac{R}{S}\right|=n. $$ Thus the group $A=R/S$ has $n$ elements, which means that $nx=0$, for any $x\in A$ (I'm using additive notation, of course): this is just the same as saying that $nr\in S$ for all $r\in R$.

egreg
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  • I guess I'm unclear about why the index being n means $nR$ is contained in the subgroup? – awr May 10 '13 at 21:23