R-module isomorphism $\frac{R}{x^n R} \to \frac{x^m R}{x^{m+n} R}$

Please help me understand the R-module isomorphism $$\frac{R}{x^n R} \to \frac{x^m R}{x^{m+n} R}$$ where x is not a zero divisor.

I think it might be an instance of (L/N)/(M/N) = L/M when L contains M contains N, but I can't prove it.

Subquestion: What is the obvious map from $R \to x^{m}R/x^{m+n}R$?

$\begin{array}a f : R \to x^m R \\ f(a) = x^m a \end{array}$

$\begin{array}a g : x^m R \to x^m R/ x^{m+n} R \\ g(a) = a + x^{n+m} R \end{array}$

$f$ is surjective because it's domain is basically defined as the image of the function.

$g$ is surjective because it's the inclusion of a set into its quotient.

$g \circ f$ is surjective because it's the composite of surjective maps.

To check $f$ is an $R$-module homomorphism we just see that $f(ra + sb) = r f(a) + s f(b)$.

To check that $g$ is an $R$-module homomorphism first note that $x^{m+n}R = x^n(x^m R)$ is a submodule of $x^{m}R$ so the reduction map $g$ is a homomorphism by algebra.

I found that $r g(a) + s g(b) = ra + sb + rx^{m+n}R + sx^{m+n}R$ but I can't see how this is equal to $g(ra + sb) = ra+sb + x^{m+n}R$ for example if $r = s = 0$ this equality cannot hold.

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Consider the obvious map from $R \to x^{m}R/x^{m+n}R$ and identify the kernel. – Isaac Solomon Oct 20 '12 at 18:36
I think the obvious map should be $r \to x^{m}r/x^{m+n}r$. – Isaac Solomon Oct 20 '12 at 18:47
I don't think that's what you meant. $r\mapsto x^{-n}$ isn't all that useful. Perhaps that's just notation I'm unfamiliar with.... – Cameron Buie Oct 20 '12 at 18:50
I've changed algebra tag to abstract-algebra, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 21 '12 at 7:08

I'll assume $R$ is a commutative ring with unity. Tell me if that's not the appropriate context.

We've got a ready map $R\to x^mR$ given by $a\mapsto x^ma$. Composing this with the projection map $x^mR\to x^mR/x^{m+n}R$ gives us the "obvious map" $R\to x^mR/x^{m+n}R$. It is pretty simple to see that this is surjective, and is an $R$-module homomorphism. It remains only to show that the kernel of this map is $x^nR$, so that the two quotients are isomorphic by First Isomorphism Theorem.

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Thank you very much. I am trying what you suggested but I got stuck showing that the second map is an R-module homomorphism. – sperners lemma Oct 20 '12 at 19:09
@sperners lemma: Repeat the definitions of quotient modules. If $U$ is a submodule of $M$, it comes equipped with a natural homomorphism $M \to M/U$. – Martin Brandenburg Oct 21 '12 at 7:29

We have the obvious homomorphism $\phi\colon R\to x^mR/x^{m+n}R$, $a\mapsto [x^m a]$. If $a=x^n b$ for some $b\in R$, then $a\mapsto [x^mx^nb]=[x^{n+m}b]=0$. Thus $x^n R$ is in the kernel of $\phi$ and $\phi$ factors over the quotient, giving us a homomorphism $R/x^{n}R\to x^mR/x^{m+n}R$.

We can try to give the inverse explicitly: If $a\in x^mR$, then $a=x^mb$ for some $b\in R$. If also $a=x^mb'$ then $x^m(b-b')=0$, hence $b=b'$ beacuse $x$ is not a divisor of zero. This gives us a homomorphism $\psi\colon x^mR\to R/x^nR$, $a\mapsto [b]$ (where compatibility with $+$ and $\cdot$ follows readily). If $a\in x^{m+n}R$, then $b\in x^n R$, hence $\psi$ factors and we obtain a homomorphsim $x^mR/x^{n+m}R\to R/x^nR$.

The two homomorphism are clearly invers of each other, hence isomorphisms.

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Theorem Let $R$ be a commutative ring $x \in R$, we have the isomorphism of $R$-modules $$\frac{R}{x^n R} \simeq \frac{x^m R}{x^{m+n} R}.$$

Proof

First note that $R$ is an $R$-module and $x^n R$ is an $R$-submodule, hence the inclusion map $R \longrightarrow x^n R$ is a surjective $R$-module homomorphism.

Secondly we have the inclusion map into the quotient $x^n R \longrightarrow \frac{x^n R}{x^m (x^n R)}$ which is an $R$-module homomorphism with kernel $x^n R$.

Composing the maps and applying the first isomorphism theorem (which says that $\text{image}(\varphi)\simeq\frac{\text{domain}(\varphi)}{\text{kernel}(\varphi)}$) gives the result.

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In the first paragraph the map is not an inclusion, but multiplication. In the second paragraph it is a projection onto the the quotient with kernel $x^{m+n}R$ – Julian Kuelshammer Oct 25 '12 at 17:07