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.
 A: 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.
A: 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.
A: 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.
