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.