Let $\mathfrak{m}$ be maximal in $R$. Show that $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is an $R/\mathfrak{m}$-vector space for all $n\geq 0$ Let $R$ be a commutative ring and $\mathfrak{m}$ a maximal ideal of $R$.  Show that $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is an $R/\mathfrak{m}$-vector space for all $n\geq 0$
I mostly just want to make sure I didn't miss something in my proof:
So since $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is a ring, the only property left to prove is that $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is closed under scalar multiplication.  I define scalar multiplication to be the product of elements in $R$ and then at the end I mod out by $\mathfrak{m}^{n+1}$.
So an arbitrary element of $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ looks like: $$\displaystyle\sum_{k=1}^ta_k+\mathfrak{m}^{n+1},$$ where $a_k=c_k\displaystyle\prod_{i=1}^nm_i$ for $c_k,m_i\in\mathfrak{m}$.
And an arbitrary element of $R/\mathfrak{m}$ looks like:
$$r+\mathfrak{m}.$$
Thus we take their product to obtain:
$$(r+\mathfrak{m})(\displaystyle\sum_{k=1}^ta_k+\mathfrak{m}^{n+1})=r\sum_{k=1}^ta_k+r\mathfrak{m}^{n+1}+\mathfrak{m}\sum_{k=1}^ta_k+\mathfrak{m}\cdot\mathfrak{m}^{n+1}.$$
The second term is clearly in $\mathfrak{m}^{n+1}$.  The fourth is also since $\mathfrak{m}^{n+2}\subseteq\mathfrak{m}^{n+1}$. And the third term as well since $\mathfrak{m}$ does not contain units.  Thus this sum becomes $r\sum_{k=1}^ta_k+\mathfrak{m}^{n+1}$, which is what I wanted.  Is this good, am I done?  Thanks.
 A: Generalize it to make the problem easier. If $R$ is a ring, $I$ is an ideal of $R$, and $M$ is an $R$-module, then the $R$-module $M/IM$ becomes an $R/I$-module. There are at least three possible ways to see this (increasing in abstraction):


*

*Just verify directly that $\overline{r} \cdot \overline{m} := \overline{rm}$ defines a module structure. In particular one has to prove that this is well-defined.

*$R \to \mathrm{End}(M) \to \mathrm{End}(M/IM)$ kills $I$, hence factors through the quotient $R/I$.

*$M/IM = M \otimes_R R/I$.


In your case $I=\mathfrak{m}$ and $M=\mathfrak{m}^n$. Here is another interesting example (as requested by  Math Gems): When $R=\mathbb{Z}$ and $p$ is a prime number, then for every abelian group $A$ we get the $\mathbb{F}_p$-vector space $A/pA$.
A: Clearly $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is an $R$-module, to show that it is also an $R/\mathfrak{m}$-module (i.e. a $R/\mathfrak{m}$-vector space), you must show that, as you said, scalar multiplication is well-defined. So if $\bar{a},\bar{b}$ are two elements of $R/\mathfrak{m}$, you need to show that $\bar{a} \cdot m = \bar{b} \cdot m$ in $\mathfrak{m}^n/\mathfrak{m}^{n+1}$. This is equivalent to show that $(a-b) \cdot m = 0 \in \mathfrak{m}^n/\mathfrak{m}^{n+1}$.
This is equivalent to showing that multiplication by something in $\mathfrak{m}$ is zero in $\mathfrak{m}^n/\mathfrak{m}^{n+1}$. But this is easy... (I leave it tou you to convine yourself of that...)
