Viewing $m/m^2$ as a vector space I am reading Atiyah MacDonald`s Commutative Algebra textbook. In Page 91, next paragraph, I do not understand part of the following: 

If $A$ is a local ring, $m$ its maximal ideal, $k=A/m$ its residue field, the $A-$module $m/m^2$ is annihilated by $m$ and therefore has the structure of a $k-$vector space. 

I understand that $m/m^2$ is an $A-$module. I do not get how $m/m^2$ is annihilated by $m$. Why does one need this statement (i.e. annihilation by $m$) to conclude that  $m/m^2$ is a $k-$vector space ? I mean $k$ is a field, so  $m/m^2$ is automatically a vector space as it is a $k-$ module. Finally why is $m/m^2$ Special and not $m/m^n$ for $n>2$? 
 A: $\mathfrak{m}/\mathfrak{m}^2$ is annihilated by $\mathfrak{m}$ because if $x,y\in \mathfrak{m}$ then $xy\in \mathfrak{m}^2$, hence $x\cdot(y+\mathfrak{m}^2)=0+\mathfrak{m}^2$.
Since $\mathfrak{m}/\mathfrak{m}^2$ is an $A$-module on which $\mathfrak{m}$ acts trivially, we can make it an $A/\mathfrak{m}$-module by defining the action
$$ \overline{x}\cdot(y+\mathfrak{m}^2)=x\cdot(y+\mathfrak{m}^2)$$
for $\bar{x}\in A/\mathfrak{m}$ and $x\in A$ representing $\bar{x}$.
In order for this action to be well-defined (i.e. be independent of the choice of $x\in A$ representing $\bar{x}\in A/\mathfrak{m}$), it is essential that $\mathfrak{m}$ act trivially on $\mathfrak{m}/\mathfrak{m}^2$. This is why we consider $\mathfrak{m}/\mathfrak{m}^2$ rather than say $\mathfrak{m}/\mathfrak{m}^3$.
A: For an $A$-module $M$ to be a module over $A/I$, we need that multiplication by an element of $A/I$ is actually well defined.  That is, for any $m\in M$, $a\in A$, and $i\in I$, we must have $am=(a+i)m$, or subtracting, that $im=0$.  And conversely, if $IM=0$, then the $A$-module structure of $M$ will pass to an $A/I$-module structure.
Another way of looking at it is that $M/IM$ is a module over $A/I$, but if $IM=0$, you have $M\cong M/IM$.  
As for why $I/I^2$ is annihilated by $I$ (I'm using $I$ since this doesn't depend on using maximal ideals), that just follows from the fact that an element of $I/I^2$ is of the form $i+I^2$, and if you multiply by $j\in I$, you get $ij+I^2$, but by the definition of $I^2$, $ij\in I^2$, and so $ij+I^2=0+I^2$.  
And $I/I^2$ isn't special.  You will have that $I^k/I^{k+1}$ is an $A/I$-module for all $k$.  However, $I/I^{k+1}$ is only a module over $A/I^k$.  
A: $m/m^2$ is anhilated by $m$ i.e for every $a\in m, x\in m/m^2, ax=0$ so the map $A\times m/m^2\rightarrow m/m^2, (a,x)\rightarrow ax$ defines a map $A/m\times m/m^2\rightarrow m/m^2, ([a],x)\rightarrow ax$,  where $[a]$ is the class of $a\in A$ in $A/m=k$.
