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 $n>2$?

$\mathfrak{m}/\mathfrak{m}^2$ 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 $\mathfrak{m}/\mathfrak{m}^3$.