Quotient of maximal ideals by its power giving simple modules So I understand that $R/m$ where $m$ is a maximal ideal would give simple module. My question is, would $m/m^2$ also give a simple module?
My progress thus far: My first approach was to realize that inside $R$, and only prime ideal containing $m^2$ is $m$ itself (by thinnking of the radical  of $m^2$). To show that $m/m^2$ is simple means there are no ideals (besides $m^2$ and $m$) which contain $m^2$. Then I reached a dead-end.
My second approach is as follows: $m/m^2$ can be understood as a $R/m$-module, and $R/m$ is a field, so $m/m^2$ is a vector space. So $m/m^2$ being simple is equivalent to saying that its dimension is 1. 
Is this true for a general commutative ring $R$? How about for polynomial rings such as the polynomial ring over the integers $Z[x]$? If the answers are negative, what can be said about the structure of $m/m^2$ in the general case and in $Z[x]$?
 A: It's not true in general. A counterexample is $k[x^2,x^3]$ with $m=(x^2,x^3)$. $(x^3)+m^2$ is a proper submodule of $m/m^2$. Indeed, following your second approach, $m/m^2=\{ax^2+bx^3+m^2\}$, which is dimension $2$ over $k[x^2,x^3]/m\cong k$.
If you know some differential or algebraic geometry, you can think of the above counterexample as saying that the cotangent bundle to the curve $y^2=x^3$ in $k^2$ is not $1$-dimensional at $(0,0)$, because the curve is not smooth at that point. In ring theoretic terms, the problem here is that $k[x^2,x^3]$ is not integrally closed.
In fact, one has the following theorem, useful in number theory and algebraic geometry, which you can use to cook up plenty more counterexamples: Let $R$ be a Noetherian domain of dimension $1$. Then $R$ is integrally closed if and only if for each maximal ideal $m$, $m/m^2$ is 1-dimensional over $R/m$. (A variant of this statement is proved for local rings in Atiyah-MacDonald, from which you can easily recover the statement above.)
A: Just to add to Brett Frankel's answer:
The $R$-action on $m/m^2$ factors through the field $R/m$, and so $m/m^2$ is a simple $R$-module iff it is a simple $R/m$-module iff it is zero or one-dimensional as an $R/m$-vector space.  
E.g. if $R$ is a local ring and $m$ is finitely generated (say because $R$ is Noetherian), then by Nakayama's lemma the dimension of $m/m^2$ is equal to the minimal number of generators of $m$, so $m/m^2$ is one-dimensional iff $m$ is non-zero principal, and is zero-dimensional iff $m = 0.$
