Reducibility of an $\mathbb{R}[X]$-module Let $q \in \mathbb{R}[X]$ be an irreducible polynomial of degree 2 and assume $n > 1$ is an intiger. My question is whether the $\mathbb{R}[X]$-module $\mathbb{R}[X] / (q^n)$ is reducible (i.e has a proper $\mathbb{R}[X]$-submodule).
If $q = X^2 + 1$ and $n = 2$ then the submodule generated by $X^2 + 1$ has dimension 2, hence is a proper submodule but I can't seem to crack the general case.
EDIT: I should probably also say that this is clearly equivalent to the following fact: If $T : \mathbb{R}^n \to \mathbb{R}^n$ has no eigenvalues then there is a $T$-invariant plane in $\mathbb{R}^n$. I am more interested in an "algebraic" reason for why this is true
 A: Let $A$ be a commutative ring. The sub-$A$-modules of $A$ are precisely the ideals of $A$. In particular, $A$ is irreducible iff it is a field. If $I$ is an ideal of $A$, the submodules of $A/I$ correspond to the ideals of $A$ which contain $I$. So $A/I$ is irreducible iff $I$ is maximal. 
Assume $A$ is principal ideal domain, $a$ is an irreducible element, and $n$ is a positive integer. Then the sub-$A$-modules of $A/(a^n)$ correspond to the ideals $(a^i)$, $0\le i\le n$. In particular, $A/(a^n)$ is irreducible iff $n=1$. 
EDIT 1. For the question in your Edit: $\mathbb R^n$ is an $\mathbb R[T]$-module. $\mathbb R[T]$ is isomorphic to $\mathbb R[X]/(f)$, where $(f)$ is the minimal polynomial of $T$. If $f=f_1^{m(1)}\cdots f_k^{m(k)}$ is the factorization of $f$ (the $f_i$ being the distinct irreducible divisor of $f$), then, by the Chinese Remainder Theorem, $\mathbb R[X]/(f)$ is isomorphic to the product of the $\mathbb R[X]/(f_i^{m(i)})$. This answer the question. [I can give you more details if you want.] 
EDIT 2. Here is one of the things that were implicit in Edit 1. If your ring $A$ is a product $B\times C$ of the rings $B$ and $C$, then an $A$-module $M$ is given in a "unique" way as a product $N\times P$ of a $B$-module $N$ by a $C$-module $P$. Given $M$, you define $N$ and by $N:=(1,0)M$, $P:=(0,1)M$. [Here "ring" means "commutative ring".]
