$V=\{f:\mathbb{Z}/4\mathbb{Z}\rightarrow\mathbb{R}\}$ decompose into irreducible $\mathbb{R}[x]$-modules So we define a module structure on $V$ by $xf(r)=f(r+1)$. Now if we fix a basis we can define the operator $x$ as the matrix
$$\left[\begin{array}{cccc}
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array}\right]$$
Now from what I've seen the module structure on $V$ is isomorphic to $\mathbb{R}[x]/(x^4)$. Is this incorrect? If it isn't  then it doesn't have a decomposition into irreducible modules correct?
I'm sorry my algebra is a bit rusty. For these type of problems is there a way to get the module structure in a routine manner. Namely, if I start with a vector space and prescribe it a polynomial structure $F[x]$, how do I find the $F[x]$-module structure.
SOLUTION: Computing the characteristic polynomial of our matrix we get $p(x)=x^4-1$. By Cayley Hamilton theorem evaluating our matrix by this polyonial get us zero. Also the dimension of $\mathbb{R}[x]/(p(x))$ is a four dimensional real vector space. Thus, $$V\simeq \mathbb{R}[x]/(x^4-1)\simeq \mathbb{R}[x]/(x-1)\oplus \mathbb{R}[x]/(x+1)\oplus\mathbb{R}[x]/(x^2+1)$$
 A: This answer is adding some more details to the solution given in the comments, and also shows how one can calculate an explicit decomposition of $V$ into irreducible $\mathbb{R}[x]$-submodules.
This also gives an alternative solution to the question.

With respect to the $\mathbb{R}$-basis $\mathcal{B} = (f_1, f_2, f_3, f_4)$ of $V$ given by $f_i([n]) = \delta_{[n],[4-i]}$ the action of $x \in \mathbb{R}[x]$ is given by the matrix
$$
     A
  := \begin{bmatrix}
       0 & 0 & 0 & 1 \\
       1 & 0 & 0 & 0 \\
       0 & 1 & 0 & 0 \\
       0 & 0 & 1 & 0
     \end{bmatrix}
     \in \operatorname{M}_4(\mathbb{R}).
$$
For convenience we will therefore replace the given $\mathbb{R}[x]$-module by the real vector space $V = \mathbb{R}^4$, endowed with the $\mathbb{R}[x]$-module structure given by $x{.}v = Av$ for all $v \in \mathbb{R}^4$.
There are now two possible ways to proceed:

Since the action of $x$ is given by cyclic rotation of the standard basis vectors $e_i$, it follows that $V$ is cyclic as an $\mathbb{R}[x]$-module (every basis vector $e_i$ generates it as an $\mathbb{R}[x]$-module).
It therefore follows for the minimal polynomial $m(x) \in \mathbb{R}[x]$ of $A$ that $V \cong \mathbb{R}[x]/(m(x))$ as $\mathbb{R}[x]$-modules.
Since $A^4$ is the identity matrix we find that $A$ satisfies the polynomial $x^4 - 1$, and that $m(x)$ therefore divides $x^4 - 1$.
Since $x^4 - 1$ is monic and
$$
    \deg m(x)
  = \dim \mathbb{R}[x]/(m(x))
  = \dim V
  = 4
  = \deg (x^4 - 1)
$$
it follows that $m(x)$ and $x^4 - 1$ coincide, so that $m(x) = x^4 - 1$.
Since
$$
    m(x)
  = x^4 - 1
  = (x^2 + 1)(x^2 - 1)
  = (x^2 + 1)(x + 1)(x - 1)
$$
is a decomposition of $m(x)$ into irreducible (!) polynomials it follows that
$$
        V
  \cong \mathbb{R}[x]/( x^4 - 1 )
  \cong        \mathbb{R}[x]/(x^2 + 1)
        \oplus \mathbb{R}[x]/(x + 1)
        \oplus \mathbb{R}[x]/(x - 1).
$$
is a decomposition into irreducible $\mathbb{R}[x]$-modules.

The above approach tells us how the decomposition of $V$ into irreducible $\mathbb{R}[x]$-modules looks like up to isomorphism.
But we can also calculate the actual decomposition of $V$ into irreducible submodules:
The characteristic polynomial $p(x)$ of $A$ is given by $p(x) = m(x) = x^4 - 1$ since $p(x)$ is a monic multiple of $m(x)$ (by Cayley-Hamilton) with $\deg p(x) = 4 = \deg m(x)$.
Since the complex roots of $p(x)$, namely $1$, $-1$, $i$ and $-i$, are pairwise distinct, it follows that $A$ is complex diagonalizable with these four eigenvalues.
A basis $\mathcal{C}' = (v_1, v_2, v_3, v_4)$ of $\mathbb{C}^4$ consisting of corresponding eigenvectors is given by
$$
     v'_1
  := \begin{bmatrix}
       1 \\ 1 \\ 1 \\ 1
     \end{bmatrix},
     v'_2
  := \begin{bmatrix}
       \phantom{-}1 \\ -1 \\ \phantom{-}1 \\ -1
     \end{bmatrix},
     v'_3
  := \begin{bmatrix}
       \phantom{-}1 \\ \phantom{-}i \\ -1 \\ -i
     \end{bmatrix},
     v'_4
  := \begin{bmatrix}
       \phantom{-}1 \\ -i \\ -1 \\ \phantom{-}i
     \end{bmatrix}.
$$
(Note that the eigenvectors $v'_3$ and $v'_4$ are conjugated to each other. It is possible to choose them this way because the eigenvalues $i$ and $-i$ are conjugated to each other and the matrix $A$ is real.)
The basis $\mathcal{C}'$ gives a decomposition of $\mathbb{C}^4$ into irreducible, one-dimensional $\mathbb{C}[x]$-submodules $\langle v'_i \rangle$, where $x \in \mathbb{C}[x]$ acts on $\mathbb{C}^4$ by $x{.}v' = Av$ for all $v' \in \mathbb{C}^4$.
Since we are interested in the real case, we set $v_1 := v'_1$ and $v_2 := v'_2$, and replace $v'_3$ and $v'_4$ by the two real vector
$$
     v_3
  := \operatorname{Re} v'_3
   = \frac{v'_3 + v'_4}{2}
   = \begin{bmatrix}
       \phantom{-}1 \\ \phantom{-}0 \\ -1 \\ \phantom{-}0
     \end{bmatrix}
   \quad\text{and}\quad
     v_4
  := \operatorname{Im} v'_3
   = \frac{v'_3 - v'_4}{2i}
   = \begin{bmatrix}
       \phantom{-}0 \\ \phantom{-}1 \\ \phantom{-}0 \\ -1
     \end{bmatrix}
$$
Then $\mathcal{C} := (v_1, v_2, v_3, v_4)$ is an $\mathbb{R}$-basis of $\mathbb{R}^4 = V$ such that $U_1 := \langle v_1 \rangle$, $U_2 := \langle v_2 \rangle$ and $U_3 := \langle v_3, v_4 \rangle$ are $\mathbb{R}[x]$-submodules of $V$ with $V = U_1 \oplus U_2 \oplus U_3$.
With respect to the basis $\mathcal{C}$ the action of $x \in \mathbb{R}[x]$ on $V$ is given by the following matrix (which is in real Jordan normal form):
$$
    B
  = \begin{bmatrix}
       1 &    &   &              \\
         & -1 &   &              \\
         &    & 0 &          - 1 \\
         &    & 1 & \phantom{-}0
     \end{bmatrix}
  \in \operatorname{M}_4(\mathbb{R}).
$$
That the submodules $U_1, U_2, U_3$ are irreducible can be seen in at least two ways:


*

*Similar to $V \cong \mathbb{R}[x]/(x^4 - 1)$ we find that $U_1 \cong \mathbb{R}[x]/(x-1)$, $U_2 \cong \mathbb{R}[x]/(x+1)$ and $U_3 \cong \mathbb{R}[x]/(x^2 + 1)$.

*Both $U_1$ and $U_2$ are irreducible since they are one-dimensional. That $U_3$ is irreducible can also be seen in at least two ways:


*

*Every non-trivial $\mathbb{R}[x]$-submodule of $U_3$ must be one-dimensional (since $U_3$ is $2$-dimensional), which would result in a non-existing real eigenvector for the matrix
$$
  \begin{bmatrix}
    0 &          - 1 \\
    1 & \phantom{-}0
  \end{bmatrix}
  \in \operatorname{M}_2(\mathbb{R}).
$$

*By identifying $U_3$ with the plane $\mathbb{R}^2$ and the action of $x$ on $U_3$ with the rotation by $90^\circ$, we find that for every non-zero vector $v \in U_3$ both $v$ and $x{.}v$ are linerly independent and do therefore span $U_3$.
So every non-zero $v \in U_3$ generates $U_3$ as an $\mathbb{R}[x]$-module.


