Semi-simple $\mathbb R[x]$-module structure on $\mathbb R^2$ This might just be a trivial question:

Let $R=\mathbb{R}[x]$ and define an $R$-module structure on $\mathbb{R}^2$ with $x$ acting by $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Is the module semi-simple?

So my thought is: since the matrix admits no eigenvalue in $\mathbb{R}$, we cannot find any non-trivial submodule, thus this is a simple module, thus semi-simple. Is this a correct understanding? What's the correct/better way to understand this?
 A: Let $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. The characteristic polynomial of $A$ is $x^2+1$, and thus the module is isomorphic to $\mathbb R[x]/(x^2+1)$ (since $x^2+1$ is irreducible over $\mathbb R$), hence it is simple.
A: The thought that the matrix not admitting an eigenvalue implies that the module is semi-simple is not correct if the $\mathbb{R}$-dimension of your module is greater than $2$. For example let $x$ operate on $\mathbb{R}^4$ by $$f_x = \begin{pmatrix} 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}.$$
The minimal polynomial of this matrix is its characteristic polynomial which is $(X^2+1)^2$. Hence $\mathbb{R}^4$ with the induced $\mathbb{R}[X]$-structure is isomorphic to the $\mathbb{R}[X]$-module $\mathbb{R}[X]/(X^2+1)^2$ which is indecomposable. However, $\mathbb{R}^2\times \{0_{\mathbb{R}^2}\}$ is a $\mathbb{R}[X]$-submodule, so $\mathbb{R}^4$ with this $\mathbb{R}[X]$-structure is not simple and hence not semi-simple, because indecomposable. The matrix also admits no eigenvalues in $\mathbb{R}$.
For the proof that your given module is semi-simple, see the other answer.
