I am now studying Burnside's Theorem from Lang's Algebra. It states as follow: Let $E$ be a finite-dimensional vector space over an algebraically closed field $k$, and let $R$ be a subalgebra of $End_{k}(E)$. If E is a simple $R$-module, then $R = End_{k}(E)$.
I want to construct an example in which $k$ is not algebraically closed and $E$ is a simple $R$-module when $R \neq End_{k}(E)$.
You may as well start with your favorite non-algebraically closed field like $\mathbb R$.
Let $E=\mathbb R^2$. Then $\mathrm{End}(E_\mathbb R)\cong M_2(\mathbb R)$, and $\mathbb C\cong R=\left\{\left[\begin{smallmatrix}a&b\\-b&a\end{smallmatrix}\right]\mid a,b\in \mathbb R\right\}$ is an $\mathbb R$ subalgebra of the endomorphism ring.
$E$ is simple over $R$ because given the equation
$$ \begin{bmatrix}a&b\\-b&a\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}w\\ z\end{bmatrix} $$
you always have solutions if $[x,y]^\top$ is nonzero:
$$ a=\frac{wx+yz}{x^2+y^2}\\ b=\frac{wy-xz}{x^2+y^2} $$
Therefore $R$ acts transitively on the nonzero elements of $E$, and it is a simple $R$ module.
But $\mathbb C\ncong M_2(\mathbb R)$.