Let $k$ be a field. I have to show that $\text{End}_k(k^2)$ is simple.
First of all, I don't see why this is true. For example, if $k=\mathbb{C}(x_1,x_2,\dots)$ then $\varphi\colon k^2\rightarrow k^2$ defined by $(f(x_1,x_2,\dots),g(x_1,x_2,\dots))\mapsto (f(x_2,x_3,\dots),g(x_2,x_3,\dots))$ is a homomorphism, but it is not surjective, so $\varphi\notin \text{End}_k(k^2)^*$, the $*$ denotes all the invertible elements. So the ideal generated by $\varphi$ is not trivial or the whole ring.
I don't know why this statement should be true now. Perhaps my counterexample is flawed, or something else. Could you help me solve this? Thanks.