Kernel of surjective module homomorphism

Question

Let $R$ be viewed as a left module over itself and let $S$ be a simple left $R$-module. Using simplicity of $S$ we can easily show that for $s \in S \setminus \{0 \} $ the module homomorphism $\phi_s :R \rightarrow S$, $ \phi_s (r)=r\cdot s $ is surjective. Then the next part of the proof I’m reading just says that the kernel of this map is a maximal left ideal. Obviously it’s a left ideal as the kernel is a left $R$-submodule of $R$ which is precisely a left ideal, but why is it maximal?

Answer 1

Suppose $\ker \phi_s$ isn't maximal, then exists an ideal $J$ such that $\ker \phi_s \subset J \subset R$. Applying $\phi_s$ we obtain: $$ \phi_s (\ker \phi_s) = \{ 0 \} \hspace{1 cm} \phi_s(J)=\tilde{J} \hspace{1 cm} \phi_s(R) = S, $$ where in the last equiation we use that $\phi_s$ is surjective.

As $S$ is simple, necesary we have that neither $\tilde{J}=\{0\}$, from where $J \subset \ker\phi_s$ and therefore $J = \ker \phi_s$, or $\tilde{J} = S$ from where $J=R$.

This is exactly that $\ker \phi_s$ is a maximal ideal.