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Let $M$ be a non-trivial irreducible (simple) $R$-module . Let $0 \ne m \in M$ and $A(m_0):=\{x \in R: xm_0=0\}$ , then is $A(m_0)$ a maximal left-ideal of $R$ and as $R$-modules , $M$ and $R/A(m_0)$ are isomorphic ?

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Yes and yes. You switch notation from $m$ to $m_0$, so I'm going to stick with $m\in M$ non-zero. You can deduce the maximality of the left ideal $A(m)$ from the second claim as follows. Define $\varphi:R\to M$ by $\varphi(r)=rm$. This is an $R$-module homomorphism whose image is a submodule of $M$ containing the non-zero element $m$, and hence must be all of $M$ by irreducibility, so $\varphi$ is surjective. Its kernel is $\{r\in R:rm=0\}=A(m)$. So this $\varphi$ induces an isomorphism of $R$-modules $R/A(m)\simeq M$. Because of the submodule structure of $R/A(m)$ (its submodules are in bijection with left ideals containing $A(m)$) and, on the other hand, its irreducibility, due to the isomorphism with $M$, it follows that $A(m)$ must be a maximal left ideal.

I'm assuming you are okay with the fact that $A(m)$ is a left ideal, and were primarily interested in why it was maximal.

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