# N is a simple nonzero right A-module (Equivalent Conditions)

Let $N$ be a nonzero right $A$-module. Then the following are equivalent:

(i) $N$ is simple.

(ii) $uA=N$ for all $u\in N\setminus\{0\}$

(iii) $N\cong A/M$ for some maximal right ideal $M$ of $A$.

Proof: (i)=>(ii) Let $u\in N\setminus \{0\}$. $uA$ is a submodule of $N$. (Let $ua\in uA$ and $a'\in A$, then $ua\cdot a'\in uA$, $ua_1+ua_2=u(a_1+a_2)\in uA$). Since $N$ is simple, $uA\neq 0$ implies $uA=N$.

(ii)=>(i) Condition (ii) implies that $N$ is the only nonzero submodule of $N$, thus $N$ is simple.

(ii)=>(iii) Let $\psi:A\to N=uA$, $\psi(a)=ua$. $\psi$ is an $A$-linear map that is surjective, thus $A/\ker\psi\cong N$. $M:=\ker\psi$ is a right ideal of $A$. Since (ii) implies (i), $N$ is a simple module. Thus by Correspondence Theorem, $M$ is a maximal right ideal.

(iii)=>(i) Follows from the Correspondence Theorem: The map $S\mapsto S/M$ is a bijection from the set of submodules of $A$ containing $M$ and the submodules of $A/M$. Thus if $M$ is maximal, the only submodules containing $M$ are $M$ and $A$, thus the only submodules of $N\cong A/M$ are $M/M\cong 0$ and $A/M\cong N$, i.e. $N$ is simple. ## Author: mathtuition88

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