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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