Proposition 1

Let A be a ring.

a) Let $\mathfrak{m}$ be a left ideal of A. The A-module $A_s/\mathfrak{m}$ is simple if and only if $\mathfrak{m}$ is a maximal left ideal.

b) Let M be a simple A-module, and let $x$ be a nonzero element of M. We have the equality $M = Ax$, the annihilator $\mathfrak{m}$ of $x$ is a maximal left ideal of A, and the mapping $a\mapsto ax$ defines, by passing to the quotient, an isomorphism from $A_s/\mathfrak{m}$ to M.

c) Let M be a nonzero A-module. If we have $M = Ax$ for every nonzero element $x$ of M, then M is simple.

The submodules of $A_s/\mathfrak{m}$ are of the form $\mathfrak{n}/\mathfrak{m}$, where $\mathfrak{n}$ is a left ideal of A containing $\mathfrak{m}$ (I, §4, No. 6, p. 41, Theorem 4); consequently, the A-module $A_s/\mathfrak{m}$ is simple if and only if we have $\mathfrak{m}\not= A$ and every left ideal $\mathfrak{n}$ of A containing $\mathfrak{m}$ satisfies $\mathfrak{n}/\mathfrak{m}= 0$ or $\mathfrak{n}/\mathfrak{m}= A_s/\mathfrak{m}$, that is, $\mathfrak{n}=\mathfrak{m}$ or $\mathfrak{n}= A$. This proves a).

Under the assumptions of b), $Ax$ is a nonzero submodule of M, hence is equal to M. Consequently, the mapping $a\mapsto ax$ defines, by passing to the quotient, an isomorphism from $A_s/\mathfrak{m}$ to M; the A-module $A_s/\mathfrak{m}$ is therefore simple, and the left ideal $\mathfrak{m}$ is maximal by a). This proves b).

Under the assumptions of c), let N be a nonzero submodule of M. If $x$ is a nonzero element of N, then we have $Ax\subset N$ and $Ax= M$, and therefore M = N. Hence M is simple.