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If ''I'' is a right ideal of ''R'', then ''I'' is simple as a right module if and only if ''I'' is a minimal non-zero right ideal: If ''M'' is a non-zero proper submodule of ''I'', then it is also a right ideal, so ''I'' is not minimal. Conversely, if ''I'' is not minimal, then there is a non-zero right ideal ''J'' properly contained in ''I''. ''J'' is a right submodule of ''I'', so ''I'' is not simple.
If ''I'' is a right ideal of ''R'', then ''R''/''I'' is simple if and only if ''I'' is a maximal right ideal: If ''M'' is a non-zero proper submodule of ''R''/''I'', then the preimage of ''M'' under the quotient map {{nowrap|''R'' → ''R''/''I''}} is a right ideal which is not equal to ''R'' and which properly contains ''I''. Therefore ''I'' is not maximal. Conversely, if ''I'' is not maximal, then there is a right ideal ''J'' properly containing ''I''. The quotient map {{nowrap|''R''/''I'' → ''R''/''J''}} has a non-zero kernel which is not equal to {{nowrap|''R''/''I''}}, and therefore {{nowrap|''R''/''I''}} is not simple.
Every simple ''R''-module is isomorphic to a quotient ''R''/''m'' where ''m'' is a [[maximal ideal|maximal right ideal]] of ''R''.<ref>Herstein, ''Non-commutative Ring Theory'', Lemma 1.1.3</ref> By the above paragraph, any quotient ''R''/''m'' is a simple module. Conversely, suppose that ''M'' is a simple ''R''-module. Then, for any non-zero element ''x'' of ''M'', the cyclic submodule ''xR'' must equal ''M''. Fix such an ''x''. The statement that {{nowrap begin}}''xR'' = ''M''{{nowrap end}} is equivalent to the surjectivity of the homomorphism {{nowrap|''R'' → ''M''}} that sends ''r'' to ''xr''. The kernel of this homomorphism is a right ideal ''I'' of ''R'', and a standard theorem states that ''M'' is isomorphic to ''R''/''I''. By the above paragraph, we find that ''I'' is a maximal right ideal. Therefore ''M'' is isomorphic to a quotient of ''R'' by a maximal right ideal.
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