A module annihilated by a maximal ideal is semisimple

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I want to show that if $M$ is a module over a commutative ring $R$ that is annihilated by a maximal ideal $I$ of $R$, then $M$ is a semisimple $R$-module. What I have in mind is the following: if $M$ is annihilated by $I$, then $M$ is a module over $R/I$, which is a field. Since every module over a field is semisimple, then $M$ must be semisimple as an $R$-module.

Is this argument OK? Is it also true for noncommutative rings?

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The argument is what I would have used: it’s sound.

Is it also true for noncommutative rings?

No... take any simple non Artinian ring. Then every nonzero module has annihilator $\{0\}$, which is maximal, and obviously some of the modules are not semisimple.

One cannot get anywhere replacing “maximal ideal” with “maximal right ideal”, because an annihilator if a module is always two-sided, so any such maximal right ideal would also be a maximal two-sided ideal.