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Let $M$ be a semisimple module over a ring $R$ such that for each nonzero $x\in M$ we have the annihilator $ann_R(x)=0$. Then for any nonzero $r\in R$ we have $rM=M$, because $M$ is a direct sum of simple modules $M_i$ ($i\in I$), and for any $i\in I$ we have $rM_i=M_i$, because the submodule $rM_i$ of $M_i$ would be nonzero by the hypothesis. Hence, $rM=\bigoplus_{i\in I} rM_i=\bigoplus M_i=M$.

Is the converse of the above also true, i.e., if $rM=M$ for any nonzero $r\in R$, and if $ann_R(x)=0$ for every $x\in M$, then could we deduce that $M$ is semisimple? Thanks for any help!

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    Faithfulness does not mean that all annihilators are zero: is a weaker condition, because it means $$\bigcap_{x \in M} \mathrm{ann}_R (x) = 0$$ – 2017-02-02

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The converse is false: a counterexample is $R= \Bbb{Z}$ and $M= \Bbb Q$.