If every cyclic $R$-module is projective, then $R$ is semisimple.
I know how to prove this if the definition of cyclic module is $M=Rm$ where $m$ is an element of $M$. The problem is, we defined in class that a cyclic module as $ R/aR$ for some $a\in R$. The definitions are equivalent when $R$ is a $PID$. But why is the proposition true if $R$ is arbitrary?
It isn't true that if every module of the form $R/aR$ is projective, then $R$ is semisimple. For instance, let $R=\mathbb{F}_2^X$ for some infinite set $X$. Then for any $a\in R$, the ideal $aR$ is a direct summand of $R$, with complement $(1-a)R$ (this follows from the fact that $a^2=a$). Thus $R/aR\cong (1-a)R$ is projective. But not every ideal in $R$ is a direct summand: for instance, the ideal $I\subseteq R$ of all elements which have only finitely many nonzero coordinates is not even finitely generated, so it cannot be a direct summand.