I feel like one of the hypotheses in the problems I am working on is not necessary and I will explain why.
Let $R$ be a commutative ring with no divisors of zero. Show every nonzero element of a projective $R$ module is free.
Is the conclusion still true is we do not assume the module is projective?
My reasoning is as follows. Let $M$ be a projective $R$-module and let $x\in M$ be a nonzero element. To show $x$ is free it to show the set $\{x\}$ is free as a subset so it suffices to check the cyclic module $\langle x\rangle = Rx$ is free of rank 1. But this amounts to show there is an isomorphism $Rx \cong R$. This is easy because the map is trivially surjective and injective by the fact that there are no zero divisors... Where did we need projective here?
The only thing I see projectivity giving is the fact that $M$ is a direct factor of the direct sum $\oplus_{t\in T}R$ so we can get a representation for each $x = (r_t)_{t \in T}$ but I don't see how this representation is needed.