If every projective $A$ module is free, then the only idempotents are $0$ and $1$

Question

I have to prove that if every projective $A$ module is free, then the only idempotents of $A$ are $0$ and $1$.

I know that $A$ contains an idempotent implies $A \cong G \times H$, where neither $G$ nor $H$ are zero, but i can’t do anything after that.

Answer 1

Let $A\cong G\times H$ as you said. Then both $G,H$ are projective $A$ modules, since they are summands. By hypothesis, both are free of rank greater or equal to $1$.

But now you have $A\cong A^k$ where $k\geq 2$, and that is not possible for commutative ring since they have the invariant basis number (IBN) property.

Thus the only idempotents can be trivial.

Answer 2

I am assuming that $R$ is a commutative ring such that every projective $R$-module is free. Then every idempotent $e\in R$ yields a splitting $R=eR\oplus (1-e)R$, which (by hypothesis) implies that $eR$ is free. By the invariant basis number property, it follows that the only free $R$-submodules of $R$ are $0$ and $R$ itself, yielding that $e=0$ or $e=1$ respectively.