The following question was asked in my assignment of modules and I could not solve this question despite thinking a lot.
Question: Show that $A/I$ is a projective $A$-module if and only if $I$ is a principal ideal generated by an idempotent element.
Attempt: $\implies$ To show that principal ideal is generated by an idempotent element.
I thought of using the result (I didn’t know which other result can be used): Given $A/I$ is a projective $A$-module. So, we have $F = K \oplus A/I$ such that $F$ is a free $A$-module and $K$ is also an $A$-module.
$\impliedby$ Let $I = (b)$ such that $b^2 = b$. But even here I am not able to prove that $A/I$ is projective (I tried by proving using only the definition).
Can you please shed light?
Note that we have the short exact sequence $0 \to I \to A \to A/I \to 0$, and $A$ is a free module. Thus, $A/I$ is projective iff this sequence splits. But this sequence splits iff $I$ is a direct summand of $A$. Can you finish this from here?