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.

No G and H are non trivial rings. It implies that Spec A is disconnected. — Harun rashid, Sep 15 '19 at 15:19

score 2 · Accepted Answer · answered Sep 15 '19 at 18:59

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.

score 1 · Answer 2 · answered Sep 15 '19 at 21:13

1

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.

answered Sep 15 '19 at 21:13

pre-kidney

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Stale browser/app view? I posted this same solution two hours earlier. – rschwieb Sep 15 '19 at 23:29
I dont see how invariant basis number property implies the only free r submodules being r and 0? – Harun rashid Sep 16 '19 at 10:35

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

2 Answers2