Multiplication map of an $R$-algebra $A$ is projective module iff diagonal is open?

Question

Let $R$ be a commutative ring and let $R\to A$ be a commutative $R$-algebra. Consider the canonical homomorphism $A\otimes _R A\to A$ and its opposite, the diagonal $\operatorname{Spec}A\to \operatorname{Spec}A\times_{\operatorname{Spec}R}\operatorname{Spec}A$.

Are the following equivalent?

1. $A\otimes _RA\to A$ is projective as an $A\otimes _RA$-module;
2. The diagonal $\operatorname{Spec}A\to \operatorname{Spec}A\times_{\operatorname{Spec}R}\operatorname{Spec}A$ is (Zariski) open.

Are they equivalent under some finiteness hypothesis? How to show this?

Answer 1

I am assuming that in 1. you meant $A$ is a projective $A\otimes_R A$ module. More generally, let $\pi:S\to A$ be a surjection of commutative rings with identity. If $A$ is a $S$-projective, then $\pi$ splits as algebras. This will prove what you need. So, let $\phi:A\to S$ be a splitting as $S$-modules (since $A$ is $S$-projective). Let $\phi(1_A)=e\in \phi(A)$. Then, $e(1-e)\in\phi(A)$, since $\phi(A)$ is an $S$-submodule. But $\pi(e(1-e))=0$, since $\pi(e)=1_A$ and $\pi$ is an algebra map. But, $\pi:\phi(A)\to A$ is an isomorphism and thus $e(1-e)=0$. So, $e$ is an idempotent and the rest should be clear.