# Equivalent conditions on tensor product and product of rings

( Background: These is one part of a critera for commutative rings $$f:A \rightarrow B$$ to be etale. )

It is claimed that the following two conditions 1 & 2 are equivalent.

1. The multiplication map $$p: B \otimes _A B \rightarrow B$$ is the projection on to a summand. There exists another commutative ring $$q: B \otimes _A B \rightarrow R$$ such that $$p$$ and $$q$$ induces isomorphism $$B \otimes_A B \rightarrow B \times R$$

and

1. There exists an idempotent element $$e \in B \otimes_A B$$ such that $$p$$ induces an isomorphism $$(B \otimes_A B)[1/e] \simeq B$$ under localization at $$e$$.

I can prove 1=>2. But I can't prove 2=>1. Help would be appreciated : )

• For any idempotent $e$ in any commutative ring $R$, the localization $R[1/e]$ is isomorphic to $(1-e)R$. Also, $R$ is isomorphic to $(1-e)R \times eR$. In particular, this applies for $R=B \otimes_{A} B$. Jun 12, 2020 at 0:39
• Yea. so what i'm struggling is to prove iso of $(1-e)R$ and $R[1/e]$ Jun 12, 2020 at 2:18

Let $$R$$ be a ring and $$e$$ an idempotent.
So $$e^2=e$$ or, perhaps written in a more suitable form for this question : $$e(e-1) = 0$$.
In particular if you invert $$e$$, then $$e-1 = e^{-1} 0 = 0$$ so $$e=1$$. Inverting an idempotent turns it into $$1$$.
In particular, you have a factorization of the canonical morphism as $$R\to R/(e-1)\to R[1/e]$$.
Conversely, if you kill $$e-1$$, you get $$e=1$$, so $$e$$ is invertible, so you also get a factorization of the canonical morphism as $$R\to R[1/e]\to R/(e-1)$$. It's then easy to check that these give you an isomorphism $$R[1/e]\cong R/(e-1)$$