( 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 $$ B \otimes_A B \rightarrow B \times R $$


  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 : )

  • $\begingroup$ 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 $R=B \otimes_{A} B$. $\endgroup$ Jun 12, 2020 at 0:39
  • $\begingroup$ Yea. so what i'm struggling is to prove iso of $(1-e)R$ and $R[1/e]$ $\endgroup$
    – Bryan Shih
    Jun 12, 2020 at 2:18

1 Answer 1


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 $R[1/e]\cong R/(e-1)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.