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( 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 : )

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  • $\begingroup$ 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

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

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