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From reading Atiyah and MacDonald, I know of the result that a absolutely flat commutative ring has all principal ideals idempotent.

Reading around on math reference, I think that if a commutative ring $R$ is absolutely flat (that is, all $R$-modules are flat), then every ideal $H$ is idempotent, so $H^2=H$.

However, maybe my English is not so good, but I don't fully understand the proof provided there. I don't get the language "descend into tensor algebra and imagine..." and such. Is there a better polished proof of why all ideals in a absolutely flat ring $R$ are idempotent? Thank you very much.

2 Answers 2

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I think they are saying this. Given an ideal $I$, tensor the inclusion $ 0 \to I \to R $ with $R/I$. We get (flatness of $R/I$) $ 0 \to I/I^2 \to R/I. $ Thus $I/I^2 = 0$, so $I = I^2$.

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    Why does from the exactness of $ 0 \to I/I^2 \to R/I $ follow that $I/I^2 = 0$?2018-07-02
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This holds for noncommutative rings as well. If you are aware that "absolutely flat" is sometimes called "von Neumann regular", then you know that every finitely generated ideal of $R$ is generated by an idempotent.

If $I$ were an ideal with $I^2\subsetneq I$, pick $x\in I\setminus I^2$. Then $(xR)^2=(eR)^2=eR=xR$. But this means that $x\in (xR)^2\subseteq I^2$, a contradiction. So, $I^2=I$.