# Does totally flat commutative ring imply all ideals are idempotent?

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, (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.

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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 $I = I^2$.
 (I am using the fact that whenever $I$ is an ideal in a ring $R$ and $M$ is an $R$-module, we have $M \otimes R/I = M/IM$.) – countinghaus Apr 28 '12 at 2:23 Thanks!${}{}{}$ – Camilla Vaernes Jun 9 '12 at 21:05
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, $I^2=I$.