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A commutative ring $R$ is absolutely flat if every $R$-module is flat. Prove that the following are equivalent:

1) $R$ is absolutely flat

2) Every principal ideal of $R$ is idempotent

3) Every finitely generated ideal of $R$ is a direct summand

I'm stuck on implication $1) \to 2) $

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up vote 6 down vote accepted

If $\mathfrak a$ is an ideal in a ring $A$ and $M$ is an $A$-module, then the map $$ \begin{array}{ccc} M\otimes_A A/\mathfrak a &\to& M/\mathfrak aM \\ m\otimes (a+\mathfrak a)&\mapsto& ma+\mathfrak a M \end{array} $$ is an isomorphism, which we will call the quotienting isomorphism. Proving this map is an isomorphism is exactly Exercise 2.2 in Atiyah-Macdonald.

Now, suppose $R$ is absolutely flat and let $x\in R$. Then $$ 0\to (x)\to R\to R/(x)\to 0\tag{1} $$ is exact in $_{R}\mathsf{Mod}$. Since $R/(x)$ is flat, applying $(-)\otimes_R R/(x)$ to (1) gives an exact sequence $$ 0\to (x)\otimes_R R/(x)\to R\otimes_R R/(x)\to R/(x)\otimes_R R/(x)\to 0 $$ in $_{R}\mathsf{Mod}$. One then shows that the diagram $$ \begin{array}{ccccccccc} 0&\to& (x)\otimes_R R/(x)&\to &R\otimes_R R/(x)&\to &R/(x)\otimes_R R/(x)&\to& 0 \\ &&\downarrow && \downarrow && \downarrow \\ 0&\to& (x)/(x)^2&\to& R/(x)&=& R/(x)&\to& 0 \end{array}\tag{2} $$ commutes where the vertical arrows are the quotienting isomorphisms. This implies that the bottom row of (2) is exact so $(x)/(x)^2=0$ as required.

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What happens if you tensor the short exact sequence $$0\to I\to R\to R/I\to 0$$ with $I$?

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Dear Mariano, I can't find an answer to your question which would solve the problem. Could you please add some details? – Georges Elencwajg May 8 '14 at 10:43
@GeorgesElencwajg: For any ideals $I, J$, $\operatorname{Tor}^R_1(R/I,R/J) \cong (I \cap J)/IJ$. – zcn May 26 '14 at 19:07
@zcn: what does what you write have to do with Mariano's answer, who tensors with $I$ ? – Georges Elencwajg May 26 '14 at 21:41
@GeorgesElencwajg: Ah, I missed that part. In that case, I believe Mariano means: tensor with $R/I$ – zcn May 26 '14 at 21:44

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