Exercise 2.27 Atiyah-Macdonald, absolute flatness 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) $
 A: 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.
A: What happens if you tensor the short exact sequence $$0\to I\to R\to R/I\to 0$$ with $I$?
A: I'll put in the picture because I could not write the diagram in latex
