Atiyah-Macdonald Book qn 17 Faithfully Flat I am wondering how to solve this problem: given $f:A\rightarrow B$ and $g:B\rightarrow C$ ring homomorphisms. If $g\circ f$ is flat, and $g$ is faithfully flat, then $f$ is flat.
If I am not mistaken, the question asks us to prove that $B$ is a flat $A$-module. So we want to show that if $M$ and $N$ are $A$ modules and if $M\rightarrow N$ is injective, then $M\otimes_{A}B\rightarrow N\otimes_{A}B$ is also injective.
So I proceed as follows: since $C$ is flat over $A$, so $M\otimes_{A}C\rightarrow N\otimes_{A}C$ is also injective. But I am not sure from here how to use the fact that $C$ is faithfully flat over $B$. Here are some approaches that I tried:
1) $C\cong C\otimes_{B}B$, BUT to use associativity property on $M\otimes_{A}(C\otimes_{B}B)$, I require $C$ and $B$ to be $A$ modules.
2) So I attempted to see if $C\cong C\otimes_{A}B$ as $A$ modules, but I couldn't.
Any other approaches?
 A: Since $g\circ f$ is flat you know that $0\to M\otimes _A C\to N\otimes _A C$ is injective.
 The crucial remark is that $M\otimes_A C$ is isomorphic to $ (M\otimes_A B)\otimes _B C$ and similarly for $N$, so that $0\to (M\otimes_A B)\otimes _B C \to (N\otimes_A B)\otimes _B C $ is injective.
Now faithful flatness of $C$ over $B$ implies (see auxiliary result below) that $0\to M\otimes_A B \to N\otimes_A B$ is injective , which is what you wanted.  
An auxiliary result
 I have used above that given a morphism of $B$-modules $u:P\to Q$ the fact that $u\otimes _B C:P\otimes _B C\to Q\otimes _B C$ is injective implies (if $C$ is faithfully flat over $B$) that $u:P\to Q$ is injective.    
Since Atiyah-Macdonald don't mention that result, I'll prove it:
Consider the kernel $K=Ker(u)$ and the exact sequence $0\to K\to P\stackrel {u}{\to} Q$.
By flatness of $C$ over $B$ it induces an exact sequence $0\to K\otimes _B C\to P\otimes _B C\to Q\otimes _B C$.
Since by hypothesis $u\otimes _B C: P\otimes _B C\stackrel {u\otimes _B C}{\to}Q\otimes _B C $ is injective, this yields $K\otimes _B C=0$ which finally gives $K=0$ by property iv) (one of the equivalent definitions of faithful flatness) of Exercise 16.
Saying that $K=0$ is of course equivalent to saying that $u$ is injective, which is what I promised to prove.
