# 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?

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Why do you write "qn 17L" for Exercise 17 of Chapter 3 ? (Also: the book is by Atiyah and Macdonald.) – Georges Elencwajg Nov 14 '12 at 8:28
Your 2) is completely false, as can be seen by taking $C=A$ since it would imply $A\cong A\otimes_A B\cong B$ for any $A$-algebra $B$ ! – Georges Elencwajg Nov 14 '12 at 9:11
Elencwajg: I have changed the title accordingly. Thanks! I was probably typing too fast. Authur: I just found out how to accept the answers. Thanks for reminding me! – enoughsaid05 Nov 14 '12 at 9:45
@enoughsaid05 $B$ is an $A$ - module with action given by $f$, while $C$ is a $B$ - module with action give by $g$ and hence is an $A$ - module with the action given by $g \circ f$. – user38268 Nov 14 '12 at 10:12

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.

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But I don't see how $M\otimes_{A}C$ and $(M\otimes_{A}B)\otimes_{B}C$ are isomorphic. And $M\otimes_{A}B$ is a tensor product of two $A$-modules, so how is it possible that $M\otimes_{A}B$ is a $B$-module? – enoughsaid05 Nov 14 '12 at 9:47
Okay, I guess one can go around by letting an action on $M\otimes_{A}B$ via $b\cdot (m\otimes p)=m\otimes bp$, then I will have no problems with the isomorphisms! I think your answer is correct! Thanks! – enoughsaid05 Nov 14 '12 at 9:57
Dear enoughsaid: yes, your formula is correct. The multiplication by scalars from $B$ on elements of $M\otimes _A B$ is exactly as you wrote it. – Georges Elencwajg Nov 14 '12 at 10:00
@enoughsaid05 In general because $B$ is an $A$ - $B$ bimodule you can make the tensor product $M \otimes_A B$ a right $B$ module simply by defining multiplication as $(m\otimes p) \cdot b = m \otimes (pb)$. – user38268 Nov 14 '12 at 10:10