Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.