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Let $A,B$ be commutative rings with 1, $\varphi:B\rightarrow A$ a ring homomorphism and M an $A$-module. If M is flat as an $A$-module, is it also flat as a $B$-module? (The structure of $B$-module is obviously given by $bm=\varphi(b)m$ for all $b\in B,m\in M$).

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Did you try to find examples before asking? A probably useful suggestion is to double the time spent looking for examples the next time you have such a question, and to repeat this every time; everytime you do find an example, restart. :-) –  Mariano Suárez-Alvarez Sep 12 '12 at 19:35
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up vote 3 down vote accepted

Suppose $B=\mathbb C[X]$, $A=\mathbb C$ and the map $\phi:B\to A$ is the unique $\mathbb C$-linear ring map such that $\phi(X)=0$.

Now let $M=\mathbb C$ be the free $A$-module of rank $1$, which is plainly flat. Is it $B$-flat?

Another example: $B=\mathbb Z$, $A=\mathbb Z/2\mathbb Z$ and let $M$ be again a free $A$-module of rank $1$.

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Thank you for the answer: 1) In the first example we can choose the map $\psi:\mathbb{C}[x]\rightarrow\mathbb{C}[x]$, $\psi(p(x))=xp(x)$ and then tensor it with $M=\mathbb{C}$. I then get, e.g., $1\otimes 1\mapsto x\otimes 1=1\otimes \phi(x)1=0$. 2) I can tensor the map $\mathbb{Z}\xrightarrow{\cdot 2}\mathbb{Z}$ with $A$ and get the required contradiction. –  random Sep 12 '12 at 22:47
    
So I believe the key fact here is that $A$ must be a flat $B$-module for the statement to be true. –  random Sep 12 '12 at 22:52
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