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 $b\in B,m\in M$).
Flat module over A implies flat module over B
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abstract-algebra
commutative-algebra
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4Did 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. :-) – 2012-09-12
1 Answers
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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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0So I believe the key fact here is that $A$ must be a flat $B$-module for the statement to be true. – 2012-09-12