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This is probably obvious but I am getting stuck thinking about it:

Let $A$ be a commutative ring with unity, $M$ a flat $A$-module, and $N\subset M$ a submodule.

Is $N$ necessarily flat over $A$?

I haven't found a simple counterexample, but I also haven't found more than a vague intuitive reason to think so. Thanks in advance.

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    It might be interesting to notice that if $M$ and $M/N$ are flat, then $N$ is flat.2012-11-16

3 Answers 3

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Let $A=\mathbf{Z}/4\mathbf{Z}$. Then $A$ is flat over itself, but the ideal $I:=2\mathbf{Z}/4\mathbf{Z}$ is not flat over $A$. This is because when we tensor the injection $I\hookrightarrow A$ with $I$, we get the map $I\otimes_AI\rightarrow I$ which sends $r\otimes s$ to $rs$, is visibly the zero map. The group $I\otimes_AI$ is not zero because it is isomorphic to $(\mathbf{Z}/2\mathbf{Z})\otimes_{\mathbf{Z}/2\mathbf{Z}}(\mathbf{Z}/2\mathbf{Z})\cong \mathbf{Z}/2\mathbf{Z}$. So $I\otimes_AI\rightarrow I$ is not injective.

However, if $A$ is a principal ideal domain, or more generally a Dedekind domain, then submodules of flat $A$-modules are flat, because for such a ring $A$, flat$=$torsion-free, and it is clear that submodules of torsion-free modules are torsion-free (over any domain).

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    Ok, thanks for the clarification.2012-11-15
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Let $A$ be a ring. Then $A$ is of course flat over $A$. But sub-modules of $A$ are just ideals, and these are rarely flat over $A$.