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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. – user18119 Nov 16 '12 at 16:14
up vote 10 down vote accepted

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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Is there a typo in your answer? You write $(\mathbb{Z}/2) \otimes_{\mathbb{Z}/2} (\mathbb{Z}/2)$ when you're referring to $I \otimes_A I$. Shouldn't it be $(\mathbb{Z}/2) \otimes_{\mathbb{Z}/4} (\mathbb{Z}/2)$? And in this case, is the isomorphism $(\mathbb{Z}/2) \otimes_{\mathbb{Z}/4} (\mathbb{Z}/2) \cong \mathbb{Z}/2$ easy to see? Thanks. – Fredrik Meyer Nov 15 '12 at 11:35
No, it's not a typo. I'm using the following fact: if $A$ is a ring and $M$ and $N$ are $A$-modules annihilated by some ideal, then the natural map $M\otimes_AN\rightarrow M\otimes_{A/I}N$ (which looks like the identity $m\otimes n\mapsto m\otimes n$) is an isomorphism. In the situation of the answers, $M=N=2\mathbf{Z}/4\mathbf{Z}$ is an abelian group of order $2$, $I=2\mathbf{Z}/4\mathbf{Z}$, and $A/I=\mathbf{Z}/2\mathbf{Z}$. The isomorphism I write down in my answer is the standard isomorphism $A\otimes_AA\cong A$, $a\otimes b\mapsto ab$, for any ring $A$. – Keenan Kidwell Nov 15 '12 at 15:00
Ok, thanks for the clarification. – Fredrik Meyer Nov 15 '12 at 18:15

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 $A$.

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