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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.

Watson
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Ben Blum-Smith
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    It might be interesting to notice that if $M$ and $M/N$ are flat, then $N$ is flat. –  Nov 16 '12 at 16:14

3 Answers3

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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).

Keenan Kidwell
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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
  • @KeenanKidwell Could you direct me to the proof of the fact which you are using? That, $M\otimes_AN \cong M\otimes_{A/I}N$ – Adithya S May 11 '21 at 17:12
  • @AdithyaS Keenan already gave the map; its existence is justified by the universal property of the tensor product. – Sha Vuklia Feb 20 '23 at 13:00
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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 $A$.

the L
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The property that submodules of flat modules are flat is equivalent to being locally a valuation domain (or in homological terms, having weak global dimension at most 1), see https://stacks.math.columbia.edu/tag/092S for reference. To expand on The L's answer, this is in turn equivalent to having flat ideals. Since being reduced is a local property, this shows that any ring with nilpotents will serve as a counter example to your question, and in fact will have non-flat principal ideals.

user26857
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Badam Baplan
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  • Replaced the link with one which send us to the quoted result. – user26857 Jun 26 '19 at 06:56
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    Thanks for that. Also maybe worth noting here that higher dimensional analogs of the mentioned result hold too, e.g. for non VNRs, the weak global dimension is at most $n$ iff the weak dimension of its (f.g.) ideals is at most $n-1$ iff modules of weak dimension at most $n-1$ have submodules of dimension at most $n-1$. Go-to references are chapter $8$ of Rotman's Homological Algebra and $5D$ of T.Y. Lam's Lectures on Modules and Rings. – Badam Baplan Jun 26 '19 at 07:21