# Example of a finitely generated flat module which is not free

I couldn't come up with an example of a finitely generated flat module which is not free. I know that over local rings, freeness and flatness are equivalent. So the ring cannot be a local ring.

Let $R$ be a semisimple ring which isn't a division ring, and take an idempotent $e\notin \{0,1\}$. Then we have that $R=eR\oplus(1-e)R$ is a nontrivial decomposition of $R$, and both pieces are cyclic and projective (since they are summands of $R$) hence flat.

Actually neither piece is a free module, but we'll argue here that at least one of them isn't free to simplify things. If they were both free, that would imply that $R\cong R^n$ for some $n>1$ as modules. (Each factor contributes at least one $R$, you see.) But since $R$ has the IBN property, this is impossible.

Thus at least one of the pieces is flat but not free.

If you really want to be concrete, you can use a finite semisimple ring to make things obvious: let's try $R=\Bbb F_2\times\Bbb F_2$. $\Bbb F_2$ is denoting the field of two elements.

The module $I=\Bbb F_2\times \{0\}$ is a direct summand of $R$, but it can't be free: a free module would have to have at least four elements, and this module only has two elements!

Rings over which every module is flat are called absolutely flat, or Von Neumann regular. If $I$ is any nontrivial ideal of an absolutely flat ring, then $R/I$ is finitely generated flat over $R$, but not free (since $R/I \neq 0$ and $\mathrm{Ann}(R/I)=I \neq 0$). Products of fields are absolutely flat.

Here is an example that one can come up with from algebraic number theory. Consider $K = \Bbb{Q}(\sqrt{-6})$ and the ideal $I = (2,\sqrt{-6}) \subseteq \mathcal{O}_K$. This ideal is clearly finitely generated as an $\mathcal{O}_K$ module and furthermore not only is it flat but it is a projective module: we have

$$I \oplus ( 3,\sqrt{-6}) \cong (\sqrt{-6}) \oplus \mathcal{O}_K \cong \mathcal{O}_K \oplus \mathcal{O}_K.$$

If you're interested in knowing how we have such isomorphisms you can look at my answer here. But now $I$ cannot possibly be free because the generators are linearly dependent over $\mathcal{O}_K$:

$$3 \cdot (2) + \sqrt{-6} \cdot (\sqrt{-6}) = 0.$$

• In fact, every fractional ideal of a Dedekind ring $R$ is a projective $R$-module, so in particular every ideal of a ring of integers in a number field is projective. – Nils Matthes May 23 '13 at 11:32
• @NilsMatthes Yes that is certainly the case, IIRC due to invertibility of an ideal in the ring of integers of an algebraic number field. – user38268 May 23 '13 at 13:10