Posted by Jason Polak on 16. June 2011 · 4 comments · Categories: homological-algebra · Tags: ,

I’m going to talk a bit about flatness. I created the following exercise for myself while studying flatness: describe some good examples of flat modules that are not projective.

We call a left $ R$-module $ F$ flat if the functor $ -\otimes F$ is exact, and of course the analogue can be made for right R-modules. As the tensor functor $ -\otimes F$ is always right exact, this is a rather natural definition to make. Recall that there is a similar definition for the $ \mathrm{Hom}(P,-)$ functor: we call an $ R$-module $ P$ projective if $ \mathrm{Hom}(P,-)$ is exact.

Projective modules have a rather nice characterisation. That is, a module $ P$ is projective if and only if it is the direct summand of a free module. In particular free modules are projective. Mac Lane in his book “Homology” gives an example of a projective module that is not free: take the ring $ \mathbb{Z}\oplus\mathbb{Z}$ and consider the submodule $ \mathbb{Z}\oplus 0$ of it.

Flat modules on the other hand, are a bit more tricky. One characterisation is that a (left) $ R$-module $ F$ is flat if and only if $ \mathrm{Hom}_{Ab}(F,\mathbb{Q}/\mathbb{Z})$ is injective as a right R-module — recall that $ I$ is called injective if $ \mathrm{Hom}(-,I)$ is an exact functor, and is just the dual concept to projective. However this characterisation isn’t always as useful for explicit calculations, as it’s not always easy to show that a module is injective.

Any projective module is flat. Indeed, if $ P$ is a projective left $ R$-module, then $ \mathrm{Tor}^1(A,P) = 0$ for any $ A$ by considering the projective resolution $ 0\to P\to P\to 0$, although one can easily prove this directly by using the lifting property for $ P$. However, just as there are projective modules that are not free, there are also flat modules that are not projective.

We can get a hint of how to find one because every finitely presented flat module $ F$ is projective – recall that $ F$ being finitely presented means that there is an exact sequence $ R^m \to R^n\to F\to 0$ for some natural numbers $ m$ and $ n$. So we can’t take $ F$ to be finitely presented – so why not go all out and take $ F$ to be infinitely generated? In the category of $ \mathbb{Z}$-modules, the example $ \mathbb{Q}$ is a good start, and in fact it’s a good exercise to show that $ \mathbb{Q}$ is flat. It’s not projective, however (hint: tensoring with any divisible Z-module kills torsion).

Update: The next part of this post was supposed to give another finitely generated but not finitely presented module that was flat but not projective. However, my example did not actually work as the commenter acl pointed out (see comments), so I’ll have to think for a while to figure out a correct example.

4 Comments

  1. Cool…I started reading your blog today…finally. Fun stuff.

  2. Thanks! Your comment has motivated me to attempt to write more stuff! :)

  3. This module does not seem to be flat: The multiplication by $x$ is injective on $C([0,1])$ but induces the $0$-map on $\R$.

    • It is true. My example actually does not work, and I have removed it noting the change in the post. I honestly don’t know why I thought it did at the time. Thanks for pointing this out.

      I shall have to think a bit harder for a new example.

Leave a Reply to Jason Polak Cancel reply

Your email address will not be published. Required fields are marked *