A direct product of projective modules which is not projective

Question

I am looking for an elementary example of a family $\{M_\alpha\}_\alpha$ of projective $R$-modules whose direct product is not projective. The simplest example that I know is the $\Bbb{Z}$-modules, $\Bbb{Z}, \Bbb{Z}, \Bbb{Z}, \cdots$ whose product is not free and hence not projective (as $\Bbb{Z}$-module) but this example uses the fact that $\Bbb{Z}$ is a hereditary ring or something equivalent to that which (in my opinion) is not that elementary. So, I would be grateful if somebody can come up with a simpler example I mean an example that requires as less as possible advanced algebra.

Answer 1

A few remarks, to be expanded below: (1) first is that the proof that $M = \prod_{i=1}^\infty\mathbb{Z}$ is not free is elementary, and (2) second is that it might be hard to find simpler examples, at least if "simple" refers to how simple the ring is itself.

(1) In fact, here's a proof that I learned from Kaplansky's book "Infinite Abelian Groups": Assume $M$ is free. Pick a prime $p$. Let $S\subseteq M$ be the submodule of sequences such that the power of $p$ dividing each term goes to infinity. $S$ is a submodule of $M$ so is free as well. Multiplication by $(p,p^2,p^3,\dots)$ is an injection $M\hookrightarrow S$. Since $M$ is uncountably generated so is $S$.

Now by definition of $S$, we have that $S/pS$ is a vector space of countable dimension dimension over $\mathbb{Z}/p$ as each term here as a representative with finitely many terms. Yet, if $S$ is free and uncountably generated then $S/pS$ should have uncountable dimension over $\mathbb{Z}/p$.

(2) I have a feeling that finding "easier" examples will be hard. Here are some reasons:

Note that if you did want to try and find an example, you should just experiment with products of the ring $R$ itself. Indeed, S. Chase proved that any product of projectives being projective is equivalent to any arbitrary product of copies of $R$ is projective. (Examples like $\mathbb{Z}/4$ won't work either, since they satisfy the descending chain condition on ideals and all f.g. ideals are finitely related. Any example you are looking for in a ring cannot have these properties.)

Moreover, homologically $\mathbb{Z}$ (and PIDs that are not fields) are pretty simple in that they have global dimension one and they are commutative. Global dimension zero rings won't work since they are semisimple hence quasifrobenius, or equivalently, projectives and injectives coincide. So in this case a product of projectives is a product of injectives, hence injective, hence projective.

I would be interested to see an example however where the ring is more complicated but the proof is even simpler. In fact, I would be interested to see any example at all with a proof significantly different than the one above.