Question

This semester I am teaching a graduate course in commutative algebra, and I have been taking the occasion to try to look at the proofs of some the results in my basic source material (Matsumura, Eisenbud, Bourbaki...) which I skipped as a little too complicated the first $N$ times around.

Recently I got the chance to read and understand I. Kaplansky's big theorem on projective modules, i.e., that a(n even infinitely generated) projective module over a local ring is free. En route to establishing this, he proves another result which is interesting but rather technical:

Theorem (Kaplansky, 1958): Every projective module is a direct sum of countably generated projective submodules.

For my take on this result, see $\S 3.10$ of these notes. In particular, it raises two natural questions:

Question 1: Is there a ring $R$ and an $R$-module $M$ which is not a direct sum of countably generated submodules?

Question 2: Is there a ring $R$ and a projective $R$-module $P$ which is not a direct sum of finitely generated submodules?

I was able to look up that the answer to Question 1 is "yes". In particular, I found work of L. Fuchs which says that for every infinite cardinal $\kappa$ there is an indecomposable (i.e., not expressible as a nontrivial direct sum) commutative group $G$ of cardinality $\kappa$. I would however be interested in hearing other examples or other takes on Question 1.

My real question is Question 2: presumably the answer is either yes or unknown, or people would mention the stronger result when Kaplansky's Theorem is discussed. A theorem of Bass that M. Reyes pointed out to me in his answer to another recent question of mine on modules is relevant in this regard: obviously an affirmative answer to Question 2 must involve an infinitely generated projective module, and if $R$ is Noetherian and connected then every infinitely generated projective module is free, hence a direct sum of singly generated submodules!

Answer 1

For question two the example that is given most frequently seems to be that of the ring $R$ of continuous real valued functions on $[0,1]$ and the ideal of all functions $f$ which vanish on some interval $[0,\epsilon(f)]$ where $\epsilon(f)\in (0,1)$. This ideal is countably generated and projective but not a direct sum of finitely generated submodules. You might also want to take a look at the article "When every projective module is a direct sum of finitely generated modules" by W. McGovern, G. Puninski and P. Rothmaler.

Answer 2

[Warfield, Robert B., Jr. Rings whose modules have nice decompositions. Math. Z. 125 1972 187--192. MR0289487 (44 #6677)] shows that over commutative Artinian rings with non-principal ideals, there exist indecomposable modules which are not countably generated. This answers (1).

Answer 3

I just wanted to add a little bit to the argument given in Lam's book (as communicated by Gjergji Zaimi). Let $R = C([0,1],\mathbb{R})$ be the ring of real-valued continuous functions on the closed unit interval, and let $I$ be the ideal of all functions $f$ which vanish identically on some neighborhood of $0$. Lam shows that $I$ is projective (a nice application of the Dual Basis Lemma) and also not free: indeed, he remarks that every element $f$ of $I$ has a nontrivial annihilator -- namely any nonzero function with support contained in the zero set of $f$ -- whereas for a free $R$-module $\bigoplus_{i \in I} R$ any standard basis element $e_i$ clearly has zero annihilator.

What Lam does not address -- as far as I can see -- is why $I$ is moreover not a direct sum of finitely generated submodules. But here is a nice argument for this using Swan's Theorem: we are asking whether the projective module $I$ is a direct sum of finitely generated projective modules. But every finitely generated projective module over $R$ corresponds to a vector bundle over $[0,1]$. However, since $[0,1]$ is contractible, every vector bundle over $[0,1]$ is trivial, and thus every finitely generated projective $R$-module is free. Thus, if $I$ were a direct sum of finitely generated submodules, it would itself be free, which we previously saw is not the case.

I'm sure there's also a purely algebraic proof of this, but I am very fond of Swan's Theorem...