# Example of a projective module which is not a direct sum of f.g. submodules?

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!

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.

• The example I mentioned above is also example 2.12D in T-Y Lam's "Lectures on Modules and Rings". – Gjergji Zaimi Feb 17 '11 at 7:09
• @Gjergji: thanks. I recently ordered Lam's book from amazon. It sounds like I made a good choice. – Pete L. Clark Feb 17 '11 at 7:20

[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).

• He mentions that the problem of knowing which rings have arbitrarily large indecomposable modules is hard and open, at the time of writing, even for the integers---Fuchs proposed a construction which was later shown to only work up to the first inaccessible cardinal... (talk about monstrous groups...) – Mariano Suárez-Álvarez Feb 17 '11 at 7:14
• Ah, thanks for the tip. I was taking the MathSciNet review of Fuchs's 1959 paper at its word. I will reference Warfield's paper instead of Fuchs's. – Pete L. Clark Feb 17 '11 at 7:52

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...

Pete, here is another nice example (shown to me by Bergman many years ago) answering question 1 in another way.

Let $F$ be a field, let $R=F[x]$, and let $M=F[[x]]$ (which is an $R$-module in the obvious way). If $M$ were a direct-sum of countably generated $R$-submodules, there would necessarily be infinitely many components (by an $F$-dimension argument). The element $1\in R\subseteq M$ would lie inside a finite subsum, so we could write $M=M_1\oplus M_2$ with $R[x]\subseteq M_1$ and $M_2\neq 0$. Thus, $M_2\cong M/M_1$ has each of its elements infinitely divisible by $x$, which only holds for the 0 element in $M$, yielding the necessary contradiction.

Let me give one more (elementary) example for Question 1. Let $k$ be an uncountable field (e.g. $\mathbb R$ or $\mathbb C$ will do) and $R = k[x]$ the ring of polynomials. Let $M = k(x)$, i.e. the field of fractions of $R$, which is naturally an $R$-module.

$M$ is indecomposable, since even more holds: Every two non-trivial submodules intersect non-trivially (every two non-zero elements have a common non-zero $R$-multiple).

Finally, why is $M$ not countably generated (in a slightly informal way): Each element of $M$ has only finitely many distinct irreducible polynomials in its denominator (in lowest terms), so among countably many elements, only countably many distinct irreducible polynomials appear in the denominators. Addition and multiplying by elements of $R$ cannot produce "new" irreducible factors of denominators. However, you have to generate (uncountably many) fractions of the form $1/(x-a)$ for each $a \in k$ and since the polynomials $x-a$ are clearly irreducible, we are done.