TITLE: Example of a projective module which is not a direct sum of f.g. submodules?
QUESTION [13 upvotes]: 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!

REPLY [2 votes]: 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.