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

[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).

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...