Let $R$ a ring with trivial Picard group, so every rank $1$ projective module is free. What does that tell me about the structure of projective modules of rank $n$?
For starters, if $P$ is any projective module of rank $n$ then $P$ will have trivial determinant, because $\bigwedge^n P$ will have constant rank $1$ and be trivial by assumption.
Furthermore, the relation $Pic(A \times B) \cong Pic(A) \times Pic(B)$ forces any projection of $R$ to have trivial Picard group, so when we form the componentwise determinant of any finitely generated projective module it will be also be trivial.
But this is as far as I get... lots of trivial determinants, but to what end?
I know that there are examples of rings with $Pic(R) = 0$ and even non-free rank $2$ projective modules, such as mentioned here. On the other hand, semi-local rings (or more generally rings which are $0$-dimensional modulo their Jacobson radicals) are examples of rings over which any rank $n$ projective modules are free.
To give some context, I am trying to show that every projective rank $n$ module of some special class of rings are free (I'm pretty sure they are). I am able to show it for $n=1$ by constructing explicit (and fairly elaborate) isomorphisms, but I have no idea how to generalize the construction, because my knowledge of invertible modules doesn't extend to higher ranks. My hope is that there is some theory I am missing that will bridge the gap!
How does one go about generalizing knowledge about invertible modules to higher ranks?
EDIT: Actually I think there might be a 'dimension-shifting' approach that I had missed the first time around. Namely, if we know that rank $n$ projectives are free, then consider a rank $n+1$ projective and the evaluation homomorphism $P \otimes_R \operatorname{Hom}(P, R) \rightarrow R$ which one checks locally must be surjective. This gives a short exact sequence $0 \rightarrow K \rightarrow P \otimes_R \operatorname{Hom}(P, R) \rightarrow R \rightarrow 0$ which splits, so $K$ is seen to be a rank $n$ projective, and by hypothesis free. Thus we have $P \otimes_R \operatorname{Hom}(P, R) \cong R^{n+1}$. Of course this doesn't finish the argument, but it puts us in good position to generalize a construction from invertible modules to higher ranks. Still curious to hear if anyone else has ideas! EDIT2: That last edit didn't make any sense because actually $K$ would then have rank $n^2 - 1$ which is useless, but maybe there's still some merit in the approach.
