Let $R$ be a commutative ring and let $M,N$ be two finitely generated projective $R$modules which have equal rank (not necessarily constant). What kind of general results are there concerning the question of determining whether $M$ and $N$ are isomorphic or not? This is certainly a nontrivial question (e.g., consider the Picard group of a ring = isomorphism classes of f.g. projective modules of constant rank 1), but I'm looking for techniques for proving the (non)existence of isomorphisms for certain modules of the same rank.

1$\begingroup$ One criteria is: If there is an epimorphism $f: M \to N$, then $M, N$ are isomorphic. $\endgroup$– RalphSep 15, 2011 at 18:24

$\begingroup$ Could you give a reference for that? $\endgroup$– NikitasSep 11, 2020 at 12:07
1 Answer
If $R$ is noetherian of dimension d, then we have:
The Bass Cancellation Theorem: If $M$ has rank $\ge d+1$ after localizing at any prime, and if there exists a finitely generated projective module $Q$ such that $Q\oplus M\approx Q\oplus N$, then $M\approx N$. (You probably don't even need to assume $N$ projective for this, as long as you know that $M$ is.)
Plumstead's Theorem: If $R=A[t]$ for some noetherian ring $A$, then you can replace $d+1$ with $d$.
In the nonnoetherian case you can work with jdimension instead of dimension (the jdimension of $R$ is the length of a maximal chain of ideals, each of which is an intersection of maximal ideals).

$\begingroup$ Thanks for the reply! If I may ask a followupquestion: I'm in the setting where I know that the two modules are projections of free modules of different rank. Thus, finding a module such that the sums are equal (as you described above) may not be convenient. Any ideas on how to go about this? $\endgroup$ Sep 16, 2011 at 7:52

$\begingroup$ Joakim: Of course every f.g. projective module is the surjective image (I assume this is what you mean by "projection") of some f.g. free module, so this is not useful information. Of course you can make both free modules have the same rank by adding a free direct summand to one of them, but that still doesn't tell you anything useful. On the other hand, if you can say more about R, there is hope; if, for example, R is a polynomial ring over a field, then all fg projectives are free. $\endgroup$ Sep 16, 2011 at 19:10

$\begingroup$ I have just learned that Bart Plumstead passed away just a few weeks before I posted this reference to his theorem. The (quite beautiful) paper is at jstor.org/pss/2374448 . $\endgroup$ Oct 2, 2011 at 5:57