# How to prove that a projective module is not free?

Let $A$ be a noncommutative (perhaps $\ast$-) algebra (over $\mathbb{C}$) and let $M$ be a projective module defined via a projector $P\in M_n(A)$; i.e. $M=P(A^n)$. Furthermore, assume that all objects are given "explicitly"; the algebra by (a finite number of) generators and relations, and $P$ as a specific $(n\times n)$-matrix with entries in $A$.

I'm looking for different ways of proving that $M$ is NOT a free module, and I'd be interested in all you favourite ways of doing this! (I'm not only looking for methods that work in general, but also tests that may not be conclusive in every case). Thanks!

• Is A a finite dimensional algebra over a field or at least finitely generated over a field or over Z? – Benjamin Steinberg Apr 23 '15 at 14:25
• I edited the question to include the assumptions that A is an algebra over the complex numbers and that the number of generators are finite, but the algebra is not necessarily finite dimensional. – Joakim Arnlind Apr 24 '15 at 6:18
• One technique is to prove that $A$ has invariant basis number, and $M^{k}=A^{\ell}$ with $\ell\nmid k$. There are other similar arguments along these lines, which use various formulations of "dimension". – Pace Nielsen Apr 28 '15 at 4:53