Long exact sequence of free modules and rank inequality Given a long exact sequence of free (left) modules $M_i$ of finite rank $k_i$ over possibly non-commutative ring $R$:
$\dots \to M_{i-1}\to M_i \to M_{i+1}\to \dots$ 
What is the condition on $R$ which allows to deduce that $k_i\leq k_{i-1}+k_{i+1}$? 
The mentioned property certainly holds if one assumes that $R$ admits a representation into $M_m(F)$ for some $m\in \mathbb N$, where $F$ is a field. I also think that the same will hold if $R$ is a semisimple algebra. I am looking for some other classes of rings which satisfy this property. I was thinking about represenations in von Neuman algebras and to make use of von Neuman rank, but I am not sure about that. Any help will be highly appreciated
 A: Here is a result in the right direction, organizing the comments (some of which were unjustified, e.g., use of IBN). It applies to finite type von Neumann algebras and their regular rings, as well as all stably finite regular rings, and free rings too.
A (unital) ring is left semihereditary if every finitely generated left ideal is projective. This is a Morita invariant; in particular, this implies that every finitely generated submodule of a finitely generated free module is projective. Examples include von Neumann regular rings, rings all of whose matrix rings are Baer (thus including AW*-algebras), free rings, .... The class of semihereditary rings is closed under direct limits with one to one maps. 
Recall that a ring is stably finite if all right invertible square matrices are invertible. This includes all AW*-algebras of finite type, their regular rings, free rings, and many others. The class of stably finite rings is closed under direct limits with one to one maps, and subrings.
Proposition Let $R$ be a left semihereditary stably finite ring. Then the condition $k_i \leq k_{i+1} + k_{i-1}$ is satisfied.
Proof. Let $P_1$ be the image of $M_i$ in $M_{i+1}$. As $M_{i+1}$ is free and $P_1$ is finitely generated (since $M_i$ is), we have that $P_1$ is projective. Hence $M_i \to P_1$ splits, and thus $M_i $ is isomorphic to $P_1 \oplus P_2$, where $P_2$ is the kernel of $M_i \to M_{i+1}$. We also have $P_1$ is the kernel of $M_{i+1} \to M_{i+2}$, so by the same argument, $P_1$ is a direct summand of $M_{i+1}$.
By exactness, $P_2$ is the image of $M_{i-1}$ in $M_i$, and freeness of $M_i$  (and $P_2$ finitely generated) implies $P_2$ is projective, and thus $P_2$ is isomorphic to a direct summand of $M_{i-1}$.
We have $M_i$ isomorphic to $P_1 \oplus P_2$, and  $P_1$ is isomorphic to a direct summand of $M_{i+1}$ and $P_2$ is isomorphic to a direct summand of $M_{i-1}$. Hence $M_i$ is isomorphic to a direct summand of the free module $M_{i-1} \oplus M_{i+1}$. If $k_{i} > k_{i-1} + k_{i+1}$, we would obtain a contradiction to stable finiteness.\qed
Since semihereditariness is fairly strong, we should be able to deal with subrings, say $S \subset R$, where $R$ is at least semihereditary. Unfortunately, tensoring on the left with ${}_S R_S$ need not yield exactness of the sequence of free $R$-modules,  $\cdots \to R \otimes_S M_{i} \to \cdots$ (although there may be ways around this). However, if $R_S$ is flat, then exactness is preserved. Hence:
Corollary Suppose $S\subset R$ is an inclusion of rings such that $R_S$ is flat, and $R$ is a left semihereditary stably finite ring. Then $k_i \leq k_{i+1} + k_{i-1}$.
It's been a long time since I did ring theory (25 years?).
Edit The corollary covers semiprime (right) Goldie rings, since their (classical) ring of quotients is flat. Also, if $K$ is a two-sided ideal such that $K^2 = K$, then $R/K$ is flat, covering some additional cases. In addition, a direct limit of rings with one to one maps each of which satisfies the inequality, also satisfies it.
