# 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

• Are you sure you want to include non-commutative rings? For modules over a non-commutative ring, the "rank" may not even be well-defined (rings without IBN, for instance). – Arturo Magidin Dec 16 '16 at 22:37
• @ArturoMagidin This can be avoided by saying $M_i$ is free on $k_i$ generators (rather than of rank $k_i$). In that case, stable finiteness (right inverses in matrix rings are left inverses) is likely sufficient, .... This is a slightly stronger hypothesis than invariant basis number. – David Handelman Dec 17 '16 at 0:33
• And for stably finite von Neumann regular rings, the inequality is true. The image of $M_i$, $P_2$, is a fg submodule of $M_{i+1}$, hence is projective, and the same for the image of $M_{i-1}$, (denoted $P_1$), so splits, and we thus have $M_i$ isomorphic to $P_1 \oplus P_2$. Splitting of $M_{i-1} \to P_{1}$ ensures that $P_1$ is iso to a direct summand of $M_{i-1}$, so $M_i$ is iso to a direct summand of $M_{i-1} \oplus M_{i+1}$. Stable finiteness then guarantees $k_i \leq k_{i+1} + k_{i-1}$ (actually, so does IBN, in this case). – David Handelman Dec 17 '16 at 0:42
• I should have added, under the assumption that each $M_i$ is free on $k_i$ generators, in the proof that the inequality is correct for sf (or IBN) vn regular rings. – David Handelman Dec 17 '16 at 0:44
• And (possibly answering the question), it presumably is the case that if $S \subset R$ is an inclusion of unital rings such that $R$ satisfies the property on all such sequences, then so does $S$. I guess you just tensor on the right with ${}_SR_S$. – David Handelman Dec 17 '16 at 0:54

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.

• David Handelman, Thank you very much for the answer. I really appreciate it. I am not very familiar with left semihereditary stably finite rings. Are there any known examples of rings of this type which do not admit linear representations in $M_m(F)$ for some $m\in \mathbb N$, where F is a field? I am simply trying to compare the condition you suggest with the condition mentioned in my question. In addition, it looks like the condition you mention can be substituted with the condition that R admits a homomorphism into a left semihereditary stably finite ring. Am I wrong? – RAG Dec 17 '16 at 22:02
• There are lots of examples of (semihereditary) rings which admit no representations in finite dimensional matrix rings (over a division ring); for example, any direct limit of semisimple algebras which has no bounds on the index of nilpotents; von Neumann algebra factors of finite type; and many of others. Unfortunately, the subring argument does not work unless the overring is flat (and requires the map to be an embedding). But this still covers a lot of cases, such as noncommutative orders in semisimple algebras. – David Handelman Dec 17 '16 at 22:19