## Rank and the IBN Property

**Point of Post: **In this post we discuss the notion of rank for free modules, in particular when it is and is not defined, and discuss the notion of the invariant basis number for rings.

*Motivation*

Ok, so fine, we have defined the idea of free modules in the hope that their simple attributes shall be useful to us, that they shall somehow be “the simplest case”. We motivated this by pointing out how nice vector spaces were, how the existence of bases made everything “nice’, and in particular pointed out the existence of dimension. Ok, so we showed that free modules (those modules with bases) do have the nice ability of being able to define transformations on their bases and then extend by linearity (that is actually how we abstractly defined them, and then showed that this was equivalent to the existence of a basis). That said, we made no mention of dimension and the pursuant isomorphism theorem we are damn hoping it brings (). Why is this? Why would I motivate the notion of free module with such a tantalizing carrot and then not discuss it? Well, perhaps because technically I made a lie of omission. Namely, this isn’t always true. Wait, what? It would seem last time that if one reads closely that I did prove a result looking like , just not in the language of dimension. Namely, we defined a free module to be any unital module which is is isomorphic to a free module over a set . Ok, but we proved that any two free modules over a given set are isomorphic, and since is a free module over the set we know that . Yay, right? Well, yes and no. It may seem like a great result, and it seems so intuitive that defining this says that . The problem is, if we are going to define dimension in this way we better be sure that we can’t find another set with and . This is bound to be true, yeah? Well, no. Unfortunately we can actually find rings such that which completely dashes, at least in those cases, being able to define a “dimension function” on modules over those rings. That said, all is not lost. Most “tame” (e.g. commutative) rings do satisfy the property that the notion of dimension is well-defined. In fact, if a ring satisfies the so-called *invariant basis number *(IBN) property, that implies for finite then the notion of “dimension” is well-defined for free left -modules. That said, we should be aware of these “crazy” examples, so as to not lose perspective and assume everything is always nice (even though 99% of the time, it will be).

*Rank of a Free Module*

Before we begin actually discussing notions of how big bases are, we need the following characterization of bases which should be familiar from linear algebra:

**Theorem: ***Let be a ring and a unital left -module. Then, a subset is a basis for if and only if it is a maximally linearly independent set (i.e. no linearly independent set properly contains it) and is a minimal generating set (no proper subset generates ).*

**Proof: **Suppose first that is a basis. To see that is a maximally linearly independent set we suppose that there exists some element with is linearly independent. Clearly (since adding zero to any set makes it linearly dependent), and and so and since not all the coefficients are zero (since ) it follows that is linearly dependent contradictory to assumption. Suppose now that was not a minimal generating set. In particular, suppose there was a proper subset which generates . We note, then choosing we see that, by assumption that generates , we can write but this contradicts the uniqueness of representation for the basis . Thus, there does not exist a proper subset of that generates .

The converse is obvious.

We begin by proving a somewhat interesting theorem. Namely, we claimed in the above motivation that we can find free -modules with bases of different cardinalities. The following theorem says that this is a somewhat finite dimensionally fueled problem in the sense that if a module possesses an infinite basis then all the bases for the space must be the same cardinality. Indeed:

**Theorem: ***Let be a free module over a ring . If possesses an infinite basis then every basis for is equipotent to .*

**Proof: **We begin by showing that cannot have a finite basis. This is simple enough, for if is another finite basis of we know that for each we can find finitely many elements in which span to contain . Clearly then but since this right hand set contains in its span a basis it clearly must generate and since it is clearly a proper subset of this contradicts the minimal generating set property of the basis . Thus, we see that every possible basis for is infinite.

We now show that they all possible infinite bases for have the same cardinality. Indeed, suppose that is another basis for , which by the previous paragraph, is infinite. So, consider the map (where denotes the set of finite subsets of ) defined by sending to the unique finite set of elements of that generate it.

We claim that each element of has finite preimage. Indeed, we note that , but since the number of elements of is finite we can find a finite subset (similar to how we constructed one in the previous paragraph) of which generates and so , but clearly then by linear independence this allows us to conclude that and so is finite as desired.

From this it’s easy to see that since identifying elements of into finite equivalence classes under the relation if and only if gives us a set which is equipotent to (since is infinite) and injects . But, since is infinite it is basic set theory that and so by the above analysis we may conclude that . But, by symmetry we may also conclude that and so by the Schroeder-Bernstein theorem we may conclude that .

Since was arbitrary the conclusion follows.

This result tells us that the failure for a ring to have all free modules over that ring have well defined “dimension” is a finitary obstruction. In particular, suppose that a ring is such that there exists a unital left free module with not well-defined “dimension”. This should mean that we can find two bases which aren’t equipotent, and by the above theorem it must be the case that both these bases are finite. By noting that we see that a ring having a not well-defined “dimension” for its modules is the existence of with yet . Thus, if we define rings with the property that for implies as *invariant basis number rings *(IBN-rings for short) or rings satisfying the *invariant basis number property* (IBN property for short) and the *rank *of a free module to be the size of any of its bases (the generalization of dimension) we have the following theorem

**Theorem: ***Every left -module has a well-defined rank if and only if is an IBN-ring.*

**References:**

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

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