## Rank and the IBN Property (Pt. II)

**Point of Post: **This is a continuation of this post

If is an IBN-ring we denote the rank of left free modules over by . We now finally have the following theorem which we have strived for:

**Theorem: ***Let be an IBN-ring. Then, two free -modules are isomorphic if and only if . In particular, every free module is isomorphic to and implies .*

At this point, the existence of rings not satisfying the IBN property seems a little fishy, and so we shall now give the classic example of one. Let be an contably infinite dimensional vector space over some field (e.g. or for example) and consider the ring . We claim that as left -modules. Why is this true? Well, for define to be the function which carries an ordered basis to and does the analogous things, except exchanging the role of even and odd. We then define by . One can easily verify that is an isomorphism of left -modules and so our desired claim follows. So, we can be sure that there do exist rings which don’t satisfy the IBN property. But, as the motivation indicated most “nice” rings do. For example:

**Theorem: ***Every division ring is an IBN-ring.*

**Proof: **It suffices to show that if a left -module has two finite bases and then . The idea for this is simple enough, we shall assume without loss of generality that and show that we can successively replace the ‘s by ‘s and still get generating sets, and if then after steps we shall have the set is a proper generating subset of contradictory to the assumption that is a basis. Ok, so now that we have laid out the basic idea let’s get to it.

As stated, we assume without loss of generality that , and assume for a contradiction that, in fact, . Since generates we can find elements such that . Since at least one of the is nonzero we may assume without loss of generality (in any case, we could just reorder) that . Note then that since is a division ring this tells us that (upon division by ) . This clearly implies that every element of is in the span of and thus this latter set is a generating set. Thus, we can find such that . Now, we know that one of is nonzero and so if then and which contradicts that is linearly independent. Thus, one of is nonzero, and we may assume without loss of generality that . Thus, we see that and so by similar reasoning the previous case we have that is a generating set for . Continuing in this way we clearly arrive at the fact that is a generating set for , and since (by assumption) this is a proper subset of we’ve arrived at a contradiction. Thus, as desired.

This theorem really allows us to conclude that modules over division rings are very nice. Namely, since every module over a division ring has a basis we know that every module over a division ring is free, and since division rings are IBN rings we see then that every module over a division ring has well-defined notion of rank, and that this determines entirely the isomorphism type of the module. Of course, since every field is a division ring, this reproves the theorem for vector spaces.

This theorem actually allows us to prove that a certain class of rings, very arguably “larger” than the class of division rings. To do this we first make an observation of how to create submodules of a given -modules from ideals of . Namely, suppose that is a left -module and is an ideal of . We define . Note that it is obvious that . Of course then the quotient can be given the structure of an -module. But, what we want to show is that can be given the structure of a module in a meaningful way (i.e. in such a way that structure of still transfers to this module). Indeed, what we have is the following:

**Theorem: ***Let be a ring and a left -module. Then, for any left ideal of the multiplication defines a left -structure on which is unital if is. Moreover, if is free with basis then is a basis for the left -module .*

**Proof: **To prove that this defines a left -structure on we must merely check well-definedness, associativity, and both distributivities. We show only the first two, since the distributivities are done similarly. To see that this operations is well-defined we assume for a second that , we see then that there exists for which . We see then that

where, of course, the operative observation is that an so . To see that this multiplication is associative we merely note that

To see that is unital if is we merely note that .

It remains to show the basis property. Clearly generates and so it suffices to show that it is linearly independent. To see this we make the obvious observation that is a free module over with basis . Thus, if then we see that . By definition, since we can write it as where and . But, by definition we can find such that and so

and since is a basis for we may conclude that . But, note that the right hand side is in (since each is, and is left ideal). But, this implies that each and so . The conclusion follows.

With this we have the following, very nice, theorem:

**Theorem: ***Every commutative unital ring is an IBN-ring.*

**Proof: **Suppose that for some . By Krull’s theorem we can find some maximal ideal of . Note then that the isomorphism descends naturally to an isomorphism of left -modules. But, by the previous theorem we know that is an -dimensional vector space and an -dimensional vector space. Since vector spaces have well-defined dimension we may conclude that . Since was arbitrary, the conclusion follows.

In particular, we see that the notion of the rank of abelian groups makes sense.

**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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