Abstract Nonsense

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.

$\text{ }$

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 ($F\cong R^{\oplus \dim F}$). 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 $F\cong R^{\oplus \dim F}$, just not in the language of dimension. Namely, we defined a free module $F$ to be any unital module $F$ which is is isomorphic to a free module over a set $S$. Ok, but we proved that any two free modules over a given set $S$ are isomorphic, and since $R^{\oplus S}$ is a free module over the set $S$ we know that $F\cong R^{\oplus S}$. Yay, right? Well, yes and no. It may seem like a great result, and it seems so intuitive that defining $\#(S)=\dim F$ this says that $F\cong R^{\oplus\dim F}$. The problem is, if we are going to define dimension in this way we better be sure that we can’t find another set $E$ with $F\cong R^{\oplus E}$ and $\#(E)\ne \#(S)$. This is bound to be true, yeah? Well, no. Unfortunately we can actually find rings such that $R\cong R^2$ 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 $R^n\cong R^m$ implies $n=m$ for finite $n=m$ then the notion of “dimension” is well-defined for free left $R$-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).

$\text{ }$

Rank of  a Free Module

$\text{ }$

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:

$\text{ }$

Theorem: Let $R$ be a ring and $M$ a unital left $R$-module. Then, a subset $\left\{x_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a basis for $M$ 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 $M$).

Proof: Suppose first that $\mathscr{B}=\left\{x_\alpha\right\}_{\alpha\in\mathcal{A}}$ is a basis. To see that $\mathscr{B}$ is a maximally linearly independent set we suppose that there exists some element $x\in M-\mathscr{B}$ with $\mathscr{B}\cup\{x\}$ is linearly independent. Clearly $x\ne 0$ (since adding zero to any set makes it linearly dependent), and $\displaystyle x=\sum_{\alpha\in\mathcal{A}}r_\alpha x_\alpha$ and so $\displaystyle x+\sum_{\alpha\in\mathcal{A}}(-r_\alpha)x_\alpha=0$ and since not all the coefficients are zero (since $x\ne 0$) it follows that $\mathscr{B}\cup\{x\}$ is linearly dependent contradictory to assumption. Suppose now that $\mathscr{B}$ was not a minimal generating set. In particular, suppose there was a proper subset $\{x_\beta\}_{\beta\in\mathscr{B}}\subsetneq\{x_\alpha\}_{\alpha\in\mathcal{A}}$ which generates $M$. We note, then choosing $x_\alpha\notin\{x_\beta\}$ we see that, by assumption that $\{x_\beta\}$ generates $M$, we can write $\displaystyle \sum_{\beta\in\mathscr{B}}r_\beta x_\beta=x_\alpha$ but this contradicts the uniqueness of representation for the basis $\mathscr{B}$. Thus, there does not exist a proper subset of $\mathscr{B}$ that generates $M$.

$\text{ }$

The converse is obvious. $\blacksquare$

We begin by proving a somewhat interesting theorem. Namely, we claimed in the above motivation that we can find free $R$-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:

$\text{ }$

Theorem: Let $F$ be a free module over a ring $R$. If $F$ possesses an infinite basis $\mathscr{B}$ then every basis for $F$ is equipotent to $\mathscr{B}$.

Proof: We begin by showing that $F$ cannot have a finite basis. This is simple enough, for if $\{x_1,\cdots,x_n\}$ is another finite basis of we know that for each $x_k$ we can find finitely many elements $\{x_{k,1},\cdots,x_{k,m_k}\}$ in $\mathscr{B}$ which span to contain $x_k$. Clearly then $\{x_1,\cdots,x_n\}\subseteq\{x_{k,i}:k\in[n]\text{ and }i\in[m_k]\}$ but since this right hand set contains in its span a basis it clearly must generate $M$ and since it is clearly a proper subset of $\mathscr{B}$ this contradicts the minimal generating set property of the basis $\mathscr{B}$. Thus, we see that every possible basis for $F$ is infinite.

$\text{ }$

We now show that they all possible infinite bases for $F$ have the same cardinality. Indeed, suppose that $\mathscr{B}'$ is another basis for $F$, which by the previous paragraph, is infinite.  So, consider the map $f:\mathscr{B}\to \mathscr{P}_0(\mathscr{B}')$ (where $\mathscr{P}_0(\mathscr{B}')$ denotes the set of finite subsets of $\mathscr{B}'$) defined by sending $x\in\mathscr{B}$ to the unique finite set of elements of $\mathscr{B}'$ that generate it.

$\text{ }$

We claim that each element of $S\in\mathscr{P}_0(\mathscr{B}')$ has finite preimage. Indeed, we note that $f^{-1}(S)\subseteq\text{span }(S)$, but since the number of elements of $S$ is finite we can find a finite subset (similar to how we constructed one in the previous paragraph) $T$ of $\mathscr{B}$ which generates $S$ and so $f^{-1}(S)\subseteq\text{span }(T)$, but clearly then by linear independence this allows us to conclude that $f^{-1}(S)\subseteq T$ and so $f^{-1}(S)$ is finite as desired.

$\text{ }$

From this it’s easy to see that $\#(\mathscr{B})\leqslant \#(\mathscr{P}_0(\mathscr{B}'))$ since identifying elements of $\mathscr{B}$ into finite equivalence classes under the relation $x\sim y$ if and only if $f(x)=f(y)$ gives us a set $\mathscr{B}/\sim$ which is equipotent to $\mathscr{B}$ (since $\mathscr{B}$ is infinite) and injects $\mathscr{B}/\sim\;\hookrightarrow\mathscr{P}_0(\mathscr{B}')$. But, since $\mathscr{B}'$ is infinite it is basic set theory that $\#(\mathscr{P}_0(\mathscr{B}'))=\#(\mathscr{B}')$ and so by the above analysis we may conclude that $\#(\mathscr{B})\leqslant\#(\mathscr{B}')$. But, by symmetry we may also conclude that $\#(\mathscr{B}')\leqslant\#(\mathscr{B})$ and so by the Schroeder-Bernstein theorem we may conclude that $\#(\mathscr{B})=\#(\mathscr{B}')$.

$\text{ }$

Since $\mathscr{B}'$ was arbitrary the conclusion follows. $\blacksquare$

$\text{ }$

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 $R$ is such that there exists a unital left free module $M$ with not well-defined “dimension”. This should mean that we can find two bases $\mathscr{B},\mathscr{B}'$ which aren’t equipotent, and by the above theorem it must be the case that both these bases are finite. By noting that $R^{\oplus\mathscr{B}}\cong M\cong R^{\oplus\mathscr{B}'}$ we see that a ring having a not well-defined “dimension” for its modules is the existence of $m,n\in\mathbb{N}$ with $m\ne n$ yet $R^m\cong R^n$. Thus, if we define rings $R$ with the property that $R^m\cong R^n$ for $m,n\in\mathbb{N}$ implies $m=n$ 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

$\text{ }$

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

$\text{ }$

$\text{ }$

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.

November 17, 2011 -

1. […] Rank and the IBN Property (Pt. II) Point of Post: This is a continuation of this post […]

Pingback by Rank and the IBN Property (Pt. II) « Abstract Nonsense | November 17, 2011 | Reply

2. […] (making the whole pursuit, while educational, moot).  Since is commutative we know that it is an IBN-ring and so for to be a free -module it needs to be isomorphic to precisely one of the […]

Pingback by Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt.I) « Abstract Nonsense | November 21, 2011 | Reply

3. […] a basis for . In particular, the arbitrary coproduct of free modules is free. Moreover, if is an IBN-ring then […]

Pingback by Coproduct of Free Modules are Free, but not Arbitrary Products « Abstract Nonsense | November 23, 2011 | Reply

4. […] a left ideal in . Define for each the module to be equal to (where this notion has been defined before). We define mappings for each by the rule . Note that this actually makes sense since (since ). […]

Pingback by Inverse Limits of Modules (Pt. I) « Abstract Nonsense | December 9, 2011 | Reply

5. […] of scalars on a free module results in a free module, and in the case that (and thus ) have the IBN property they have the same […]

Pingback by Extension of Scalars(Pt. II) « Abstract Nonsense | January 25, 2012 | Reply

6. […] and so, in particular, is a -map. Since then is a -isomorphism and commutative rings have the IBN property we may conclude that as […]

Pingback by Q and Q^2 are not Isomorphic as Groups « Abstract Nonsense | January 31, 2012 | Reply

7. […] be a field and an extension. We define the index of , denoted , to be the dimension where we are thinking about as an […]

Pingback by The Degree of an Extension « Abstract Nonsense | March 8, 2012 | Reply

8. […] Theorem: Commutative rings have the IBN property. […]

Pingback by The Tensor Algebra and Exterior Algebra (Pt. V) « Abstract Nonsense | May 10, 2012 | Reply

9. […] Theorem: Commutative rings have the IBN property. […]

Pingback by Tensor Algebra and Exterior Product (Pt. VI) « Abstract Nonsense | May 21, 2012 | Reply

10. […] be a lattice in (i.e. a discrete subgroup of of rank ), then we know that for -independent complex numbers and (if this isn’t known to you, […]

Pingback by Riemann Surfaces (Pt. II) « Abstract Nonsense | October 2, 2012 | Reply