# Abstract Nonsense

## Rank and the IBN Property (Pt. II)

Point of Post: This is a continuation of this post

$\text{ }$

If $R$ is an IBN-ring we denote the rank of left free modules $F$ over $R$ by $\text{rank}(F)$. We now finally have the following theorem which we have strived for:

$\text{ }$

Theorem: Let $R$ be an IBN-ring. Then, two free $R$-modules $F,K$ are isomorphic if and only if $\text{rank}(F)=\text{rank}(K)$. In particular, every free module $F$ is isomorphic to $R^{\oplus\text{rank}(F)}$ and $F\cong R^{\oplus \lambda}$ implies $\lambda=\text{rank}(F)$.

$\text{ }$

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 $V$ be an contably infinite  dimensional vector space over some field $k$ (e.g. $\mathbb{R}[x]$ or $L^1([a,b])$ for example) and consider the ring $R=\text{End}_k(V)$. We claim that $R\cong R^2$ as left $R$-modules. Why is this true? Well, for $\varphi\in\text{End}_k(V)$ define $\varphi_e$ to be the function which carries an ordered basis $(x_1,x_2,x_3,x_4\cdots)$ to $(0,\varphi(x_2),0,\varphi(x_4),\cdots)$ and $\varphi_0$ does the analogous things, except exchanging the role of even and odd. We then define $f:\text{End}_k(V)\to\text{End}_k(V)^2$ by $\varphi\mapsto (\varphi_e,\varphi_0)$. One can easily verify that $f$ is an isomorphism of left $\text{End}_k(V)$-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:

$\text{ }$

Theorem: Every division ring $R$ is an IBN-ring.

Proof: It suffices to show that if a left $R$-module $F$ has two finite bases $\{x_1,\cdots,x_n\}$ and $\{y_1,\cdots,y_m\}$ then $m=n$. The idea for this is simple enough, we shall assume without loss of generality that $m\geqslant n$ and show that we can successively replace the $x_i$‘s by $y_i$‘s and still get generating sets, and if $m>n$ then after $n$ steps we shall have the set $\{y_1,\cdots,y_n\}$ is a proper generating subset of $\{y_1,\cdots,y_m\}$ contradictory to the assumption that $\{y_1,\cdots,y_m\}$ is a basis. Ok, so now that we have laid out the basic idea let’s get to it.

$\text{ }$

As stated, we assume without loss of generality that $m\geqslant n$, and assume for a contradiction that, in fact, $m>n$. Since $\{x_1,\cdots,x_n\}$ generates $F$ we can find elements $r_1,\cdots,r_n\in R$ such that $y_1=r_1x_1+\cdots+r_nx_n$. Since at least one of the $r_i$ is nonzero we may assume without loss of generality (in any case, we could just reorder) that $r_1\ne 0$. Note then that since $R$ is a division ring this tells us that  (upon division by $r_1$) $x_1=r_1^{-1}y+r_1^{-1}r_2x_2+\cdots+r_1^{-1}r_nx_n$. This clearly implies that every element of $\{x_1,\cdots,x_n\}$ is in the span of $\{y_1,x_2,\cdots,x_n\}$ and thus this latter set is a generating set. Thus, we can find $s_1,\cdots,s_n\in R$ such that $y_2=s_1y_1+s_2x_2+\cdots+s_nx_n$. Now, we know that one of $s_1,\cdots,s_n$ is nonzero and so if $s_2=\cdots=s_n=0$ then $s_1\ne 0$ and $y_2=s_1y_1$ which contradicts that $\{y_1,\cdots,y_m\}$ is linearly independent. Thus, one of $s_2,\cdots,s_n$ is nonzero, and we may assume without loss of generality that $s_2\ne 0$. Thus, we see that $x_2=s_2^{-1}y_2+s_2^{-1}s_1+\cdots+s_2^{-1}s_nx_n$ and so by similar reasoning the previous case we have that $\{y_1,y_2,\cdots,x_n\}$ is a generating set for $F$. Continuing in this way we clearly arrive at the fact that $\{y_1,\cdots,y_n\}$ is a generating set for $F$, and since (by assumption) this is a proper subset of $\{y_1,\cdots,y_m\}$ we’ve arrived at a contradiction. Thus, $m=n$ as desired. $\blacksquare$

$\text{ }$

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.

$\text{ }$

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 $R$-modules from ideals of $R$. Namely, suppose that $M$ is a left $R$-module and $\mathfrak{a}$ is an ideal of $R$. We define $\displaystyle \mathfrak{a}M=\left\{\sum_{\text{finite}}a_im_i:a_i\in\mathfrak{a}\text{ and }m_i\in M\right\}$. Note that it is obvious that $\mathfrak{a}M\leqslant M$. Of course then the quotient $M/\mathfrak{a}M$ can be given the structure of an $R$-module. But, what we want to show is that $M/\mathfrak{a}M$ can be given the structure of a $R/\mathfrak{a}$ module in a meaningful way (i.e. in such a way that structure of $M$ still transfers to this module). Indeed, what we have is the following:

$\text{ }$

Theorem: Let $R$ be a ring and $M$ a left $R$-module. Then, for any left ideal $\mathfrak{a}$ of $R$ the multiplication $(r+\mathfrak{a})(m+\mathfrak{a}M)=rm+\mathfrak{a}M$ defines a left $R/\mathfrak{a}$-structure on $M/\mathfrak{a}M$ which is unital if $M$ is. Moreover, if $M$ is free with basis $\{x_\alpha\}_{\alpha\in\mathcal{A}}$ then $\{x_\alpha+\mathfrak{a}M\}_{\alpha\in\mathcal{A}}$ is a basis for the left $R/\mathfrak{a}$-module $M/\mathfrak{a}M$.

Proof: To prove that this defines a left $R/\mathfrak{a}$-structure on $M/\mathfrak{a}M$ 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 $x+\mathfrak{a}=y+\mathfrak{a}$, we see then that there exists $a\in\mathfrak{a}$ for which $x=y+a$. We see then that

$\text{ }$

\begin{aligned}(x+\mathfrak{a})(m+M/\mathfrak{a}M) &=xm+M/\mathfrak{a}M\\ &=(ym+am)+M/\mathfrak{a}M\\ &=ym+M/\mathfrak{a}M\\ &=(y+\mathfrak{a})(m+M/\mathfrak{a}M)\end{aligned}

$\text{ }$

where, of course, the operative observation is that $am\in \mathfrak{a}M$ an so $am+M/\mathfrak{a}M=0$. To see that this multiplication is associative we merely note that

$\text{ }$

\begin{aligned}((x+\mathfrak{a})(y+\mathfrak{a}))(m+M/\mathfrak{a}M) &= (xy+\mathfrak{a})(m+M/\mathfrak{a}M)\\ &= (xy)m+M/\mathfrak{a}M\\ &= x(ym)+M/M\mathfrak{a}M\\ &= (x+\mathfrak{a})(ym+M/\mathfrak{a}M)\\ &= (x+\mathfrak{a})((y+\mathfrak{a})(m+M/\mathfrak{a}M))\end{aligned}

$\text{ }$

To see that $M/\mathfrak{a}M$ is unital if $M$ is we merely note that $(1+\mathfrak{a})(m+M/\mathfrak{a}M)=1m+M/\mathfrak{a}M=m+M/\mathfrak{a}M$.

$\text{ }$

It remains to show the basis property. Clearly $\{x_\alpha+M/\mathfrak{a}M\}_{\alpha\in\mathcal{A}}$ generates $M/\mathfrak{a}M$ and so it suffices to show that it is linearly independent. To see this we make the obvious observation that $\mathfrak{a}M$ is a free module over $\mathfrak{a}$ with basis $\{x_\alpha\}_{\alpha\in\mathcal{A}}$. Thus, if $\displaystyle \sum_{\alpha\in\mathcal{A}}(r_\alpha+\mathfrak{a})(x_\alpha+M/\mathfrak{a}M)=0$ then we see that $\displaystyle \sum_{\alpha}r_\alpha x_\alpha\in M/\mathfrak{a}M$. By definition, since $\displaystyle \sum_\alpha r_\alpha x_\alpha\in M/\mathfrak{a}M$ we can write it as $\displaystyle \sum_i a_i m_i$ where $a_i\in \mathfrak{a}$ and $m_i\in M$. But, by definition we can find $s_{i,\alpha}\in R$ such that $\displaystyle m_i=\sum_{\alpha}s_{i,\alpha}x_\alpha$ and so

$\text{ }$

$\displaystyle \sum_\alpha r_\alpha x_\alpha=\sum_i a_i m_i=\sum_\alpha (\sum_i a_i s_{i,\alpha})x_\alpha$

$\text{ }$

and since $\{x_\alpha\}$ is a basis for $M$ we may conclude that $\displaystyle r_\alpha=\sum_i a_i s_{i,\alpha}$. But, note that the right hand side is in $\mathfrak{a}$ (since each $a_i$ is, and $\mathfrak{a}$ is left ideal). But, this implies that each $r_\alpha\in\mathfrak{a}$ and so $r_\alpha+\mathfrak{a}=0$. The conclusion follows. $\blacksquare$

$\text{ }$

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

$\text{ }$

Theorem: Every commutative unital ring $R$ is an IBN-ring.

Proof: Suppose that $R^n\cong R^m$ for some $m,n\in\mathbb{N}$. By Krull’s theorem we can find some maximal ideal $\mathfrak{m}$ of $R$. Note then that the isomorphism $R^n\cong R^m$ descends naturally to an isomorphism $R^n/\mathfrak{m}R^n\cong R^m/\mathfrak{m}R^m$ of left $R/\mathfrak{m}$-modules. But, by the previous theorem we know that $R^n/\mathfrak{m}R^n$ is an $n$-dimensional vector space and $R^m/\mathfrak{m}R^m$ an $m$-dimensional vector space. Since vector spaces have well-defined dimension we may conclude that $m=n$. Since $m,n$ was arbitrary, the conclusion follows. $\blacksquare$

$\text{ }$

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

$\text{ }$

$\text{ }$

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 -