# Abstract Nonsense

## Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt.I)

Point of Post: In this post we show that submodules of free modules need not be free, and give some intuition why this should be true. We then show that, in fact, submodules of $R$-modules hold (for a unital commutative ring $R$) precisely when $R$ is a PID.

$\text{ }$

Motivation

Now that we have discussed free modules we’d like to discuss some of the properties that hold in these free modules. In particular, while tempting it is, in general, wrong to take theorems about the “nicest free modules” (i.e. vector spaces/modules over division rings) and transfer these to general theorems about free modules. This post shall be devoted to resolving one such “inaccurate transfer”. Namely, the “theorem” that every submodules of a free module is free. For example, everyone knows that submodules of vector spaces, or more generally submodules of modules over division rings, are free. Why? Well, the obvious reason–they’re still vector spaces/modules over division rings which we know are always free! Ok, now try to make an analogous proof for a general free left $R$-module $F$ and some submodule $N\leqslant F$. Since we know that, in general, all modules over a given ring are not free (in fact, we shall eventually (and this is a real eventually) show that the only rings for which this statement is true are division rings) we can’t rely on this argument. So, one starts to try to create a basis for $F$ in terms of a basis for $N$. Hmm, a little thought shows that there is no natural way of doing this. We could “hope” that we can nab a basis for $F$ which contains elements of $N$, but we can’t really see a way of necessarily doing this (one may be inclined to, hopefully to some mild embarrassment, try the old linear algebra trick of  “extend a basis from $N$ to $F$“, which of course has a little logical flaw). Ok, fine, so at this point we’ve seen that there is really no natural way of showing that a submodule of a free module isn’t free, but perhaps this is because we aren’t being clever enough, the nonexistence of a proof is not a disproof. As I said though, this is, in fact, a non-theorem. Intuitively, there is an example that typifies the reason why this “theorem” sometimes fails. We know that if we are given a ring $R$ and a left ideal $\mathfrak{a}$ of $R$ then $\mathfrak{a}$ is a submodule of the free $R$-module $R$ and so if our “theorem” was true this would imply that $\mathfrak{a}\cong R^{\oplus \lambda}$ for some cardinal $\lambda$. That’s a pretty steep condition, no? We see that our good friends fields and division rings side-step this issue all together being precisely the rings for which the only ideals of $R$ are $\{0\}$ and $R$, or $R^0$ and $R^1$ respectively. But, if we take, for example some finite ring $R$ which is not a division ring then there exists some proper non-trivial ideal $\mathfrak{a}$. Clearly then we can’t have that $\mathfrak{a}\cong R^{\oplus\lambda}$ for some $\lambda$ by merely appealing to a cardinality argument, and so $\mathfrak{a}$ cannot be a free $R$-module. So, this immediately gives us a whole class of examples. For instance, $\mathbb{Z}_6$ is a free $\mathbb{Z}_6$-module and $2\mathbb{Z}_6$ is an ideal, but is surely not a free $\mathbb{Z}_6$ modules since any such (finite) module would have order $6^n$ for some $n\in\mathbb{N}\cup\{0\}$ and $|2\mathbb{Z}_6|=3$. This is a typifying example precisely because, in the category of commutative unital rings, this is the only possible obstruction in the sense that if a ring does not have this obstruction (i.e. every ideal is a free module) then one can generally state that submodules of free modules over that ring are free.

$\text{ }$

So, when does a commutative unital ring $R$ have this property, that every ideal ideal of $R$ is a free $R$-module? Well, let’s see if we can get our hands a little dirty and look at some classic examples of rings (which aren’t division rings) and see when this property sticks. As always, we got to our two fallback rings $\mathbb{Z}$ and $R[x]$ for some ring $R$. Well, clearly $\mathbb{Z}$ does satisfy this property since any non-trivial ideal (recalling that $\mathbb{Z}$ is a PID) is of the form $n\mathbb{Z}$ for $n\in\mathbb{N}$ and so trivially isomorphic to $\mathbb{Z}$ as $\mathbb{Z}$-modules–both are infinite cyclic. So, now let’s look at one of our favorite polynomial rings, $\mathbb{Z}[x]$. I claim that the ideal $(2,x)$ is not a free $\mathbb{Z}[x]$-module. Let’s try to figure out why without just initially ruining the surprise by just telling you why, let’s go through the (at least to me) obvious progression which will eventually lead to “the correct answer” (making the whole pursuit, while educational, moot).  Since $\mathbb{Z}[x]$ is commutative we know that it is an IBN-ring and so for $(2,x)$ to be a free $\mathbb{Z}[x]$-module it needs to be isomorphic to precisely one of the following

$\text{ }$

$\mathbb{Z}[x]^0=\{0\},\mathbb{Z}[x]^1=\mathbb{Z}[x],\mathbb{Z}[x]^2,\mathbb{Z}[x]^3,\cdots,\mathbb{Z}[x]^{\oplus\aleph_0},\cdots$

$\text{ }$

Thus, one’s first intuition is to try to eliminate these one-by-one. It clearly can’t be isomorphic to $\{0\}$ for cardinality reasons, it can’t be isomorphic to $\mathbb{Z}[x]$ because then $(2,x)$ is $1$-generated, but this is impossible since it is a (classic) example of a non-principal ideal. Similarly, $(2,x)$ can’t be isomorphic to $\mathbb{Z}[x]^n$ or any cardinal $n>2$ then since any isomorphism $(2,x)\to\mathbb{Z}[x]^n$ would allow us to show that $\mathbb{Z}[x]^n$ is $2$-generated (since $(2,x)$ is $2$-generated) and since, as noted, $\mathbb{Z}[x]$ is an IBN-ring this is impossible. Thus, our only chance for $(2,x)$ to be free is it to be isomorphic to $\mathbb{Z}[x]^2$. So, why is this impossible? Since they are both $2$-generated (minimally) this argument is no longer applicable, and so we need to be more inventive. Having used up the one characteristic of a basis, it seems natural to try to use the other. Namely, we know we can find two $\mathbb{Z}$-independent elements of $\mathbb{Z}[x]^2$, can we say the same of $(2,x)$? No! Indeed, we strike upon the “real” reason why $(2,x)$ can’t be free! Any two elements $t(x),s(x)$ of $(2,x)$ are $\mathbb{Z}[x]$-dependent since $t(x)s(x)-s(x)t(x)=0$!  Thus, $(2,x)$ can’t be isomorphic to $\mathbb{Z}[x]^2$ and since we have exhausted our list, can’t be free.

$\text{ }$

More generally, if our ring is commutative we have that, in general, $xy-yx=0$ for all $x,y\in R$ and so if we can take $\alpha=x$ and $\beta=-y$ and see that $x,y$ are not $R$-independents since $\alpha x+\beta y=0$ and $\alpha,\beta\ne 0$. Thus, we see that for an ideal of $R$ to be free it must be $1$-generated (since there are no linearly dependent sets of size greater than one), or said in different language, it must be a principal ideal. Moreover, we can clearly see that the proof that $\mathbb{Z}$ does satisfy the property that every ideal is free was precisely the fact that $\mathbb{Z}$ is a PID. Thus, it’s not hard to conclude (although we shall give a formal proof) that a (commutative unital) ring satisfying the property that every ideal is free are precisely the PIDs.

$\text{ }$

Rings for Which all Submodules are Free

$\text{ }$

What we now propose is the following (the proof $\mathbf{(3)}\implies\mathbf{(1)}$ is based off the proof in [4] on page 218)

$\text{ }$

Theorem: Let $R$ be a commutative unital ring. Then, the following are equivalent

$\text{ }$

\begin{aligned}&\mathbf{(1)}\quad\textit{Every submodule of every free left }R\textit{-module is free}\\ &\mathbf{(2)}\quad\textit{Every ideal of }R\textit{ is free}\\ &\mathbf{(3)}\quad R\text{ is a PID}\end{aligned}

Proof:

$\mathbf{(1)}\implies\mathbf{(2)}:$ This is immediate.

$\text{ }$

$\mathbf{(2)}\implies\mathbf{(3)}:$ Since $R$ is already commutative and unital it suffices to show that $R$ has no zero divisors and is a principal ideal ring. To do the former, let $a\in R$ be arbitrary. By assumption $(a)$ is a free $R$-module. Since $(a)$ is $1$-generated, nonzero, and $R$ is an IBN-ring we know that $(a)$ is rank one and so has a basis of the form $\{ra\}$ for some $r\in R$. Note that this implies that $sra=0$ implies $s=0$. So, if $a$ were a zero divisor then there would exists nonzero $s\in R$ with $sa=0$ and since $R$ is commutative we’d then have that $sra=rsa=0$ contradictory to assumption. Thus, $a$ is not a zero divisor, and since $a$ was arbitrary we see that $R$  has no zero divisors. Now, to prove that $R$ is a principal ideal ring we merely note that if $\mathfrak{a}$ is a nonzero ideal of $R$, then by assumption $\mathfrak{a}$ is free $\mathfrak{a}$ has a basis. But, since $rs-sr=0$ for all $s,r\in R$ we see that any subset of $\mathfrak{a}$ containing more than one element is $R$-dependent and so the basis for $\mathfrak{a}$ must be size one (it can’t be size zero, because then $\mathfrak{a}$ is trivial) but this says that $\mathfrak{a}$ is $1$-generated, or in other words, principal. Since $\mathfrak{a}$ was arbitrary the conclusion follows.

$\text{ }$

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.

[4]  Hungerford, Thomas W. Algebra. New York: Holt, Rinehart and Winston, 1974. Print.