# Abstract Nonsense

## Free Modules (Pt. I)

Point of Post: In this we lay out the definition of free modules in terms of their universal characterization and then show this is equivalent to the existence of a basis.

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Motivation

I said, in my first post discussing modules, that the idea behind module theory was that it was, in a very real sense, a generalization of linear algebra. Well, in a sense, no one can argue with this based on what we have been doing.  Every topic we’ve discussed up until this point has been “linear algebraic” in its feel and applications. Indeed, we could very well define a lexicon transfer from the language of linear algebra to the more general language of module theory: submodules are subspaces, $R$-maps are linear transformations, direct sums are direct sums, etc. That said, perhaps one of the most fundamental concepts in linear algebra, the idea that makes linear algebra (at least structurally) so simple–the idea of bases has been entirely absent from our conversation. One is sure to remember that the existence of bases for vector spaces made working with vector spaces so incredibly easy–if one wanted to define a linear transformation on a vector space one must define it only a basis and “expand by linearity”, for finite dimensional spaces we get (in fact from the previous idea) the duality between $\text{End}_k(V)$ and $\text{Mat}_n(k)$,  we have the fact that structurally vector spaces have only one invariant (in the sense that it’s the only thing keeping two vector spaces from either being isomorphic or not), dimension. In fact, this last statement about dimension is so strong that it, in a sense, reduces studying vector spaces over a field $k$ to studying $k^{\oplus\lambda}$ for cardinals $\lambda$ (i.e. direct sums of $k$ with itself, arbitrary [perhaps infinitely] many times) since one knows that any vector space $V$ over $k$ is isomorphic to a vector space of that form, and even uniquely (in the sense that the cardinal $\lambda$ is unique). If you stop and think about it, this is a statement that vector spaces are, in a very concrete sense, as simple of a structure as one could hope.

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Ok, fine, so vector spaces are extremely (structurally) simple, so why don’t we define the analogous notion for modules and reap the same benefits! Simply said, we cannot. The notion of “bases” and “dimension” breaks down when we start doing module theory over rings that are not fields (or, more generally, division rings). In fact, I claim that this should be immediately obvious, for a perhaps stupid reason. Namely, we know that $\mathbb{Z}$-modules are just abelian groups, right? Well, if the notion of basis transferred to the notion of $\mathbb{Z}$-modules then we’d have the amazing statement that every abelian group is isomorphic to $\mathbb{Z}^{\oplus\lambda}$ for some cardinal $\lambda$! Quite an amazing fact considering the existence of finite (nontrivial) abelian groups.

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Ok, so it should be now clear that we’re not going to be able to transfer the notion of bases to module theory. That said, let me repeat it, the existence of bases is a damn nice property for a given module to have. So much so, that the modules that do have this property, the existence of a basis, shall play a pivotal role in the general theory–they are the “nicest” examples. So, the question remains how precisely to define when a module should have this property of “having a basis”, called free modules, but in more ‘adult’ language. How do we phrase the definition of  free modules in a way that immediately tells us what makes them “free”, what is the operative part of their definition. Well, the first inkling of how to do this comes from a feeling most people reading this may have had at one point or another when doing vector space linear algebra. In particular, sometimes it really felt like doing things with vector spaces was really just doing things to a particular set, namely a (fixed) set of basis vectors. We do whatever we want with the set, and we know that we can extend it in a way that always “works out”.  To belabor the point, at the center of everything is some set $S$. This set $S$ sits inside our vector space $V$, in other wordsthere exists a map $i:S\to V$. And we know that anything we do to $S$ can be extended uniquely to something we do on $V$, or more seriously given any set map $f:S\to W$ where $W$ is some vector space, we can extend $f$ uniquely to some linear transformation $\widetilde{f}:V\to W$ in the sense that $\widetilde{f}\circ i=f$. This tells us that abstractly $V$ depends on this set $S$ in such a way that in terms of doing things ( i.e. defining functions)  $S$ is the only thing that matters. This is precisely how we shall abstract the notion of “having a basis”, of being “free” to more general $R$-modules. Of course, this formulation is precisely (although maybe not so apparently) equivalent to the existence of a basis (defined appropriately) for our space.

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Free Modules

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Let $S$ be a set and $R$ a unital ring. The ordered pair $\left(M,i\right)$ is a free left $R$-module over $S$ if $M$ is a left $R$-module and $i:S\to M$ is a set map which satisfies the following universal property: for any left $R$-module $N$ and and set map $f:S\to N$ there exists a unique $R$-map $g:M\to N$ such that the following diagram commutes

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$\begin{matrix}S & \overset{i}{\longrightarrow} & M\\ & _{f}\searrow & \big\downarrow{^g}\\ & & N\end{matrix}$

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We say that an $R$-module $F$ is free if there exists some free module $(M,i)$ over a set $S$ with $F\cong M$.

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The first natural thing to ask is whether, as in the case of products and coproducts whether this map $i$ has to have any special properties that, while not present in the definition, are consequences of the universal properties. In fact, this is true, and if we hope for free modules to be like vector spaces they are imperative qualities:

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Theorem: Let $(F,i)$ be a free left $R$-module over $S$. Then, $i$ is injective and $\text{im }i$ generates $F$.

Proof: Choose any $R$-module with at least $\#(S)$ many elements, the ring $R^S$ for example (assuming $R\ne\{0\}$, but this case is uninteresting an trivial anyways). Consider then any injection $f:S\to R^S$ and note that by assumption we can find an an $R$-map $g:F\to R^S$ with $g\circ i=f$. Now, this implies that $g\circ i$ is an injection, and from basic set theory we may conclude that $i$ is an injection.

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Now, we must merely prove that $\text{im }i$ generates $F$. To do this we just do some magic with the universal property again. Namely, we consider the obvious maps $S\xrightarrow{f}\text{im }\xrightarrow{\iota}\langle\text{im }a\rangle\xrightarrow{\iota_A}F$. By definition we can find some $g:F\to A$ such that $\iota\circ f=g\circ f$. Consider the map $\iota_A\circ g:F\to F$, this is an $R$-map and it’s not hard to see that $(\iota_A\circ g)\circ f=\iota_A\circ \iota\circ f$, but since evidently this is true if we replace $\iota_A\circ g$ by $\text{id}_F$ we may conclude by uniqueness that $\iota_A\circ g=\text{id}_F$ and so evidently $\iota_A$ is surjective, which implies the result. $\blacksquare$

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What we’d now like to verify is that free modules are unique up to isomorphism. Indeed:

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Theorem: Let $(F,i)$ and $(K,j)$ be any two free modules over the set $S$. Then, $F\cong K$.

Proof: Since $i:S\to F$ we know that there exists a unique morphism $f:K\to F$ with $f\circ j=i$. Similarly, we can product a unique morphism $g:K\to F$ with $g\circ i=j$. But, clearly then we see that $f\circ g\circ i=f\circ j=i$ and $g\circ f\circ j=g\circ i=j$. But, since $\text{id}_K$ and $\text{id}_F$ are clearly maps which do this, and by uniqueness we can conclude that $f\circ g=\text{id}_K$ and $g\circ f=\text{id}_F$ and so $f$ is an isomorphism. $\blacksquare$

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The only thing left to verify is that free modules exist on any given set. This is easy enough since, following our intuition gained from vector spaces, all free modules should just be direct sums. Thus, what we claim is that $\displaystyle \left(R^{\oplus S},i\right)$, where $i:S\to R^{\oplus S}$ is defined by having the $s^{\text{th}}$ coordinate of $i(s)$ be $1$ and $0$ elsewhere, is a free module on $S$. Indeed, given any left $R$-module $N$ and any map $f:S\to N$ we can define a natural map $g:R^{\oplus S}\to N$ by defining

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$\displaystyle g((x_s))=\sum_{s\in S}x_sf(s)$

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(i.e. we just extended by linearity). It is not hard to verify that $g$ is, in fact, a well-defined $R$-map $R^{\oplus S}\to N$ with $g\circ i=f$, and moreover $g$ is unique with respect to this last property. Thus, $R^{\oplus S}$ is a free module over $S$, and so we at least know free modules always exist.

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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 16, 2011 -

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