# Abstract Nonsense

## Free Modules (Pt. II)

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

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Bases and their Relation to Free Modules

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We now define the notion of basis for a module and show that the existence of a basis is equivalent to a module being free. So, we begin by making the obvious definition. We call a finite set $\{m_1,\cdots,m_n\}$  elements of a left $R$-module $M$ linearly independent if whenever $\displaystyle \sum_i r_im_i=0$ implies that $r_1=\cdots=r_n=0$. We say that an arbitrary subset $\left\{m_\alpha\right\}_{\alpha\in\mathcal{A}}$ is linearly independent if every finite subset is linearly independent. A basis for $M$ is a set $\mathscr{B}$ of elements of $M$ which both generates $M$ and is linearly independent.

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Of course, the more standard way to think about a basis for a module is the following:

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Theorem: Let $M$ be a left $R$-module. Then, $\mathscr{B}$ is a basis for $M$ if and only if every element of $M$ can be written uniquely as an $R$-linear combination of elements of $\mathscr{B}$.

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which is trivial to prove.

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The obvious connection between our previously defined notions of free module and modules with bases are that $\text{im }i$ should clearly be a basis for our free module and conversely taking $S$ to be a basis and defining $i$ to be inclusion should give us a free module. Indeed, we prove the first of these statements:

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Theorem: Let $(F,i)$ be a free module over a set $S$. Then, $\text{im }i$ is a basis for $F$.

Proof: We have already proven that $\text{im }_i$ generates $F$ and so it suffices to prove that every element of $F$ cannot be written in two different ways with elements of $\text{im }_i$. To see this, suppose that $\displaystyle \sum_{s\in S}r_s i(s)=\sum_{s\in S}t_s i(s)$. Define the maps $\zeta,\eta:S\to F$ by $\zeta(s)=r_s$ and $\eta(s)=t_s$. Clearly then the guaranteed extension maps are equal (since they are just the two sums, which by assumption are equal) and so $\eta=\zeta$ and so $r_s=t_s$. The conclusion follows. $\blacksquare$

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With this we can prove what intuition tells us, that free modules are precisely those modules with a basis:

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Theorem: Let $R$ be a unital ring and $F$ a unital left $R$-module. Then, $F$ is free if and only if $F$ admits a basis.

Proof: By the above theorem we know that if $F$ is free then $F$ is isomorphic to a free module $(K,i)$ over some set $S$, let’s call the isomorphism $\phi:K\xrightarrow{\approx}F$. We know $\text{im }i$ is a basis for $K$ and it’s not hard to see that $\phi(\text{im }i)$ is a basis for $F$.

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Conversely, suppose that $F$ admits a basis $\mathscr{B}$. We claim that $(F,i)$, with $i:\mathscr{B}\hookrightarrow F$ the usual inclusion, is a free module over $\mathscr{B}$. Indeed, suppose that $N$ is any left $R$-module and $f:\mathscr{B}\to N$ is any set map. We then define the map $\displaystyle g:F\to N$ by writing every element of $F$ in its unique form $\displaystyle \sum_{s\in S}r_s i(s)$ and defining $g$ of this element to be $\displaystyle \sum_{s\in S}r_s f(s)$. This is easily seen to be a well-defined $R$-map on $F$ which satisfies $g\circ i=f$. Moreover, it’s clear that any $R$-map satisfying this composition equality would have to be this map as can be seen by just expanding by linearity. Thus, $(F,i)$ really is a free module on $\mathscr{B}$. $\blacksquare$

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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.