Abstract Nonsense

Crushing one theorem at a time

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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[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 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , ,

1 Comment »

  1. […] Let be a ring and a unital left -module. Then, a subset is a basis for if and only if it is a maximally linearly independent set (i.e. no linearly independent set […]

    Pingback by Rank and the IBN Property « Abstract Nonsense | November 17, 2011 | Reply

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