## Free Modules (Pt. II)

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

*Bases and their Relation to Free Modules*

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 elements of a left -module *linearly independent* if whenever implies that . We say that an arbitrary subset is *linearly independent *if every finite subset is linearly independent. A *basis *for is a set of elements of which both generates and is linearly independent.

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

**Theorem: ***Let be a left -module. Then, is a basis for if and only if every element of can be written uniquely as an -linear combination of elements of .*

which is trivial to prove.

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

**Theorem:*** Let be a free module over a set . Then, is a basis for .*

**Proof: **We have already proven that generates and so it suffices to prove that every element of cannot be written in two different ways with elements of . To see this, suppose that . Define the maps by and . Clearly then the guaranteed extension maps are equal (since they are just the two sums, which by assumption are equal) and so and so . The conclusion follows.

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

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

**Proof: **By the above theorem we know that if is free then is isomorphic to a free module over some set , let’s call the isomorphism . We know is a basis for and it’s not hard to see that is a basis for .

Conversely, suppose that admits a basis . We claim that , with the usual inclusion, is a free module over . Indeed, suppose that is any left -module and is any set map. We then define the map by writing every element of in its unique form and defining of this element to be . This is easily seen to be a well-defined -map on which satisfies . Moreover, it’s clear that any -map satisfying this composition equality would have to be this map as can be seen by just expanding by linearity. Thus, really is a free module on .

**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.

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