## Free modules

Following yesterday’s examples of module constructions, we consider a ring with unit. Again, is a left and a right module over itself by multiplication.

We can form the direct sum of a bunch of copies of over any (finite or infinite) index set : . Every element of this module is a list of elements of indexed by — — and all but a finite number of them are zero. The ring acts from the left by .

One special thing about this module is that any element can be written as a sum — — where is the element with a in the slot indexed by and in all the other slots. This sum makes sense because there are only a finite number of nonzero terms to consider for any given module element. Since any element can be written as an -linear combination of these , we say they “span” the module.

Even better, there’s no way of writing any of the as an -linear combination of the others. More specifically, if we have some -linear combination , the only way for it to be the zero element of the module is for *all* of the to be zero. Since there are no -linear relations between the , we say that they are “linearly independent” (over ).

These two conditions — span and linear independence — show up all the time. Whenever we have a linearly independent collection of module elements that span a module, we say that they form a “basis” of the module. By the spanning property, every module element can be written as a linear combination of basis elements. The linear independence tells us that this expression is unique.

Now it’s important to note that not all modules even have a single basis. As an example of a module without a basis, consider the abelian group as a -module. Now no element of this module is even linearly independent on its own! Clearly , even when is nonzero, so is not linearly independent. Also whenever is even, so can’t be linearly independent either. There are no linearly independent sets, so no basis.

On the other hand, if an -module *does* have a basis , I claim that it’s isomorphic to a direct sum of copies of , as above. Just take the index set to index the basis itself and try to find an isomorphism . Construct the function by sending to and extend by -linearity. Since is a basis we can write an element of as , which must be sent to . Since is a basis of , the only way an element of gets sent to zero is if all the are zero already, and every element in the target gets hit at least once. Thus the function is an isomorphism.

Now, by the way direct sums interact with , we see that for any left module we have

thus if we pick a list of elements of indexed by — no restriction on how many nonzero elements we pick — we get a unique homomorphism from to sending to . This justifies calling a “free” left -module, analogously to free groups, free rings, and so on.

The upshot of this property is that when we’re dealing with two free modules and and we have a basis in hand for each, then we have a nice way of writing down homomorphisms from to . Let’s use as our basis for and as our basis for . Then we can specify any homomorphism by saying where sends the basis of . We write . But then since has a basis we can write the in terms of the , getting .

What if we have another homomorphism , where is free on ? If we write then we compose homomorphisms to get

If this looks familiar, it’s because we’re getting the coefficients of the composite homomorphism on the right by matrix multiplication! That’s right: we’re finally getting to high school algebra II here. One thing I’ll point out here that your teacher probably didn’t tell you is that we only wrote down a matrix for a homomorphism *after picking a basis for each free module*. A free module may have many different bases, and it requires a choice to pick one or another to write down a matrix. This choice may lead to all sorts of artifacts in the matrix that really have nothing to do with the homomorphism itself and everything to do with the basis. Thus we’ll try everywhere to avoid using a specific basis unless one clearly stands out as useful.

[…] I know I usually go light on Sundays, but I want to finish off what I started yesterday. Remember that we’re considering free modules over a ring with unit. A free module has a […]

Pingback by Vector spaces « The Unapologetic Mathematician | May 6, 2007 |

[…] concretely, now: we know that every vector space over is free as a module over . That is, every vector space has a basis — a set of vectors so that every […]

Pingback by Matrices I « The Unapologetic Mathematician | May 20, 2008 |

[…] Let’s use universal properties to prove this! We consider the direct sum , and we have a basis for and a basis for . But remember that the whole point of a basis is that vector spaces are free modules. […]

Pingback by The Sum of Subspaces « The Unapologetic Mathematician | July 21, 2008 |

[…] key point is to realize that is a free right module over . That is, we can find some collection of vectors in so that any other one can […]

Pingback by Induced Matrix Representations « The Unapologetic Mathematician | November 25, 2010 |

[…] suitable map such that is a coordinate patch, it turns out that we can actually give an explicit basis of the module of vector fields over the ring […]

Pingback by Coordinate Vector Fields « The Unapologetic Mathematician | May 24, 2011 |