## The Tensor Algebra and Exterior Algebra (Pt. IV)

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

**Exterior Products of Free Modules**

We’d now like to discuss how everything works out when we are talking about exterior products of free modules.

In particular, we know that if is free, then by virtute of the fact that the tensor product commutes with coproducts we know that is free. But, it is rookie-mistake number one to think that quotients of free modules (even by free submodules) are necessarily free (is free?!) and so there is no a priori reason to think that freeness of should imply the freeness of .

Now, while not explicitly stated, it should be fairly obvious that if is a spanning set for then is a spanning set for since that set, with wedges replaced with tensors, is a spanning set for of which is a quotient. Of course, there is a huge amount of redundancies, in fact, most of these wedges are zero. For example, we can immediately replace this set with where now the indices still live in but now we require that for any . But, once again this is still redundant for we have indexed by injections , but by virtue of anticommutativity, we can index by combinations of elements of . In particular, spans where . In particular, we see that if is spanned by elements then is spanned by elements. In particular, we see the following:

**Theorem: ***If is spanned by elements and then *

This makes sense with our above interpretation of being zero. Indeed, this should say that there are no nonzero alternating -linear -maps out of . But, it of course suffices to check that if is such a map that for the in the spanning set of . But , our old pal The Pigeonhole Principle (easily one of the most under appreciated theorems in mathematics!) tells us that since that at least two of these ‘s must be equal and so .

What we’d like to show is that if is indeed free, so that we can take this set of ‘s to be a basis, then this generating set is also a basis. The key to this is that when is free we shall be able to embed back into –a strange concept, embedding a quotient back into the space. The embedding should look familiar either if you are familiar with the more classic approach to multilinear algebra (via the alternating map) or if you know Leibniz expansion of the determinant. Namely, we are going to define an embedding in such a way that

To see that such a map even exists we note that the mapping defined as above (except we are mapping to this sum) is alternating. Indeed, it’s easy to see that this map is multilinear (being a sum of multilinear maps) and is alternating for

Now, we write (recalling that is a group map ) and since is idempotent this is just . So, we see that is mapped to

Thus, we are afforded an -linear map with holding. What we’d like to now show is that if is free this is actually an embedding. Indeed, let be a basis for . Let where runs over indices and evidently means if . Now, if this goes to zero we have the relation

That said, from the basic theory of tensor products we know that , as and run over their respective indices, form a basis for the free -module which tells us that for each and each –but this in particular tells us that for all . Thus, we see that the kernel of our map is trivial and thus we have an honest to god embedding. Now, note that the image of the vectors are linearly independent (for if and another index aren’t equal we can see that the image of and will contain different basis vectors!) and thus since we are dealing with an embedding this tells us that that are linearly independent in . Thus, we may conclude:

**Theorem: ***Let be a free -module with basis . Then, is a free -module with basis as the indices are chosen arbitrarily according to the condition . In particular, we see that is .*

This is actually enables us to evaluate the full exterior algebra, as a module. Indeed:

**Theorem: ***Let be a f.g. free -module of rank . Then, is a free -module of rank .*

**Proof: **We know that is . Now, each is free of rank and thus is free of rank

While this embedding seemed incidental it shall come back to important for us when we finally get to the differential forms stuff that we are working towards.

**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.

[4] Lang, Serge. *Algebra*. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. *Abstract Algebra*. New York: Springer, 2007. Print.

[6] Conrad, Keith. “Exterior Powers.” *Www.math.uconn.edu/~kconrad*. Web. <http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf>.

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