## Tensor Algebra and Exterior Product (Pt. VI)

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

A cool corollary of this pointed out in [6] is the following:

**Theorem: ***Let be an -map (with commutative!). Then, if is mono then and if is epi then . *

**Proof: **Suppose that is epi. then, is epi. But, and thus we see that which by previous proof means that we can’t have and thus . If is mono then since these are free modules we have that is mono. Since this implies that and thus similar logic leads us to .

I’m not one-hundred percent sure that there isn’t any circularity to this argument (point it out if there is!) but to me it seems that combining these two implies:

**Theorem: ***Commutative rings have the IBN property.*

The standard proof of this, if you recall, is that if as -modules then (where is some maximal ideal afforded to us by Krull’s theorem) as modules, but from basic module theory this just says that as -modules. This has reduced us to having to verify that fields have the IBN property which is the standard “exchange lemma” (apparently this lemma is actually named) proof. My point in bringing this up is that to do this proof quickly, and well, one needs to have a working knowledge of both tensor products and maximal ideals, AND requires one to already know the proof for fields. Thus, if there is no circularity in the above argument it is much more technically simple than the “standard” one.

**Relation to Determinants**

What I’d now like to discuss is how, in particular, exterior powers of maps on f.g. free modules relate to determinants.

Recall that if is an f.g. free -module with an ordered basis there is an association that takes an -map and associates an matrix defined by making the entry of the unique such that . Moreover the association is actuall an -algebra isomorphism (the proof is the same as that for the case of fields). Moreover we know that if we define to be the unique -isomorphism with then where acts on in the usual way (once again, this is the same as the proof in the case of fields).

Thus, just in the case of vector spaces, we have that we can think about endmorphisms as being matrices, as long as we keep track of bases. Moreover, we can define the determinant of a matrix in by the rule

Moreover, we can still define the adjugate matrix to be the matrix whose entry is gotten by deleting the row and column of , taking the determinant, and multiplying by . We see then, same as the proof from basic linear algebra, that

And thus, we can see that (i.e. ) if and only if .

What we’d like to show is that if is an ordered basis for and then

Indeed, suppose that then

Now, any with repeated numbers makes and so we may consider our sum only over indices which are bijections of . Thus, we see that

But, and so

**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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