The Tensor Algebra and Exterior Algebra (Pt. V)
Point of Post: This is a continuation of this post.
Exterior Powers of Maps
Not surprisingly, the construction is functorial. In other words, we should not only be able to be able to take exterior powers of modules but also take exterior powers of . So, if is an -map we should be able to construct some -map mapping which respects compositions and identities.
The idea of this map is simple, but powerful. Namely, we know that from our map we get a map such that . Now, considering the composition where is the universal -linear map and is the quotient map, we get an alternating -linear map which satisfies . Thus, by the universal characterization of the exterior power we are afforded a map
as we would expect such a map to satisfy. I think it’s fairly obvious that and that . Thus, we may conclude that
Theorem: The association of to is a functor .
In fact, it’s pretty easy to see that:
Theorem: Let be a surjective -map. Then, is surjective.
Proof: Let be a generating set for . We know then that there is some such that . But, it’s trivial that is a generating set for , and moreover that if then . Thus, hits a generating set of and thus is surjective.
What is also true is that:
Theorem: Let be an injective -map with a direct summand of . Then, is injective. Moreover, is a direct summand of .
Proof: Since is a direct summand we have a split short exact sequence
where is the complement of in . We know that is injective from the simple fact that functors take retractions to retractinos. Or, said differently, since our sequence splits we get a retraction map such that . Then, and thus is injective. Now, to prove that we actually have that is a direct summand of we merely note that we get the exact sequence
and since we have a retraction for the conclusion follows from the splitting lemma.
Let us define a ring to be semisimple if every submodule of a left -module is a direct summand. For example division rings, finite dimensional matrix rings over division rings, and finite products of finite dimensional matrix rings over division rings are semisimple. It is the content of the famed Wedderburn-Artin theorem that these precisely the semisimple rings (i.e. that the rings described are semisimple, and they are the only semisimple rings). Anyways, we are getting too far off course. The above allows us to conclude the following:
Theorem: Let be a semisimple ring and a map of left -modules. Then, is surjective (resp. injective) if is surjective (resp. injective).
Note that this does NOT say that if is a field then is exact! The reason it should be “obvious” that our functor isn’t even right exact is a simple fact about binomial coefficients. Indeed, functors that are right exact commute with quotients, and so if was right exact then where are -vector spaces. But, counting dimensions both sides (saying that and ) this says that
a decidedly untrue fact.
Now, back to our main discussion about facts concerning the exterior power of maps. The above told us that we need, in general, the image of our map to be a direct summand to say that the induced exterior power map is injective. That said, if we assume that both are free then this assumption is not necessary. Indeed:
Theorem: Let be an -map with free. Then, is injective.
Proof: Let be the embeddings and as we constructed earlier. Then, we have the obvious relation . That said, since free modules are flat we know that is injective since is the composition of the maps
all of which are injective.
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 Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.
 Blyth, T. S. Module Theory. Clarendon, 1990. Print.
 Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.
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 Conrad, Keith. “Exterior Powers.” Www.math.uconn.edu/~kconrad. Web. <http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf>.