Homomorphism Group (Module) of Two Modules (Pt. I)
Point of Post: In this post we discuss how given two left -modules one can construct the set of all -morphisms and how this is naturally an abelian group, and how under the right conditions (commutativity of ) it is also a left -module. Also, we define the notion of an -algebra (a -module which is also a ring, with multiplication compatible with scalar multiplication) and show how under the same assumptions the set is naturally a -algebra.
We know from the study of vector spaces that an extremely important object associated to every -vector space was the dual space of all linear functionals . Of course, after a little notation one is able to renotate the dual space as the set of all -morphisms when is considered as a one-dimensional vector space over itself. In fact, not long after this one realizes that given any other -space one can form the set of -morphisms , which turns out to be a vector space of dimension when . We saw that the studying of these spaces were not only useful for outward looking facts about linear algebra (e.g. direct applications of them, using them for their own sake) but also gave us structural information about as a vector space. Indeed, one can recall that had the ability to detect the finite-dimensionality of since if and only if . This shall be a recurring fact that we shall be discussing extensively in the math-to-come, to be particular, we will see that a general -module can be studied quite extensively by watching how the “function” (functor) as varies over other left -modules. Consequently, we begin (just the very beginning) looking at for two left -modules. We show that it’s always an abelian group, under certain conditions a left -module, and when insist that so that we get the set of endomorphisms it’s (under the same previously mentioned conditions) an -algebra (module which is also a ring). We also then discuss (practically just define) the “function” (functor) we previously mentioned and show some of its most basic properties (i.e. that it is a functor).
Homomorphism Group (Module)
Let be two left -modules. We denote the set of all -morphisms by . We call this the set of homomorphisms from or more often just the hom set between and . We begin by noting that naturally has the structure of an abelian group, if we define (for ) pointwise: . The fact that this is an abelian group is fairly trivial, but we prove it for good measure:
Theorem: Let be two left -modules. Then, the definition for defines an abelian group structure on .
Proof: Clearly this addition is associative because
for all . To see that this is actually a binary operation (i.e. that the sum of two -morphisms is really an -morphism) we note that the sum is a group homomorphism because
(note the blatant use of the fact that is an abelian group). Now, respects -multiplication since
Note that the function is a -morphism and clearly satisfies . Moreover, it’s clear that if is defined by then evidently . From this it clearly follows that is a group, and since for all it is in fact evident that it is an abelian group.
It now seems desirable, taking the case of a field as motivation, that we would like to define a left -module structure on in the obvious way. Namely, given it seems sensible to define for to be the mapping given by . Of course, we run into an obvious problem in proving that is actually a -morphism. Namely, we’d like to show that for all and . This is not hard to do if we assume that is commutative since . But, if is not commutative there is no obvious way to do this. For example, if is a faithful module then we see that being a -morphisms implies that is in the center of , etc. The point though is that we would uniformly like to define a left -module structure on for any left -modules and this forces (assuming is unital) to be commutative:
Theorem: Let be a unital ring. Then, the operation makes into a left -module for every left -modules if and only if is commutative.
Proof: Suppose first that is commutative. We have already shown that is an abelian group in the obvious way for any two left -modules . Now, we claim that the multiplication makes the abelian group into a left -module which is unital if are. The first step in this is showing that for any and . To see this we merely note that for any and one has
from where it follows that . Now, note that
so that and
so that . Lastly, note that so that . Lastly, if are unital then we see that and so so that is unital. The first part then follows.
Conversely, suppose that is a left -module under for every two left -modules . Taking in the obvious way we note that is a -morphism and so by assumption is also a -morphism for any . Choose then any and note that by assumption . Since were arbitrary the conclusion follows.
 Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.
 Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.
 Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.