## Module Homomorphisms

**Point of Post: **In this post we discuss the notion of module homomorphism.

**Motivation**

As always, when we define new structures our first job is to inspect the structure preserving homomorphisms associated with them. Fortunately for us, the morphisms for -modules turn out to be precisely what we’d hope they’d be– “linear transformations“, or at least the analogue of them for modules (i.e. group homomorphisms that respect scalar multiplication).

*Module Homomorphisms*

Let be two left -modules. We say that a map is a –*homomorphism *(or just homomorphism, or -morphsim for short) if for every and . Since is, in particular, a group homomorphism between it enjoys all the pursuant benefits. We define, as per usual, the kernel of a -morphism to be group-theoretic kernel of (i.e. the preimage of ) and the image to be the set theoretic image.

Ok, so fine, these -morphisms are nothing but linear transformations from linear algebra! Do they have the enjoy the same properties? Well, the answer is, for the most part yes. For exampe:

**Theorem: ***Let be left -modules and be a -morphism. Then, for any submodules and one has that and .*

**Proof: **We know that and are additive subgroups of their respective modules and so it just suffices to check the stability under -multiplication. To see this we note that if then since . Moreover, if then since we have that .

In particular, the above gives us that the kernel of a morphism is a submodule of the domain space.

We define a -morphism to be a *monomorphism *if whenever are any -morphisms (where is some left -modul)e such that then . We can come up with a different, more manageable, characterization of monomorpisms for unital modules though:

**Theorem: ***Let be unital left -modules and a -morphism. Then, is a monomorphism if and only if is injective.*

**Proof: **Suppose first that is a monomorphism, but there exists with . Define by and with . Clearly is a -morphisms since and and similarly for . That said, note that so that . But, this is a contradiction since being a monomorphism should imply but this is untrue since . Thus, is injective as claimed.

Conversely, if is injective then is left cancellative for any set map (this is just basic set theory) and so the conclusion obviously follows for morphisms .

We define an epimorphism to be a map such that given any two morphisms out of , i.e. two morphisms , such that one can conclude that . Just as for monomorphisms, for unital modules there is a nice way to characterize epimorphisms:

**Theorem: ***Let be unital left -modules and a morphism. Then, is an epimorphism if and only if is surjective.*

Sadly, we are not yet in a position to prove this, mostly because we haven’t developed enough definitions/machinery. So, until the point that we have and we can state the theorem fully we prove it for vector spaces–we do this because the proof is almost word-for-word the same, but it’s not clear we can do everything we are about to do for general left -modules (e.g. the existence of quotient spaces):

**Theorem: ***Let be two vector spaces over some field . Then, a linear transformation is an epimorphism if and only if is surjective.*

**Proof: **Suppose first that is an epimorphism. Consider then the quotient space and consider the two maps where is the zero map and is the canonical projection. Note then that and so . But, this implies by assumption that is an epimorphism that and so is zero-dimensional, which implies that as desired.

The converse follows because surjections are right-cancellative for any set map, let alone any morphism.

Of course, after monomorphism and epimorphism comes the stronges morphism, the isomorphism. In particular if are left -modules we say that morphism is an *isomorphism *if there exists a morphism such that and . Considering the following:

**Theorem: ***Let be left -modules and a bijective morphism. Then, is also a morphism.*

**Proof: **It’s basic group theory that is a group homomorphism, to see that it respects -multiplication we merely note that says , or letting (which we can uniquely since is bijective) .

From this it easily follows, as it did quite nicely for epi and monomorphisms that a morphism is an isomorphism if and only if it’s bijective. If there exists an isomorphism between two modules we call them *isomorphic *and denote this by saying

*Remark: *The reason I am making such a big fuss over these seeming trivialities is that the category of left -modules is going to serve as a prime example of a concrete category when I get around to discussing such categorical matters–and so the above gives a nice example of a concrete category where the epis and monos are just the surjections and injections respectively.

Now, it’s handy to note that since every -morphism is a group homomorphism we have that a -morphism will be a monomorphism if and only if it has trivial kernel.

Of course, it’s clear from the above that the composition of epi,mono,iso-morphisms are epi-mono,iso-morphisms. Moreover, if is an epimorphism then is an epimorphism and if is a monomorphism then is a monomorphism.

**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] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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