# Abstract Nonsense

## Module Homomorphisms

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

$\text{ }$

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 $R$-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).

$\text{ }$

Module Homomorphisms

$\text{ }$

Let $M,N$ be two left $R$-modules. We say that a map $f:M\to N$ is a $R$homomorphism (or just homomorphism, or $R$-morphsim for short) if $f(rx+sy)=rf(x)+sf(y)$ for every $x,y\in M$ and $s,r\in R$. Since $f$ is, in particular, a group homomorphism between $M\to N$ it enjoys all the pursuant benefits. We define, as per usual, the kernel $\ker f$ of a $R$-morphism to be group-theoretic kernel of $f$ (i.e. the preimage of $0_M$) and the image to be the set theoretic image.

$\text{ }$

Ok, so fine, these $R$-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:

$\text{ }$

Theorem: Let $M,N$ be left $R$-modules and $f:M\to N$ be a $R$-morphism. Then, for any submodules $A\leqslant M$ and $B\leqslant N$ one has that $f^{-1}(B)\leqslant M$ and $f(A)\leqslant N$.

Proof: We know that $f(A)$ and $f^{-1}(B)$ are additive subgroups of their respective modules and so it just suffices to check the stability under $R$-multiplication. To see this we note that if $f(a)\in f(A)$ then $rf(a)=f(ra)\in f(A)$ since $ra\in A$. Moreover, if $x\in f^{-1}(B)$ then since $f(rx)=rf(x)\in B$ we have that $rx\in f^{-1}(B)$. $\blacksquare$

$\text{ }$

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

$\text{ }$

We define a $R$-morphism $f:N\to M$ to be a monomorphism if whenever $g_1,g_2:L\to N$ are any $R$-morphisms (where $L$ is some left $R$-modul)e such that $f\circ g_1=f\circ g_2$ then $g_1=g_2$. We can come up with a different, more manageable, characterization of monomorpisms for unital modules though:

$\text{ }$

Theorem: Let $M,N$ be unital left $R$-modules and $f:M\to N$ a $R$-morphism. Then, $f$ is a monomorphism if and only if $f$ is injective.

Proof: Suppose first that $f$ is a monomorphism, but there exists $x\ne y\in M$ with $f(x)= f(y)$. Define $g_1:R\to M$ by $g_1(r)=rx$ and $g_2:R\to M$ with $g_2(r)=ry$. Clearly $g_1$ is a $R$-morphisms since $g_1(r+r')=(r+r')x=rx+r'x=g_1(r)+g_1(r')$ and $g_1(rr')=rr'x=r(r'x)=rg_1(r')$ and similarly for $g_2$. That said, note that $f(g_1(r))=f(rx)=rf(x)=rf(y)=f(ry)=f(g_1(r))$ so that $f\circ g_1=f\circ g_2$. But, this is a contradiction since $f$ being a monomorphism should imply $g_1=g_2$ but this is untrue since $g_1(1)=1x=x\ne y=1y=g_2(1)$. Thus, $f$ is injective as claimed.

$\text{ }$

Conversely, if $f$ is injective then $f$ is left cancellative for any set map $L\to M$ (this is just basic set theory) and so the conclusion obviously follows for morphisms $L\to M$. $\blacksquare$

$\text{ }$

We define an epimorphism $f:M\to N$ to be a map such that given any two morphisms out of $N$, i.e. two morphisms $g)1,g_2N\to L$, such that $g_1\circ f=g_2\circ f$ one can conclude that $g_1=g_2$. Just as for monomorphisms, for unital modules there is a nice way to characterize epimorphisms:

$\text{ }$

Theorem: Let $M,N$ be unital left $R$-modules and $f:M\to N$ a morphism. Then, $f$ is an epimorphism if and only if $f$ is surjective.

$\text{ }$

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 $R$-modules (e.g. the existence of quotient spaces):

$\text{ }$

Theorem: Let $V,W$ be two vector spaces over some field $k$. Then, a linear transformation $T:V\to W$ is an epimorphism if and only if $T$ is surjective.

Proof: Suppose first that $T$ is an epimorphism. Consider then the quotient space $W/T(V)$ and consider the two maps $0,\pi: W\to W/T(V)$ where $0$ is the zero map and $\pi$ is the canonical projection. Note then that $\text{im }T\subseteq=\ker\pi$ and so $T\circ\pi=T\circ 0=0$. But, this implies by assumption that $T$ is an epimorphism that $0=\pi$ and so $W/T(V)$ is zero-dimensional, which implies that $T(V)=W$ as desired.

$\text{ }$

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

$\text{ }$

Of course, after monomorphism and epimorphism comes the stronges morphism, the isomorphism. In particular if $M,N$ are left $R$-modules we say that morphism $f:M\to N$ is an isomorphism if there exists a morphism $g:N\to M$ such that $g\circ f=\text{id}_M$ and $f\circ g=\text{id}_N$. Considering the following:

$\text{ }$

Theorem: Let $M,N$ be left $R$-modules and $f:M\to N$ a bijective morphism. Then, $f^{-1}$ is also a morphism.

Proof: It’s basic group theory that $f^{-1}$ is a group homomorphism, to see that it respects $R$-multiplication we merely note that $f(rx)=rf(x)$ says $rx=f^{-1}(rf(x))$, or letting $f(x)=y$ (which we can uniquely since $f$ is bijective) $rf^{-1}(y)=f^{-1}(ry)$. $\blacksquare$

$\text{ }$

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 $M\cong N$

$\text{ }$

Remark: The reason I am making such a big fuss over these seeming trivialities is that the category $R-\bold{Mod}$ of left $R$-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.

$\text{ }$

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

$\text{ }$

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

$\text{ }$

$\text{ }$

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.

October 28, 2011 -

## 8 Comments »

1. […] 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 […]

Pingback by Homomorphism Group (Module) of Two Modules (Pt. I) « Abstract Nonsense | November 1, 2011 | Reply

2. […] Where in the above we have fixed a particular ring. We then define, obviously, to be the class of all left -modules, where we require the modules to be unital if is. Not shockingly is just the category of all right -modules. In both cases the morphisms are just -morphisms. […]

Pingback by Examples of Categories (Revisited) « Abstract Nonsense | November 2, 2011 | Reply

3. […] a ring and left -modules we say that the “chain” of -morphisms is exact at if . A “chain” of maps and […]

Pingback by Exact Sequences of Modules « Abstract Nonsense | November 3, 2011 | Reply

4. […] As an initial application we can finally finish our proof about the equivalence of epimorphisms and surjective morphisms: […]

Pingback by Quotient Modules « Abstract Nonsense | November 7, 2011 | Reply

5. […] clear how to do this, namely defining –since this map is well-defined. Moreover, since is epi (so that implies ) we have that is the unique map with this property. From this we may conclude […]

Pingback by The Module Isomorphism Theorems « Abstract Nonsense | November 11, 2011 | Reply

6. […] and the ‘s being the canonical projections, is that we do not require the ‘s to be epis. That said, it’s not hard to see that they must, in fact, be epis. In particular, taking and […]

Pingback by Product of Modules « Abstract Nonsense | November 11, 2011 | Reply

7. […] -algebra to the -algebra to be a set function which is both a unital ring homomorphism and an -map. If we think about the -algebras and as just rings and with unital ring homomorphism and . […]

Pingback by R-Algebras « Abstract Nonsense | January 10, 2012 | Reply

8. […] Let’s first show that is an injection. This is easy though since we know every injective -map is a monomorphism, and so if then necessarily . So we must now show that the […]

Pingback by The Hom Functor is Left Exact « Abstract Nonsense | January 26, 2012 | Reply