Abstract Nonsense

Homomorphism Group (Module) of Two Modules (Pt. I)

Point of Post: In this post we discuss how given two left $R$-modules one can construct the set $\text{Hom}_R(M,N)$ of all $R$-morphisms $M\to N$ and how this is naturally an abelian group, and how under the right conditions (commutativity of $R$) it is also a left $R$-module. Also, we define the notion of an $R$-algebra (a $R$-module which is also a ring, with multiplication compatible with scalar multiplication) and show how under the same assumptions the set $\text{End}_R(M)$ is naturally a $R$-algebra.

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Motivation

We know from the study of vector spaces that an extremely important object associated to every $k$-vector space $V$ was the dual space $V^\ast$ of all linear functionals $V\to k$.  Of course, after a little notation one is able to renotate the dual space as the set $\text{Hom}(V,k)$ of all $k$-morphisms $V\to k$ when $k$ is considered as a one-dimensional vector space over itself. In fact, not long after this one realizes that given any other $k$-space $W$ one can form the set $\text{Hom}(V,W)$ of $k$-morphisms $V\to W$, which turns out to be a vector space of dimension $\dim_k(V)\dim_k(W)$ when $\dim_k(V),\dim_k(W)<\infty$. 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 $V$ as a vector space. Indeed, one can recall that $\text{Hom}(V,F)$ had the ability to detect the finite-dimensionality of $V$ since $V\cong \text{Hom}(V,F)$ if and only if $\dim_k(V)<\infty$. 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 $R$-module $M$ can be studied quite extensively by watching how the “function” (functor) $\text{Hom}_R(M,\bullet)$ as $\bullet$ varies over other left $R$-modules. Consequently, we begin (just the very beginning) looking at $\text{Hom}_R(M,N)$ for two left $R$-modules. We show that it’s always an abelian group, under certain conditions a left $R$-module, and when insist that $M=N$ so that we get the set $\text{End}_R(M)$ of endomorphisms it’s (under the same previously mentioned conditions) an $R$-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).

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Homomorphism Group (Module)

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Let $M,N$ be two left $R$-modules. We denote the set of all $R$-morphisms $M\to N$ by $\text{Hom}_R(M,N)$. We call this the set of homomorphisms from $M\to N$ or more often just the hom set between $M$ and $N$. We begin by noting that $\text{Hom}_R(M,N)$ naturally has the structure of an abelian group, if we define $f+g$ (for $f,g\in\text{Hom}_R(M,N)$) pointwise: $(f+g)(x)=f(x)+g(x)$. The fact that this is an abelian group is fairly trivial, but we prove it for good measure:

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Theorem: Let $M,N$ be two left $R$-modules. Then, the definition $(f+g)(x)=f(x)+g(x)$ for $f,g\in\text{Hom}_R(M,N)$ defines an abelian group structure on $\text{Hom}_R(M,N)$.

Proof: Clearly this addition is associative because

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\begin{aligned}((f+g)+h)(x) &=(f+g)(x)+h(x)\\ &=f(x)+g(x)+h(x)\\ &=f(x)+(g+h)(x)\\ &=(f+(g+h))(x)\end{aligned}

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for all $x\in M$. To see that this is actually a binary operation (i.e. that the sum of two $R$-morphisms is really an $R$-morphism) we note that the sum is a group homomorphism because

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\begin{aligned}(f+g)(x+y) &=f(x+y)+g(x+y)\\ &=f(x)+f(y)+g(x)+g(y)\\ &=f(x)+g(x)+f(y)+g(y)\\ &=(f+g)(x)+(f+g)(y)\end{aligned}

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(note the blatant use of the fact that $N$ is an abelian group). Now, $f+g$ respects $R$-multiplication since

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$(f+g)(rx)=f(rx)+g(rx)=rf(x)+rg(x)=r(f(x)+g(x))=r(f+g)(x)$

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Note that the function $0(x)=0_N$ is a $R$-morphism and clearly satisfies $f+0=0+f=0$. Moreover, it’s clear that if $-f$ is defined by $(-f)(x)=-f(x)$ then evidently $f+-f=-f+f=0$. From this it clearly follows that $\text{Hom}_R(M,N)$ is a group, and since $(f+g)(x)=f(x)+g(x)=g(x)+f(x)=(g+f)(x)$ for all $x\in M$ it is in fact evident that it is an abelian group. $\blacksquare$

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It now seems desirable, taking the case of $R$ a field as motivation, that we would like to define a left $R$-module structure on $\text{Hom}_R(M,N)$ in the obvious way. Namely, given $f\in\text{Hom}_R(M,N)$ it seems sensible to define $rf$ for $r\in R$ to be the mapping given by $(rf)(x)=rf(x)$. Of course, we run into an obvious problem in proving that $rf$ is actually a $R$-morphism. Namely, we’d like to show that $(rf)(sx)=s(rf)(x)$ for all $s\in R$ and $x\in M$. This is not hard to do if we assume that $R$ is commutative since $(rf)(sx)=rf(sx)=rsf(x)=srf(x)=s(rf)(x)$. But, if $R$ is not commutative there is no obvious way to do this. For example, if $\text{im }f$ is a faithful module then we see that $rf$ being a $R$-morphisms implies that $r$ is in the center $\mathcal{Z}(R)$ of $R$, etc. The point though is that we would uniformly like to define a left $R$-module structure on $\text{Hom}_R(M,N)$  for any left $R$-modules $M,N$ and this forces $R$ (assuming $R$ is unital) to be commutative:

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Theorem: Let $R$ be a unital ring. Then, the operation $(rf)(x)=rf(x)$ makes $\text{Hom}_R(M,N)$ into a left $R$-module for every left $R$-modules $M,N$ if and only if $R$ is commutative.

Proof: Suppose first that $R$ is commutative. We have already shown that $\text{Hom}_R(M,N)$ is an abelian group in the obvious way for any two left $R$-modules $M,N$. Now, we claim that the multiplication $(rf)(x)=rf(x)$ makes the abelian group $\text{Hom}_R(M,N)$ into a left $R$-module which is unital if $M,N$ are. The first step in this is showing that $rf\in\text{Hom}_R(M,N)$ for any $r\in R$ and $f\in\text{Hom}_R(M,N)$. To see this we merely note that for any $x,y\in M$ and $s\in R$ one has

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$(rf)(x+y)=rf(x+y)=r(f(x)+f(y))=rf(x)+rf(y)=(rf)(x)+(rf)(y)$

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and

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$(rf)(sx)=rf(sx)=rsf(x)=srf(x)=s(rf)(x)$

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from where it follows that $rf\in\text{Hom}_R(M,N)$. Now, note that

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$((r+s)f)(x)=(r+s)f(x)=rf(x)+sf(x)=(rf)(x)+(sf)(x)$

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so that $(r+s)f=rf+sf$ and

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$(r(f+g))(x)=r(f+g)(x)=r(f(x)+g(x))=rf(x)+rg(x)=(rf)(x)+(rg)(x)$

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so that $r(f+g)=rf+rg$. Lastly, note that $(rs)f(x)=rsf(x)=r(sf(x))=r((sf))(x)=(r(sf))(x)$ so that $(rs)f=r(sf)$. Lastly, if $M,N$ are unital then we see that $(1f)(x)=1f(x)=f(x)$ and so $1f=f$ so that $\text{ Hom}_R(M,N)$ is unital. The first part then follows.

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Conversely, suppose that $\text{Hom}_R(M,N)$ is a left $R$-module under $(rf)(x)=rf(x)$ for every two left $R$-modules $M,N$. Taking $M=N=R$ in the obvious way we note that $\text{id}_R$ is a $R$-morphism $R\to R$ and so by assumption $r\text{id}_R$ is also a $R$-morphism for any $r\in R$. Choose then any $r,s\in R$ and note that by assumption $rs=r\text{id}_R(s)=(r\text{id}_R)(s)=s(r\text{id}_R)(1)=sr$. Since $s,r$ were arbitrary the conclusion follows. $\blacksquare$

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[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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