# Abstract Nonsense

## Homomorphism Group (Module) of Two Modules (Pt. II)

Point of Post: This is a continuation of this post.

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What we’d now like to discuss is the special case when $M=N$ in $\text{Hom}_R(M,N)$. Indeed, an $R$-morphism $M\to M$ is called anendomorphism on $M$. The set of all such endomorphisms will be denoted $\text{End}_R(M)$. Now, by virtue of $M$ being an abelian group we know that $\text{End}_R(M)$ is a ring under pointwise addition and composition of functions. But, we know that if we assume that $R$ is commutative then $\text{End}_R(M)$ also has a left $R$-module structure. This is a somewhat new idea now, up until this point we haven’t really cared whether our module had “more structure” (e.g. was naturally a ring in addition to being an abelian group). What makes $\text{End}_R(M)$ so nice though, is not only is it simultaneously a ring and a left $R$-module (two enriched versions of abelian groups) but these structures live in harmony. By this, I mean that the $R$-multiplication and the $\text{End}_R(M)$-multiplication respect each other in the sense that the $\text{End}_R(M)$-multiplication, thought of as a map $\text{End}_R(M)\times\text{End}_R(M)\to\text{End}_R(M)$, is bilinear with respect to the left $R$-module structure. To be more precise we have, using $\circ$ to denote composition (which is the $\text{End}_R(M)$-multiplication),  one sees that

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$(rf+sg)\circ h=r(f\circ h)+s(g\circ h)\;\text{ and }\; h\circ (r f+s g)=r(h\circ f)+s(h\circ g)$

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for all $r,s\in R$ and $f,g,h\in\text{End}_R(M)$. This type of structure has already shown up before when $R$ is a field under the name of an associative algebra (over a field) in the exact same context (i.e. considering $\text{End}_k(V)$ for a $k$-space $V$). We now extend this definition to encompass $R$-modules. Indeed, if $M,N,L$ are left $R$-modules we call a map $b:M\times N\to L$ bilinear if it’s linear in each coordinate, just as the case for vector spaces. We then defined an $R$-algebra to be a left $R$-module $M$ with a distinguished bilinear form $b:M\times M\to M$.

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With this definition and the above observation about $\text{End}_R(M)$ we have the following theorem:

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Theorem: Let $M$ be a left $R$-module. Then, $\text{End}_R(M)$ is a ring under pointwise addition and composition. Moreover, if $R$ is commutative then $\text{End}_R(M)$ is a unital $R$-algebra.

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The Functors $\text{Hom}_R(M,\bullet)$ and $\text{Hom}_R(\bullet,M)$

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What we shall now discuss shall serve to be one of the most fruitful tools in the math to come (e.g. it shall be how we discuss injective modules, it plays an absolutely pivotal role in the homological agebra I hope to soon discuss, etc.) but for right now shall look mostly like petty formalism. In particular, we wish to discuss how given a left $R$-module $M$ we have a map from the “set” of all left $R$-modules into the “set” of all abelian groups defined by $N\mapsto \text{Hom}_R(M,N)$. Moreover, we’d like to discuss how “diagrams” in the “set” of all  modules transfer to diagrams in the “set” of all abelian groups. Put out there in lingo that not every reader may be aware of (although I shall soon enough discuss category theory) we have from the basic category theory that after fixing $M$ there is the natural Hom functor $\text{Hom}(M,\bullet):R\text{-}\bold{Mod}\to\bold{Set}$, but unsurprisingly based on the above discussion we actually can do one better, we actually have that the Hom functor can easily be put as a function $R\text{-}\bold{Mod}\to\bold{Ab}$ and in the case when $R$ is commutative we actually have an endofunctor on $R\text{-}\bold{Mod}$. For those who don’t know category theory, the terminology is unimportant at this point, we just prove that the “function” $\text{Hom}_R(M,\bullet)$ transfers maps nicely in the following sense:

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Theorem: Let $R$ be a ring and $M$ a left $R$-module. Then, for any other two left $R$-modules $N,L$ and any $R$-morphism $N\xrightarrow{f}L$ there is an induced group homomorphism $\text{Hom}_R(M,N)\xrightarrow{f^\ast}\text{Hom}_R(M,L)$ given by $f^\ast(g)=f\circ g$. This association $^\ast$ has the property that if $N_1,N_2,N_3$ are left $R$-modules with $N_1\xrightarrow{f_1}N_2\xrightarrow{f_2}N_3$, so that $N_1\xrightarrow{f_2\circ f_1}N_3$ then $(f_2\circ f_1)^\ast=f_2^\ast\circ f_1^\ast$. Moreover, if $L=N$ then $\text{id}_N^\ast=\text{id}_{\text{Hom}_R(M,N)}$. Moreover, if $R$ is commutative then $\ast$ carries $R$-morphisms to $R$-morphisms.

Proof: We first check that given $N\xrightarrow{f}L$ that $\text{Hom}_R(M,N)\xrightarrow{f^\ast}\text{Hom}_R(M,L)$ really is a group homomorphism. To see this we merely note that if $g,h\in\text{Hom}(M,N)$ and $x\in M$ then

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

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so that $f^\ast(g+h)=f^\ast(g)+f^\ast(h)$. Now, to see that $\ast$ respects compositions suppose that we have the setup in the theorem statement and let $g\in\text{Hom}_R(M,N_1)$. We see then that

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$(f_2\circ f_1)^\ast(g)=(f_2\circ f_1)\circ g=f_2\circ(f_1\circ g)=f_2\circ (f_1^\ast(g))=f_2^\ast(f_1^\ast(g))=(f_2^\ast\circ f_1^\ast)(g)$

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so that $(f_2\circ f_1)^\ast=f_2^\ast\circ f_1^\ast$ as desired. Now, to see the last claimed property of this $^\ast$ association we merely check that for any $g\in\text{Hom}_R(M,N)$ one has that $(\text{id}_N^\ast)(g)=\text{id}_N\circ g=g=\text{id}_{\text{Hom}_R(N,M)}(g)$ and so the conclusion follows.

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It remains to show that in the case that $R$ is commutative that the $^\ast$ operation respects $R$-morphisms. Clearly all that needs to be checked is that $(f^\ast(g))(rx)=r(f^\ast(g(x))$ for all $g\in\text{Hom}_N(M,R)$. But, this is clear since

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$(f^\ast(g))(rx)=f(g(rx))=f(rg(x))=rf(g(x))=r(f^\ast(g))(x)$

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$\blacksquare$

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Remark: The map denoted by $f^\ast$ is sometimes denote $\text{Hom}_R(M,f)$.

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One can easily show the following (going through the same basic idea)

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Theorem: Let $R$ be a ring and $M$ a fixed left $R$-module. Then, for any two left $R$-modules $N,L$ and any map $N\xrightarrow{f}L$ there is an induced map $f^\ast:\text{Hom}_R(L,M)\to\text{Hom}_R(N,M)$ given by $f^\ast(g)=g\circ f$. This association has all the same properties as $\text{Hom}_R(M,f)$ except that it reverses compositions $(f_2\circ f_1)^\ast =f_1^\ast\circ f_2^\ast$

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Remark: The above says that the map $\text{Hom}_R(f,M)$ is a contravariant functor $R\text{-}\bold{Mod}\to\bold{Ab}$ and in the case that $R$ is commutative it is actually a contravariant functor $R\text{-}\bold{Mod}\to R\text{-}\bold{Mod}$.

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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.

November 1, 2011 -