# The Unapologetic Mathematician

## Homomorphisms of modules

Again, once we have a new structure it’s important to understand what sort of functions connect different instancees of the structure. For modules we have module homomorphisms.

By now this should be pretty straightforward. A left $R$-module $M$ is an abelian group equipped with an action of $R$. A homomorphism of $R$-modules is a homomorphism of abelian groups $f:M\rightarrow N$ that also preserves the action of $R$. That is, it’s a linear function satisfying $f(r\cdot m)=r\cdot f(m)$ for all $m\in M$. We call such a function $R$-linear.

In terms of the tensor product picture, any linear function $f:M\rightarrow N$ gives us a linear function $1_R\otimes f:R\otimes M\rightarrow R\otimes N$. The condition of $R$-linearity is that the following diagram commute: where the horizontal arrows are the actions of $R$ on $M$ and $N$, respectively.

Now if we pick two left $R$-modules $M$ and $N$ we have a bunch of different $R$-module homomorphisms from $M$ to $N$. We’ll call the set of them $\hom_{R{\rm -mod}}(M,N)$, sometimes shortened to $\hom_R(M,N)$, or even just $\hom(M,N)$. The important thing about this set is that it inherits the structure of an abelian group from the one on $N$.

There’s always a homomorphism $0:M\rightarrow N$ sending every element of $M$ to the zero element of $N$. Also, given homomorphisms $f$ and $g$ we can define $\left[f+g\right](m)=f(m)+g(m)$. It’s straightforward to check that this preserves the action of $R$ as long as $f$ and $g$ both do. Finally, there’s a homomorphism defined by $\left[-f\right](m)=-f(m)$. We can easily see that these operations on $\hom(M,N)$ satisfy the axioms of an abelian group.

In fact it gets even better. Not only are the homomorphism sets abelian groups, but composition is bilinear! Let’s consider four homomorphisms between three modules $f_1:A\rightarrow B$, $f_2:A\rightarrow B$, $g_1:B\rightarrow C$, and $g_2:B\rightarrow C$. Then we build up the composition $(g_1+g_2)\circ(f_1+f_2)$. Let’s see what it does to an element $a\in A$. $\left[(g_1+g_2)\circ(f_1+f_2)\right](a)=\left[g_1+g_2\right](\left[f_1+f_2\right](a))=$ $\left[g_1+g_2\right](f_1(a)+f_2(a))=g_1(f_1(a)+f_2(a))+g_2(f_1(a)+f_2(a))=$ $g_1(f_1(a))+g_1(f_2(a))+g_2(f_1(a))+g_2(f_2(a))=$ $\left[g_1\circ f_1+g_1\circ f_2+g_2\circ f_1+g_2\circ f_2\right](a)$

This gives us a linear function $\hom(A,B)\otimes\hom(B,C)\rightarrow\hom(A,C)$ for every three modules $A$, $B$, and $C$.

What happens if we pick all three modules to be the same one? Each homomorphism set is $\hom_R(M,M)$, which we’ll call ${\rm End}_R(M)$. Then we get a linear function ${\rm End}_R(M)\otimes{\rm End}_R(M)\rightarrow{\rm End}_R(M)$. This is a ring structure! We call it the ring of $R$-endomorphisms of $M$. If the ring $R$ is the ring of integers and $A$ is an abelian group, then ${\rm End}_\mathbb(A)$ is just the endomorphism ring ${\rm End}(A)$ we considered earlier. This is an example of how the theory of modules naturally extends the theory of abelian groups.

April 23, 2007 Posted by | Ring theory | 1 Comment