# The Unapologetic Mathematician

## Some examples of modules

Today I want to run through a bunch of examples of the constructions we’ve been considering for modules. I’ll restrict to the case of a ring $R$ with unit.

One easy example of an $R$-module that I’ve mentioned before is the ring $R$ itself. We drop down to the underlying abelian group and then act on it using the ring multiplication. There are both left and right actions here: $r\cdot x=rx$ and $x\cdot r=xr$ where $r$ and $x$ are ring elements, $x$ considered as an element of the module. We’ll start off by taking this module and sticking it into some of the constructions.

When we consider $\hom_{R-{\rm mod}}(R,M)$ for some left $R$-module $M$ the left module structures on $R$ and $M$ will get eaten and the right module structure on $R$ will get flipped over, leaving us a left $R$-module. We can pick an element $f\in\hom_{R-{\rm mod}}(R,M)$ by specifying $f(1)\in M$. Then $f(r)=f(r\cdot1)=r\cdot f(1)$, telling us where everything else goes. If we write $f_m$ for the homomorphism with $f_m(1)=m$, then the left action of $R$ on homomorphisms says
$\left[r\cdot f_m\right](1)=f_m(r\cdot1)=r\cdot f_m(1)=r\cdot m$
Thus $r\cdot f_m=f_{r\cdot m}$. This means that $\hom_{R-{\rm mod}}(R,M)\cong M$ as left $R$-modules.

On the other hand, if we consider $\hom_{R-{\rm mod}}(M,R)$ we get a right $R$-module. This consists of all $R$-linear functions from $M$ to the ring $R$ itself. We call this the “dual” module to $M$, and write $M^*=\hom_{R-{\rm mod}}(M,R)$. Elements of the dual module are often called “linear functionals” on $M$.

Tensor products are even easier. When we consider $R\otimes_RM$ for a left $R$-module $M$ we can use the construction of tensor products to write an element as a finite sum: $\sum r_i\otimes m_i$. But then we can use the middle-linear property to write $r_i\otimes m_i=(1\cdot r_i)\otimes m_i=1\otimes(r_i\cdot m_i)$, and then the linearity to collect all the terms together, giving $1\otimes m$. The tensor product eats the module structure on $M$ and the right module structure on $R$, leaving a left $R$-module structure. We calculate
$r\cdot(1\otimes m)=(r\cdot1)\otimes m=r\cdot m=(1\cdot r)\otimes m=1\otimes(r\cdot m)$
so $R\otimes_RM\cong M$ as left $R$-modules.

Now let’s take two left $R$-modules $M$ and $N$ and make $\hom(M,N)$. This is an abelian group — a $\mathbb{Z}$-module — as is $M$. Let’s write $M$ as $\hom(R,M)$ as above and then tensor over $\mathbb{Z}$ with $\hom(M,N)$. Then we can compose homomorphisms
$\hom(M,N)\otimes M\cong\hom(M,N)\otimes\hom(R,M)\rightarrow\hom(R,N)\cong N$
This is the “evaluation” homomorphism that takes an element $m\in M$ and a homomorphism $f\in\hom(M,N)$ and gives back $f(m)\in N$.

As a special case, we can take $R$ itself in place of $N$. We get an evaluation homomorphism $M^*\otimes M\rightarrow R$. This “canonical pairing” we often write as $\langle\mu,m\rangle=\mu(m)$ for a linear functional $\mu$ and module element $m$.

What if we composed with an element of $N^*=\hom(N,R)$ instead of $M\cong\hom(R,M)$? We use the evaluation homomorphism to get
$N^*\otimes\hom(M,N)=\hom(N,R)\otimes\hom(M,N)\rightarrow\hom(M,R)=M^*$
So given a homomorphism $f:M\rightarrow N$ we get a homomorphism $f^*:N^*\rightarrow M^*$

Of course, all this goes through suitably changed by swapping “right” for “left”. For example, given a right $R$-module $M$ we have a dual left $R$-module $M^*=\hom_{{\rm mod}-R}(M,R)$.

What do we get if we start with a left module $M$, dualize it, then dualize again to get another left module $M^{**}=\left(M^*\right)^*$? Following the definitions we see $M^{**}=\hom_{{\rm mod}-R}(\hom_{R-{\rm mod}}(M,R),R)$. I claim that there is a natural morphism of left $R$-modules $M\rightarrow M^{**}$. That is, a special element of
$\hom_{R-{\rm mod}}(M,\hom_{{\rm mod}-R}(\hom_{R-{\rm mod}}(M,R),R)$
but we know that this is isomorphic to
$\hom_{R-{\rm mod}}(\hom_{R-{\rm mod}}(M,R)\otimes M,R)$
which we write as
$\hom_{R-{\rm mod}}(M^*\otimes M,R)$
so we’re really looking for a special homomorphism from $M^*\otimes M$ to $R$. And we’ve got one: the canonical pairing! So we take the canonical pairing as a homomorphism from $M^*\otimes M\rightarrow R$ and pass it through this natural isomorphism to get a homomorphism $d:M\rightarrow M^{**}$. In case this looks completely insane, here it is in terms of elements: $d(m)$ takes a linear functional $\mu$ and gives back an element of the ring by the rule $\left[d(m)\right](\mu)=\mu(m)$.