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 $\left[d(m)\right](\mu)=\mu(m)$ .

About these ads

May 4, 2007 - Posted by John Armstrong | Ring theory

1 Comment »

[...] Following yesterday’s examples of module constructions, we consider a ring with unit. Again, is a left and a right module over itself by [...]

Pingback by Free modules « The Unapologetic Mathematician | May 5, 2007 | Reply

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects
Archives

May 2007

M T W T F S S

« Apr Jun »

1 2 3 4 5 6

7 8 9 10 11 12 13

14 15 16 17 18 19 20

21 22 23 24 25 26 27

28 29 30 31
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider