The Unapologetic Mathematician

Mathematics for the interested outsider

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).

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May 4, 2007 - Posted by | Ring theory

1 Comment »

  1. [...] 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


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