Abstract Nonsense

Crushing one theorem at a time

Modules (Pt. II)

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

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So, now that we have defined modules, let’s see if we can’t explain a glaring technicality. As of yet we have only really defined left R-modules. Ok, fine, so what are ‘right’ R-modules and what is the difference? Well, why what we have defined are called left R-modules should be clear–we multiplied on the left. Ok, so clearly then if we replace the above definition with the exact same one, except now our map should look like \mu:M\times R\to M we should get the notion of right R-modules. So, what’s the big fuss, why make a distinction between the two. Well, it should be the burden of any skeptic to produce a reason why the theory of left and right R-modules should be the same. Well, unsurprisingly, there is an almost valid argument between the two theories. To be more explicit, there should be, given a ring R and an abelian group M, a natural structure preservingmap which takes any left R-module structure on M to a right R-module structure on M and conversely. As stated, there is almost a natural way to do this. Indeed, suppose that we have defined a left R-module structure (written by concatenation as usual), we can then define a right R-module structure on M by defining xr for r\in R and x\in M to be rx (i.e. the predefined multiplication on the left R-module structure on M). So, what’s the problem with this? Well, the problem is that the correspondence has the unfortunate property that multiplication gets reversed. If one wants to evaluate rsx one finds that it’s equal to xsr which is the opposite of what it “should be”. In fact, it’s pretty clear that what we’ve done is define a right R^{\text{op}}-module (where R^{\text{op}} is the opposite ring). So, it’s clear that if R is a commutative ring then there really is no such distinction, but for noncommutative rings this is not the case.

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Note that unlike the case when we are dealing with vector spaces there can be redundancy in the multiplication for modules. For example, as we noted earlier every abelian group A is naturally a \mathbb{Z}-module by n-fold summation. Consider then A=\mathbb{Z}_2^n as a \mathbb{Z}-module, note then that the multiplication maps \mu_{2n} for each n\in\mathbb{Z} are identically zero (since 2a=0 for every a\in A). In the same vein we see that \mu_{2n+1}=\text{id} for every n\in\mathbb{Z}. Ever wonder why this seemingly important concept has never come up when studying vector spaces? Well, the redundancy of the multiplication is pretty easily measured by how ‘big’ the annihilator \text{Ann}_R(M)=\left\{r\in R:rm=0\text{ for all }m\in M\right\} is. But think about it, \text{Ann}_R(M) is nothing more than the kernel of the associated ring action R\to\text{End}(M) and so if we are dealing with R being a field (as in the case of previous linear algebra) we must have that \text{Ann}_R(M) is either (0) or R and since we were dealing with unital modules (if you remember correctly we did require this in linear algebra) we have that 1\notin\text{Ann}_R(M) (except for M trivial, but that is a degenerate case) and so \text{Ann}_R(M)=(0) which explains why this subtlety never popped up in linear algebra.  In fact, it’s clear that if are dealing with simple unital rings and unital modules this never happens. If M is a module such that the associated ring action is a monomorphism (i.e. if \text{Ann}_R(M) is trivial) we call the module a faithful R-module. Faithful modules are those which best reflect the structure of the associated ground ring.

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We now record for later convenience the following collection of disparate and clear facts (the proofs of which are left to the reader):

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Theorem: Let R be a ring, S a surbring of R, and M a left R-module then the following is true:

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\begin{aligned}&\mathbf{(1)}\quad 0_Rx=0_M\quad\text{for all }x\in X\\ &\mathbf{(2)}\quad M\text{ is a left }S\text{-module in the obvious way}\\&\mathbf{(3)}\quad -1_Rx=-x\quad\text{ if }R,M\text{ are unital}\\ &\mathbf{(4)}\quad\text{The induced action of }U(R)\text{ on }M\text{ is a group action}\end{aligned}

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[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.



October 27, 2011 - Posted by | Algebra, Ring Theory | , , , , , , ,

1 Comment »

  1. […] since . But, if is not commutative there is no obvious way to do this. For example, if is a faithful module then we see that being a -morphisms implies that is in the center of , etc. The point though is […]

    Pingback by Homomorphism Group (Module) of Two Modules (Pt. I) « Abstract Nonsense | November 1, 2011 | Reply

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