# Abstract Nonsense

## 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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References:

[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.