Modules

Definition of module

The quote above pretty much says it all. Have you ever wondered what would happen if you had a vector space $V$, but instead of a field of scalars you had a ring of scalars^[1]? If so, you just stumbled upon the definition of a module.

You could reasonably complain about the the use of the word "action" in the above definition, since it hasn't been defined here and you possibly only encountered it previously in the context of a group acting on a set. However, it is being used here very much in the same spirit as in group theory, in that it is a set map $R\times M\to M$ that satisfies the expected compatibility conditions between the algebraic structures on $R$ and $M$ (detailed explicitly above). With a bit more category theory under our belts, we could reasonably consider defining the general notion of an action of an object of one category on an object of another. Something to keep in the back of our minds...

Modules over commutative rings

Suppose $M$ is a left module over a commutative ring $R$. For clarity, let's denote the left action of a ring element $r$ on a module element $m$ by $r\cdot m$. Now define a right action of $R$ on $M$ by setting $m\star r=r\cdot m$. You can quickly check that this does indeed define a right action of $R$ on $M$, giving (the underlying abelian group of $M$) the structure of a right $R$-module. The only property that requires the commutativity of $R$ is the "associativity" of the action. Indeed, on the one hand we have

while on the other hand we have

Thus, we need the original left $R$-action on $M$ to satisfy $(rs)\cdot m=(sr)\cdot m$ for every $r,s\in R$ and $m\in M$. Fortunately this is true when $R$ is assumed to be commutative.^[2]

For a general (possibly noncommutative) ring, the phrase "$R$-module" is usually meant to mean "left $R$-module."

The action of zero

It will be occasionally useful to have the following simple result:

Here's a quick proof, directly analogous to the similar statement for vector spaces (a special case of modules!). For every $m\in M$ we have $0_R\cdot m=(0_R+0_R)m=0_R\cdot m+0_R\cdot m$. By additive cancellation (in the abelian group $M$) we then have $0_M=0_R\cdot m$. Similarly, $r\cdot 0_M=r(0_M+0_M)=r\cdot 0_M+r\cdot 0_M$, and additive cancellation once again gives $0_M=r\cdot 0_M$.

Examples

Given the common quote about modules, our first example of a module really must be:

Vector Spaces

If $R=k$ is a field then $k$-modules are exactly the same as $k$-vector spaces. What do we really mean when we say "are exactly the same as"? In this case we mean "literally the same, up to naming conventions". A $k$-vector space is usually defined to be a set $V$ together with a binary operation $V\times V\to V$ and a map $k\times V\to V$ satisfying a list of conditions. The first of those conditions are exactly the conditions for $V$ to be an abelian group; the rest are the same as our conditions for $V$ to be a $k$-module. So the only differences are that we call $V$ an $k$-vector space (instead of an $k$-module) and call $k$ the field of scalars (instead of simply the ring acting on $V$).

Rings as modules

Every ring $R$ can be given the structure of a left module over itself, using the action of left multiplication on itself. In this context, the submodules of $R$ (when considered as a left $R$-module) are exactly the left ideals of $R$ (when considered as a ring).

Once again, we should ask what we mean when we say "are exactly the left ideals." Here we mean that there is a (natural) bijection between the set of left ideals of the ring $R$ and the set of submodules of $R$.

In theory, one could consider $R$-module structures on $R$ defined by other (left) actions of $R$ on itself, but in practice I have never seen that done. Perhaps it is because of the inherent notational issues, as you would need to be very careful to distinguish between when the operation is the internal multiplication of the ring and when it is the ring action; perhaps one could write $rs$ for the former and $r\bullet s$ for the latter? If I come across a discussion or example of this in action (pun intended), I'll be sure to link it here.

(This statement can be made categorically precise, but it requires some setup.)

Abelian groups are $\mathbf{Z}$-modules

It is simple to show^[3] that $\mathbf{Z}$-modules are "exactly the same" as abelian groups.

Same group, different modules

A given abelian group can even have different module structures, even for the same ring $R$. Being specific about the type of module (i.e., the ring doing the acting) can be important when talking about module morphisms.

For example, the field $\mathbf{R}$ can be viewed as an $\mathbf{R}$-module, a $\mathbf{Q}$-module, or a $\mathbf{Z}$-module. (As usual, what we really mean is that the abelian group consisting of the real numbers under addition can be endowed with the structure of a module over various rings.)

Vector spaces with linear endomorphisms

Suppose $k$ is a field. If $V$ is an $k$-vector space and $T:V\to V$ is a $k$-linear transformation, then we can define an $k[x]$-module structure on $V$ by $p(x)v=p(T)v$ for every $p(x)\in k[x]$ and $v\in V$. It is straightforward to show^[4] the converse is also true, and that we have a bijection between the collection of $k[x]$-modules and pairs $(V,T)$ of consisting of a $k$-vector space and a $k$-linear endomorphism of that vector space. One can then also show that under this bijection the $k[x]$-submodules correspond to $T$-stable subspaces. We'll explore this more fully later.

Passing to quotient rings

Suppose $M$ is an $R$-module and $I\subseteq R$ is a two-sided ideal of $R$ that annihilates $M$, i.e., $im=0$ for all $i\in I$ and $m\in M$. Then we can define an $R/I$-module structure on $M$ in the obvious way: $(r+I)m=rm$ for $r\in R$ and $m\in M$. (You should verify that the "annihilation" condition is needed to ensure this action is well defined.)

For example, if $M$ is a finite abelian group of order $n$, then $nm=0$ for every $m\in M$, i.e., the ideal $⟨n⟩\subseteq \mathbf{Z}$ annihilates $M$. We can therefore also view $M$ as a $(\mathbf{Z}/n\mathbf{Z})$-module.

The zero module

For every ring $R$ the trivial group $\{0\}$ is also an $R$-module (with trivial action of $R$). This module is often denoted simply $0$. It is the zero object in the category of left $R$-modules, i.e., it is both an initial and a terminal object in the category. For each $R$-module $M$, there is a unique module morphism $0\to M$ (sending the element $0$ to $0_M$) as well as a unique module morphism $M\to 0$ (sending every element $m\in M$ to $0$). A module morphism $f:M\to N$ is trivial (i.e., $f(m)=0_N$ for all $m\in M$) exactly when it factors through the 0 module.

Suggested next notes

Module morphisms
Submodules
Quotient modules