Module Theory, Part I: Introduction, Module Homomorphisms

September 14th, 2017

Definition: Let $R$ be a ring. A module $M$ over a ring $R$ is an abelian group (with operation $+$) and a map $R\times M \rightarrow M$ which satisfies distributivity and associativity. If $R$ has a $1$, then we require that $1m = m$ for each $m \in M$.

These axioms should look fairly familiar. if $R$ is a field, then a module is exactly a vector space over $R$. A module is nothing more than a generalization of vector spaces.

Definition: A submodule is a closed subgroup $N$ of an $R$-module $M$ which is closed under the action of $R$.

Example

A significant example is modules over $\mathbb{Z}$. The action of an integer on $m \in M$ is defined straightforwardly as:

$$ nm = m + m + \dots + m $$

Where we are adding $m$ to itself $n$ times. This is the only possible action of $\mathbb{Z}$ over $M$, because of associativity and distributivity. What we have from this is that:

$\mathbb{Z}$-modules are exactly abelian groups.

In particular, $\mathbb{Z}$-submodules are exactly submodules of subgroups.

Example

By associativity, we can define a module over $F[x]$, where $F$ is a field. We simply need to define how $1, x$ act on elements in the module. Let $V$ be a vector space over $F$ – we will make $V$ an $F[x]$ module, by identifying the action of $x$ with a linear transformation $T: V \rightarrow V$.

Conversely, if we have any module $V$ over $F[x]$, then in particular $V$ is a module over $F$. But we know that:

$$\begin{aligned} x(v+w) &= xv+xw \\ x(av) &= ax(v) \end{aligned}$$

So this means that indeed $x$ is a linear transformation. So there is a natural isomorphism between vector spaces $V$ over $F$ equipped with a linear transformation $T$ and modules $V$ over $F[x]$.

Consequently, the $F[x]$-submodules of $V$ are exactly vector subspaces of $V$ which are invariant under $T$.

Proposition

A nonempty $N$ of an $R$-module $M$ is a submodule iff $x+ry \in N$ for each $x,y \in N$, $r \in R$.

Let $r=-1$ and we get the subgroup criterion. Let $x=0$ and we get closure under elements of $R$. The converse case is fairly straightforward.

Quotient Modules and Module Homomorphisms

Definition: Let $R$ be a ring and $M,N$ are $R$-modules. Then an $R$-module homomorphism $\varphi$ is a map from $M$ to $N$ so that:

$$\begin{aligned} \varphi(x+y) &= \varphi(x) + \varphi(y) \\ \varphi(rx) &= r\varphi(x) \end{aligned}$$

As expected, an isomorphism is surjective as well as injective. The kernel and images are respectively submodules of $M, N$ as expected. Finally, we define $Hom_R(M,N)$ to be the set of all $R$-module homomorphisms from $M$ to $N$.

For example, $\mathbb{Z}$-module homomorphisms are simply abelian group homomorphisms (since the second criterion is implied by the first above). Over a field, the $F$-module homomorphisms are simply linear transformations between vector spaces. Note, however, that $R$-module homomorphisms where $R$ is a ring do not necessarily have any connection to ring homomorphisms – specifically because there is no requirement that a module homomorphism send identity to identity.

Proposition

$Hom_R(M, N)$ is an $R$-module.

We can define addition and multiplication in the usual way:

$$\begin{aligned} (\varphi + \psi)(m) &= \varphi(m) + \psi(m) \\ (r\varphi)(m) &= r(\varphi(m)) \end{aligned}$$

Furthermore, if $M=N$ then we can have a well-defined ring structure; multiplication is just composition. Indeed, $Hom_R(M,M)$ is a ring with identity – and indeed it has a special name.

Definition: The ring $Hom_R(M,M)$ is called the endomorphism ring of $M$ and is denoted $End(M)$ or $End_R(M)$.

Proposition

Let $R$ be a ring and let $M,N$ be $R$-modules with $N$ a submodule of $M$. Then $M/N$ (an abelian quotient group) can be made into a module over $R$ by defining:

$$\begin{aligned} r(x+N) &= rx+N \end{aligned}$$

And we have a natural projection map $\pi: M \rightarrow M/N$ with kernel $N$.

Finally, we define the sum of two modules:

$$\begin{aligned} A+B = \{a+b\ :\ a \in A, b \in B \} \end{aligned}$$

So that we can once more define the isomorphism theorems.

Theorem (Isomorphism Theorems)
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