LEC ALG3 3

Modules

The theory of modules is based on the observation that ${\text{End}}_{\text{Ab}}(G)$ is a ring for every abelian group $G$¹. Recall that we defined group actions as group homomorphisms from a fixed group to the group of automorphisms of a set. Analogously, the left-action of a ring $R$ on an abelian group $M$ is a homomorphism of rings $\sigma :R\to {\text{End}}_{\text{Ab}}(M)$. We say that $\sigma$ makes $M$ into a left $R$ module. The uncurried version $\rho$ has type $R\times M\to M$, with the relation between them given by $\rho (r,m)=\sigma (r)(m)$. $\rho (r,m)$ is denoted by $rm$. An equivalent definition using $\rho$ follows.

Definition 108.1(Module).

Let $R$ be a ring (with $1$). A left $R$ module is an additive abelian group $M$ with the operation $R\times M\to M$ , $(r,m)\mapsto rm$ with the following axioms

1. $(r+s)m=rm+sm$
2. $r(m+n)=rm+rn$
3. $rs(m)=r(sm)$
4. $1m=m$

(1) and (3) are due to $\sigma$ being a (ring) homomorphism; (2) is $\sigma (r)$ being a (group) endomorphism of $M$.

Ditto for right $R$ module. Similar to right group actions, a right $R$-module structure may be identified with a left-$R^{\circ }$ module structure, where $R^{\circ }$ is the ‘opposite ring’ obtained by reversing the order of multiplication. However, unlike the case for groups, $R$ and $R^{\circ }$ are not isomorphic in general.

Some trivial facts:

1. $0\cdot m=0$;
2. $(-1)\cdot m=-m$.

Proposition 108.2(Proposition).

Every abelian group is a $\mathbb{Z}$-module, in exactly one way.

Proof.

Let$G$ be an abelian group. Since $\mathbb{Z}$ is initial in $\text{Ring}$, there exists exactly one homomorphism

$$\mathbb{Z}\to {\text{End}}_{\text{Ab}}(G).$$ □

04c10a

Definition 108.3(Module homomorphisms).

A homomorphism of $R$-modules is a homomorphism of abelian groups which is compatible with the module structure. That is, if $M,N$ are $R$-modules and $\phi :M\to N$ is a function, then $\phi$ is a homomorphism of $R$-modules iff

• $\phi (m_1+m_2)=\phi (m_1)+\phi (m_2)$, for all $m_1,m_2\in M$;
• $\phi (rm)=r\phi (m)$, for all $r\in R$ and $m\in M$.

Given a ring $R$, the identity $M\to M$ is an $R$-module homomorphism. Clearly, the composition of two $R$-module homomorphisms is an $R$-module homomorphism. Thus, $R$-modules form a Category, denoted by $R\text{-Mod}$. The trivial group $0$ has a unique module structure over any ring $R$ and is a zero-object in $R\text{-Mod}$, that is, it is both initial and final. It can be easily verified that a bijective homomorphism of $R$-modules is an isomorphism in $R\text{-Mod}$. By Prp 2, the category $\mathbb{Z}\text{-Mod}$ is the same as the category $\text{Ab}$.

Proposition 108.4(Proposition).

Let $R,S$ be rings and $\phi :R\to S$ be a ring homomorphism. Then, $S$ is an $R$ module.

Proof.

Let$i:S\to {\text{End}}_{\text{Ab}}(S)$ be defined by $s\mapsto (r\mapsto sr)$ (it is easily seen that $r\mapsto sr$ is indeed a group homomorphism for all $s$). We need to verify that $i$ is a ring homomorphism:

$$\begin{array}{rl}i(s_1+s_2)& =r\mapsto (s_1+s_2)r\\ & =r\mapsto s_{1}r+s_{2}r\\ & =(r\mapsto s_{1}r)+(r\mapsto s_{2}r)\\ & =i(s_1)+i(s_2).\\ & \\ i(s_1{s}_2)& =r\mapsto s_1{s}_{2}r\\ & =(r\mapsto s_{1}r)\circ (r\mapsto s_{2}r)\\ & =i(s_1)\circ i(s_2).\end{array}$$

So, $i\circ \phi :R\to {\text{End}}_{\text{Ab}}(S)$ is a ring homomorphism, making (the underlying group of) $S$ an $R$-module. Concretely, the action is $(r,s)\mapsto \phi (r)s$.□

e65692

In Prp 4, if we require $R$ to be commutative and $\phi$ to map $R$ to the center of $S$, the ring operation in $S$ become compatible with the $R$-module structure:

$$(r_1{s}_1)(r_2{s}_2)=\phi (r_1)s_1\phi (r_1)s_2=\phi (r_1)\phi (r_2)s_1{s}_2=(r_1{r}_2)(s_1{s}_2).$$

Such a structure is called an $R$-algebra.

Definition 108.5($R$-algebra).

Let $R$ be a commutative ring. An $R$-algebra is a ring homomorphism $\alpha :R\to S$ such that $\alpha (R)$ is contained in the center of $S$.

419c68

Def 5 defines an $R$-algebra as a ring with compatible $R$-module structure; it can also be thought of as an $R$-module with a compatible ring structure.

Remark 108.6(Remark).

Recall that for $G,H\in \text{Ab}$, ${\text{Hom}}_{\text{Ab}}(G,H)\in \text{Ab}$, with the group operation inherited from $H$: $({\phi }_1+{\phi }_2)(g):={\phi }_1(g)+{\phi }_2(g)$. Note that commutativity is required for this prescription to yield an element in ${\text{Hom}}_{\text{Ab}}(G,H)$:

$$\begin{array}{rl}({\phi }_1+{\phi }_2)(g_1+g_2)& ={\phi }_1(g_1)+{\phi }_1(g_2)+{\phi }_2(g_1)+{\phi }_2(g_2)\\ & ={\phi }_1(g_1)+{\phi }_2(g_1)+{\phi }_1(g_2)+{\phi }_2(g_2)\\ & =({\phi }_1+{\phi }_2)(g_1)+({\phi }_1+{\phi }_2)(g_2).\end{array}$$

The remaining group axioms are easily verified.

Similarly, if $R$ is a commutative ring, each set ${\text{Hom}}_{R\text{-Mod}}(M,N)$ can be seen as an element of $R\text{-Mod}$. Indeed, let $M$ and $N$ be $R$-modules. Since homomorphisms of $R$-modules are in particular homomorphisms of abelian groups,

$${\text{Hom}}_{R\text{-Mod}}(M,N)\subseteq {\text{Hom}}_{\text{Ab}}(M,N).$$

The operation making ${\text{Hom}}_{\text{Ab}}(M,N)$ into a group clearly preserves ${\text{Hom}}_{R\text{-Mod}}(M,N)$, so the latter is an abelian group. For $r\in R$ and $\phi \in {\text{Hom}}_{R\text{-Mod}}(M,N)$, the prescription $(r\phi )(m):=r\phi (m)$ defines a function $r\phi :M\to N$. This function is an $R$-module homomorphism if $R$ is commutative, because

$$(r\phi )(am)=r\phi (am)=(ra)\phi (m)\stackrel{!}{=}(ar)\phi (m)=a(r\phi (m)).$$

Thus, we have a natural action of $R$ on the abelian group ${\text{Hom}}_{R\text{-Mod}}(M,N)$, and it is immediate to verify that the associated map $R\to {\text{End}}_{\text{Ab}}({\text{Hom}}_{R\text{-Mod}}(M,N))$ is a ring homomorphism, making ${\text{Hom}}_{R\text{-Mod}}(M,N)$ into an $R$-module.

e7f7db

Submodules and quotients

Definition 108.7(Definition).

A submodule $N$ of an $R$-module $M$ is a subgroup of $M$ preserved by the action of $R$ on $M$.

Example 108.8(Example).

1. $R$ itself is a left $R$-module; the submodules of $R$ are the left ideals of $R$.
2. Both the kernel and the image of an $R$-module homomorphism $\phi :M\to M^{\prime }$ are submodules (of $M$ and $M^{\prime }$, respectively).
3. If $r$ is in the center of $R$ and $M$ is an $R$-module, then $rM=\{rm|m\in M\}$ is a submodule of $M$. If $I$ is any (left-) ideal of $R$, then $IM=\left\{{\sum }_i{r}_i{m}_i|r_i\in I,m_i\in M\right\}$ is a submodule of $M$.

If $N$ is a submodule of $M$, then it is in particular a normal subgroup of the abelian group $(M,+)$, so $M/N$ is an abelian group. As usual, we are interested in the question: what properties must $N$ have so that the natural projection $\pi :M\to M/N$ is a $R$-module homomorphism?

If $\pi$ is a homomorphism, we have

$$r(m+N)=r\pi (m)=\pi (rm)=(rm+N)$$

for all $m\in M$. Thus, we are forced to define the action of $R$ on $M/N$ by

$$r(m+N):=rm+N.$$

It turns out (as can be easily verified) that this prescription defines a $R$-module structure on $M/N$ for all submodules $N$. This is very similar to what happens in $\text{Ab}$, where every subgroup is normal - unlike in $\text{Gp}$ or $\text{Ring}$, being a kernel poses no restriction on the relevant substructures.

The $R$-module $M/N$ is called the quotient of $M$ by $N$.

Example 108.9(Example).

If $R$ is a ring and $I$ is a two-sided ideal of $R$, then all three of $I$, $R$, and the quotient ring $R/I$ are $R$-modules. There are two ways to view the module $R/I$: As the module arising from the canonical projection $R\to R/I$ and Prp 4, or as the quotient of the $R$-module $R$ by its submodule $I$. The latter works even when $R$ is not commutative and $I$ is just a (say) left-ideal: the quotient $R/I$ is not defined as a ring, but it is defined as a left-module. The action of $R$ on $R/I$ is given by left multiplication: $r(a+I)=(ra+I)$.

The universal property and isomorphism theorems are what you’d expect; see Aluffi (2009, pp. 161, 162).

Finitely generated modules

Definition 108.10(Finitely generated module).

The left $R$-module $M$ is finitely generated if there exist $a_1,\dots ,a_n$ in $M$ such that for any $x\in M$, there exist $r_1,\dots ,r_n\in R$ with $x=r_1{a}_1+r_2{a}_2+\dots r_n{a}_n$.

db6ba2

Warning

$M$ being a finitely generated abelian group is not the same thing as it being a finitely generated $R$-module. $\mathbb{R}$ is not finitely generated as an abelian group, but is finitely generated by $\{1\}$ as an $\mathbb{R}$-module.

Footnotes

1. If $G$, $H$ are abelian groups, then ${\text{Hom}}_{\text{Ab}}(G,H)$ is also an abelian group, with addition defined pointwise. ${\text{End}}_{\text{Ab}}(G)={\text{Hom}}_{\text{Ab}}(G,G)$ further allows its constituent morphisms to be composed; it is a ring with multiplication defined to be composition. See Aluffi (2009) III.1.1. ↩

---

References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.