Chapter 22

Rigorous

Modules over Rings & Structure Theorems

00 · Symbol Glossary

$_RM$An R-module M

An abelian group $M$ equipped with scalars from a ring $R$ instead of a field — the generalization of a vector space this chapter builds.

$rm$Scalar action of r on m

The result of "scaling" $m\in M$ by $r\in R$ , defined as part of the module structure — notation deliberately mirrors vector space scalar multiplication.

$R^n$Free module of rank n

The set of $n$ -tuples of elements of $R$ , with componentwise operations — the direct module analog of $\mathbb{R}^n$ , but over any ring $R$ .

$\mathrm{Tor}(M)$Torsion submodule

$\{m\in M : rm=0 \text{ for some nonzero } r\in R\}$ (for $R$ a domain) — the elements that "vanish" under some nonzero scaling, with no vector-space analog since fields have no nonzero zero-scalars.

01 · The Definition

Vector spaces required scalars from a field — every nonzero scalar invertible. Relaxing this single requirement, allowing scalars from any ring, gives a vastly more general (and more common) structure.

Definition — Module

Let $R$ be a ring with unity. An $R$ -module $M$ is an abelian group $(M,+)$ together with a scalar action $R\times M\to M$ , $(r,m)\mapsto rm$ , satisfying for all $r,s\in R$ , $m,n\in M$ :

M1. $r(m+n) = rm+rn$ ; M2. $(r+s)m = rm+sm$ ; M3. $(rs)m = r(sm)$ ; M4. $1_R\cdot m = m$ .

A vector space is exactly a module over a field

Every vector-space axiom is M1–M4 verbatim, with $R$ specialized to a field. Modules generalize vector spaces the same way rings generalized fields (Chapter 17) — by dropping the requirement that every nonzero scalar be invertible.

02 · Examples

Example — Every abelian group is a Z-module, uniquely

For an abelian group $(M,+)$ , define $n\cdot m$ (for $n\in\mathbb{Z}$ ) by repeated addition: $n\cdot m = m+\cdots+m$ ( $n$ times) for $n>0$ , $0\cdot m=0$ , $(-n)\cdot m = -(n\cdot m)$ . Checking M1–M4 directly (distributivity of repeated addition, associativity of "repeat $n$ times then repeat $s$ times" $=$ "repeat $ns$ times") confirms this makes $M$ a $\mathbb{Z}$ -module. Conversely, any $\mathbb{Z}$ -module structure on $M$ must satisfy this same formula (M2 and M4 force $2\cdot m = (1+1)m = m+m$ , and induction extends this to every integer) — so $\mathbb{Z}$ -modules are exactly abelian groups, with no additional choice involved. This identification is used constantly for the rest of the chapter.

Example — R as a module over itself, and ideals as submodules

$R$ is an $R$ -module via its own multiplication as the scalar action (M1–M3 are exactly the ring distributive and associative laws; M4 is the unity law). A submodule of $R$ (an additive subgroup closed under the scalar action from $R$ ) is exactly an ideal — absorption (I2 from Chapter 12) is precisely M1-style closure under scalar multiplication by all of $R$ .

Example — The free module R^n

$R^n = \{(a_1,\ldots,a_n) : a_i\in R\}$ , with componentwise addition and $r(a_1,\ldots,a_n) := (ra_1,\ldots,ra_n)$ . Direct verification of M1–M4 mirrors $\mathbb{R}^n$ 's vector-space axioms exactly. Called free of rank $n$ , generalizing the standard basis picture from linear algebra.

03 · Submodules, Homomorphisms, and Quotients

Definition — Submodule and Module Homomorphism

$N\subseteq M$ is a submodule if $(N,+)\le(M,+)$ and $rn\in N$ for every $r\in R$ , $n\in N$ . A function $\varphi: M\to N$ (both $R$ -modules) is a module homomorphism if $\varphi(m_1+m_2)=\varphi(m_1)+\varphi(m_2)$ and $\varphi(rm)=r\varphi(m)$ .

Every construction from Chapters 04–06 transfers directly

$M/N$ (cosets $m+N$ , with $r(m+N):=rm+N$ ) is a module exactly when $N$ is a submodule, by the identical well-definedness argument as Theorem 5.3 and Theorem 12.4 (subtract representatives, absorb the difference using M1). $\ker\varphi$ is always a submodule (Theorem 6.3/13.3-style: additive kernel, plus $\varphi(rm)=r\varphi(m)=r\cdot0=0$ for $m\in\ker\varphi$ ). The First Isomorphism Theorem for modules — $M/\ker\varphi \cong \mathrm{im}\,\varphi$ — holds by the identical proof strategy as Theorem 6.5 and Theorem 13.6, checking the extra scalar condition $\bar\varphi(r(m+N)) = r\bar\varphi(m+N)$ alongside the additive argument already established twice before.

04 · Free Modules, Generation, and a Key Failure

Definition — Generating Set, Free Module

$M$ is generated by $S\subseteq M$ if every $m\in M$ is an $R$ -linear combination $\sum r_is_i$ ( $r_i\in R$ , $s_i\in S$ ). $M$ is free of rank $n$ if $M\cong R^n$ .

Common mistake — Not every module has a basis

Consider $\mathbb{Z}_2$ as a $\mathbb{Z}$ -module (Section 02). Is $\{1\}$ a "basis"? It generates ( $0\cdot1=0$ , $1\cdot1=1$ ), but is it independent in the vector-space sense? We'd need: $r\cdot1=0 \implies r=0$ . But $2\cdot1 = 1+1=0$ in $\mathbb{Z}_2$ , with $2\neq0$ in $\mathbb{Z}$ . A nonzero scalar kills a nonzero element — impossible for a field's scalars acting on a nonzero vector, but routine for a general ring. $\mathbb{Z}_2$ is not a free $\mathbb{Z}$ -module; no independent generating set exists (any single generator faces this same obstruction, and the module is too small for two independent generators).

05 · Torsion

Definition — Torsion Element and Torsion Submodule

Let $R$ be a domain. $m\in M$ is a torsion element if $rm=0$ for some nonzero $r\in R$ . $\mathrm{Tor}(M) := \{m\in M : m \text{ is torsion}\}$ .

Example — Z_n is entirely torsion, Z is torsion-free

In $\mathbb{Z}_n$ (as a $\mathbb{Z}$ -module): $n\cdot[a] = [na] = [0]$ for every $[a]$ , and $n\neq0$ . So $\mathrm{Tor}(\mathbb{Z}_n) = \mathbb{Z}_n$ entirely.

In $\mathbb{Z}$ (as a $\mathbb{Z}$ -module over itself): if $r\cdot a=0$ with $r\neq0$ , then $ra=0$ in the domain $\mathbb{Z}$ , forcing $a=0$ (Theorem 11.5's cancellation, or directly: no zero divisors). So $\mathrm{Tor}(\mathbb{Z})=\{0\}$ — torsion-free.

06 · The Structure Theorem for Finitely Generated Modules over a PID

Theorem 22.1 — Structure Theorem (statement)

Let $R$ be a PID and $M$ a finitely generated $R$ -module. Then:

M \;\cong\; R^r \oplus R/(d_1) \oplus R/(d_2) \oplus \cdots \oplus R/(d_k)

for some $r\geq0$ and nonzero non-units $d_1\mid d_2\mid\cdots\mid d_k$ in $R$ (the invariant factors), with $r$ and the $d_i$ (up to associates) uniquely determined by $M$ . Each summand $R/(d_i)$ is a cyclic torsion piece killed by multiplying by $d_i$ ; the divisibility chain $d_1\mid d_2\mid\cdots\mid d_k$ makes $d_k$ the largest invariant factor controlling the longest torsion cycle.

A genuinely deep classification theorem, taken as given

The full proof — typically via Smith normal form (reducing a presentation matrix for $M$ to a diagonal form by row and column operations valid in a PID) — is a substantial undertaking on its own, comparable in scope to an entire chapter of a dedicated algebra course. We state it here, in the same spirit as the Fundamental Theorem of Algebra and Galois's solvability criterion, because its consequence is one of the most satisfying classification results in this entire course.

07 · Application: The Classification of Finite Abelian Groups

Theorem 22.2 — Fundamental Theorem of Finite Abelian Groups

Every finite abelian group is isomorphic to a direct sum of cyclic groups of prime-power order:

G \;\cong\; \mathbb{Z}_{p_1^{e_1}} \times \mathbb{Z}_{p_2^{e_2}} \times \cdots \times \mathbb{Z}_{p_k^{e_k}}

and this decomposition is unique up to reordering the factors.

Proof

$\mathbb{Z}$ is a PID (Euclidean, Theorem 15.1). By Section 02's Example, $G$ (a finite abelian group) is a finitely generated $\mathbb{Z}$ -module (finite, hence generated by its own — finitely many — elements). Since $G$ is finite, the free rank $r$ in Theorem 22.1 must be $0$ (a copy of $\mathbb{Z}=R^1$ inside the decomposition would already be infinite). So $G\cong \mathbb{Z}/(d_1)\oplus\cdots\oplus\mathbb{Z}/(d_k) = \mathbb{Z}_{d_1}\times\cdots\times\mathbb{Z}_{d_k}$ .

Each $\mathbb{Z}_{d_i}$ further splits via the Chinese Remainder Theorem (Theorem 16.1): writing $d_i = p_1^{a_1}\cdots p_m^{a_m}$ (prime factorization), the corresponding ideals $(p_j^{a_j})$ are pairwise comaximal, so $\mathbb{Z}_{d_i} \cong \mathbb{Z}_{p_1^{a_1}}\times\cdots\times\mathbb{Z}_{p_m^{a_m}}$ — splitting each cyclic piece into prime-power cyclic pieces. Combining across all $i$ gives the stated decomposition. Uniqueness is inherited from Theorem 22.1's uniqueness of invariant factors (equivalently, of the resulting prime-power pieces, called elementary divisors).

Step-by-step — All abelian groups of order 12

Factor:

12=2^2\times3

. By Theorem 22.2, an abelian group of order 12 is a product of a "2-part" of order 4 and a "3-part" of order 3.

List decompositions of the 2-part (order 4): either

\mathbb{Z}_4

(single prime-power piece

2^2

) or

\mathbb{Z}_2\times\mathbb{Z}_2

(two pieces,

2^1,2^1

) — exactly the two partitions of the exponent

2

into positive integers (

2

, or

1+1

List decompositions of the 3-part (order 3): only

\mathbb{Z}_3

(3 is prime, only one partition of the exponent

1

Combine:

\mathbb{Z}_4\times\mathbb{Z}_3

\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3

— exactly two abelian groups of order 12, up to isomorphism.

Simplify using Theorem 7.3: since

\gcd(4,3)=1

\mathbb{Z}_4\times\mathbb{Z}_3\cong\mathbb{Z}_{12}

. And

\gcd(2,3)=1

, so

\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3 \cong \mathbb{Z}_2\times\mathbb{Z}_6

. The two abelian groups of order 12 are $\mathbb{Z}_{12}$ and $\mathbb{Z}_2\times\mathbb{Z}_6$ . Note the count came from the exponents alone: the

2

-part contributed two options (the two partitions of the exponent

2

) and the

3

-part exactly one, giving

2\times1=2

— the same bookkeeping that Chapter 09's order-4 example carried out by hand for a single prime, now systematic across all prime factors at once.

The number of partitions counts the possibilities

In general, the number of abelian groups of order $p_1^{e_1}\cdots p_k^{e_k}$ (up to isomorphism) is the product of the partition numbers $p(e_1)\times\cdots\times p(e_k)$ , where $p(e)$ counts the ways to write $e$ as a sum of positive integers. This single combinatorial fact — a direct reading of Theorem 22.2 — answers, completely and finally, the question "how many groups of a given order are there?" for the abelian case, a question that occupied Sylow-theoretic case analysis for nonabelian groups throughout Chapter 09.

08 · Exercises

EXERCISE 22.1

Check M1–M4 directly for the given scalar action, treating $R=\mathbb{Z}_n$ acting on itself.

$\mathbb{Z}_n$ is a ring, hence an $R$ -module over itself with $R=\mathbb{Z}_n$ (Section 02's " $R$ as a module over itself" example, specialized). M1–M4 are exactly the ring distributive/associative/unity laws already verified in Chapter 11.

Confirm that $\mathbb{Z}_n$ is a module over itself, citing which ring axioms supply M1–M4.

EXERCISE 22.2

Check whether some nonzero integer kills every element of $\mathbb{Z}_6\times\mathbb{Z}_{10}$ simultaneously.

For any $(a,b)\in\mathbb{Z}_6\times\mathbb{Z}_{10}$ : $6\cdot(a,b) = (6a,6b) = (0, 6b)$ — not obviously zero in the second coordinate unless $6b\equiv0\pmod{10}$ , which isn't automatic. But $30\cdot(a,b) = (30a,30b) = (0,0)$ since $30$ is a multiple of both $6$ and $10$ . So every element is torsion (killed by the nonzero integer 30, or more efficiently by $\mathrm{lcm}(6,10)=30$ ). $\mathrm{Tor}(\mathbb{Z}_6\times\mathbb{Z}_{10}) = \mathbb{Z}_6\times\mathbb{Z}_{10}$ entirely — consistent with Section 05's fact that finite $\mathbb{Z}$ -modules are always entirely torsion.

Is $\mathbb{Z}_6\times\mathbb{Z}_{10}$ (as a $\mathbb{Z}$ -module) entirely torsion? Find a single nonzero integer killing every element.

EXERCISE 22.3

Follow the exact template of Section 07's order-12 computation, using the partitions of the exponent in $8=2^3$ .

$8=2^3$ . Partitions of $3$ : $3$ , $2+1$ , $1+1+1$ — three partitions, giving three abelian groups: $\mathbb{Z}_8$ , $\mathbb{Z}_4\times\mathbb{Z}_2$ , $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ .

There are exactly 3 abelian groups of order 8, up to isomorphism (alongside the two nonabelian ones, $D_4$ and the quaternion group $Q_8$ , encountered only in passing in this course — for a total of 5 groups of order 8 overall).

Using Theorem 22.2, list all abelian groups of order 8, up to isomorphism.

09 · Chapter Summary

Concept	Statement
$R$ -module	Abelian group + scalar action satisfying M1–M4
Vector space	The special case $R$ = a field
Abelian groups = $\mathbb{Z}$ -modules	A unique, forced correspondence
Ideals	Exactly the submodules of $R$ as a module over itself
Free module $R^n$	Componentwise $n$ -tuples, generalizing $\mathbb{R}^n$
Not every module is free	E.g. $\mathbb{Z}_2$ as a $\mathbb{Z}$ -module
Torsion	$rm=0$ for some nonzero $r$ ; $\mathbb{Z}_n$ entirely torsion, $\mathbb{Z}$ torsion-free
Structure Theorem	Finitely generated modules over a PID decompose via invariant factors (Thm 22.1)
Finite Abelian Groups	Unique decomposition into prime-power cyclic pieces (Thm 22.2)

Course complete. This closes the Abstract Algebra sequence: from the bare axioms of Chapter 01, through the full architecture of groups (Chapters 02–10), rings (Chapters 11–16), and fields culminating in Galois theory (Chapters 17–21), to this final chapter's modules — which reveal that groups, rings, vector spaces, and ideals were, all along, instances of a single underlying idea.