---
title: "Modules: Linear Algebra Over a Ring"
description: "A module is a vector space whose scalars form a ring rather than a field. Bases may fail to exist, and what survives is a structure theorem that delivers the Jordan form and the classification of fini"
canonical: https://lightmysky.com/learn/mathematics/modules-linear-algebra-over-a-ring-mt_0GMVp-s8wD
source: https://lightmysky.com/learn/mathematics/modules-linear-algebra-over-a-ring-mt_0GMVp-s8wD.md
retrieved: 2026-09-12
---

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# Modules: Linear Algebra Over a Ring

A module is a vector space whose scalars form a ring rather than a field. Bases may fail to exist, and what survives is a structure theorem that delivers the Jordan form and the classification of finite abelian groups at the same time.

Subject: Mathematics · Area: Abstract Algebra · Ages 22 to 24
Page: https://lightmysky.com/learn/mathematics/modules-linear-algebra-over-a-ring-mt_0GMVp-s8wD

## Ready when they can

- Give a module with no basis and say which vector-space step fails
- State the structure theorem for finitely generated modules over a principal ideal domain
- Read the Jordan form of a matrix as one instance of that theorem

## Lesson: Linear algebra with ring scalars

You already know vector spaces, where scalars come from a field and every nonzero scalar has an inverse. A module keeps the same addition rules but lets scalars come from a ring, so division can fail. That failure creates torsion: a nonzero scalar times a nonzero element can be zero. Integers mod 2 as a module over integers show this, since 2 times anything is 0, so no independent set exists and the module has no basis.

**Example.** You compare this with a free case. Pairs of integers form a free module of rank 2 with basis (1, 0) and (0, 1). Vector spaces always have bases, but modules need not, and the step that breaks is invertibility of nonzero scalars. Integers mod 6 show the same failure, since 2 times 3 is 0 mod 6 with both factors nonzero, so independence fails there too.

You then use the structure theorem for finitely generated modules over a principal ideal domain. It splits each module into a free part plus torsion cyclic pieces of prime power order. So integers mod 6 split as cyclic of order 2 plus cyclic of order 3, and sizes multiply to give 4 for two order 2 pieces and 24 for orders 4 and 6. Over polynomials the same theorem gives Jordan form: eigenvalue 5 with one eigenvector means one block with 5 on the diagonal and 1 above.

**Recap.** You allow ring scalars, watch torsion block bases, and split modules into free rank plus cyclic torsion, which also gives Jordan form.

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Polynomial Rings, Irreducibility and Unique Factorisation](https://lightmysky.com/learn/mathematics/polynomial-rings-irreducibility-and-unique-factorisation-mt_m2G9PIHj6y)
- [Diagonalisation and Powers of a Matrix](https://lightmysky.com/learn/mathematics/diagonalisation-and-powers-of-a-matrix-mt_w15uCVhdhX)
- [Vector Spaces and Subspaces](https://lightmysky.com/learn/mathematics/vector-spaces-and-subspaces-mt_Zrddx-E6_n)

## Opens up

- [Group Representations and the Group Algebra](https://lightmysky.com/learn/mathematics/group-representations-and-the-group-algebra-mt_y5JlvQGKYW)