# Free module
A free module is a module over a ring that has a basis, meaning a subset such that every element of the module can be written in exactly one way as a finite linear combination of basis elements with coefficients in the ring. Equivalently, a free module is a direct sum of copies of the underlying ring. Free modules are the module-theoretic generalization of vector spaces, and they reduce to vector spaces when the ring is a field.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) • [2](https://math.hodapp.com/lectures/foundations/modules/modules.html)
| Key fact | Detail |
|---|---|
| Definition | An R-module M is free if it has a basis, equivalently if it is isomorphic to a direct sum of copies of R indexed by some set[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) • [2](https://math.hodapp.com/lectures/foundations/modules/modules.html) |
| Rank n case | The free module of rank n is Rn = R ⊕ R ⊕ · · · ⊕ R (n copies); over a field F this is exactly Fn, the standard n-dimensional vector space[3](https://ocw.mit.edu/courses/res-18-012-algebra-ii-student-notes-spring-2022/mit18_702s22_lect19.pdf) |
| Standard examples | Rn for any positive integer n; the matrix ring Mmn(R) with basis Eij; the polynomial ring R[X] with basis 1, X, X2, ...[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) |
| Zero module | By convention the zero module is free with the empty set as basis[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) |
| Rank uniqueness | Any two bases of a free module over a commutative ring have the same cardinality; over a division ring every module is free with a specific dimension or rank[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) • [4](https://www.mathreference.com/mod%2Cbasis.html) |
| Failure in general | There are rings for which Rn ≅ Rn+t with t ≠ 0, so rank need not be an invariant[5](https://sites.math.rutgers.edu/~weibel/Kbook/Kbook.pdf) |
| Quotients | Every module is a quotient of a free module[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) |
## Basis and the direct-sum characterization
A subset S of an R-module M is linearly independent if a finite linear combination of its elements equals 0 only when all coefficients are 0, and it spans M if every element of M is a finite linear combination of members of S. A basis is a linearly independent spanning set, and a module that has a basis is free. [Linear independence](https://www.edgepedia.org/linear-independence) makes each representation unique: if two different combinations produced the same element, subtracting them would give a nontrivial combination equal to 0.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) • [4](https://www.mathreference.com/mod%2Cbasis.html)
The direct-sum description follows from the basis description. If M has basis (bi, i ∈ I), the submodule spanned by a single basis element bi is the set of all r bi with r ∈ R, and the map r ↦ r bi is an R-module isomorphism from R onto that submodule because {bi} is independent. Since the basis elements span M, M is the sum of these submodules, and independence makes the sum direct. Conversely, a direct sum of copies of R is free: the elements ei with 1 in the i-th component and zeros elsewhere form a basis.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
Free on a set is a related universal formulation: an R-module M is free on a subset X if for any R-module N and any function from X to N, there exists a unique R-module homomorphism M → N extending it. This holds exactly when X generates M and every finite relation ∑ cixi = 0 with xi ∈ X forces all ci = 0. In category theory, the free object defined by this universal property, when specialized to the category of unitary modules, yields the direct-sum definition.[6](https://math.ucla.edu/~sharifi/notes/algebra-ch05.html) • [4](https://www.mathreference.com/mod%2Cbasis.html)
## Examples
The characterization as a direct sum of copies of R makes many standard objects recognizable as free modules:[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
1. For any positive integer n, Rn is a free R-module, the canonical example of a based free module, consisting of n-tuples of elements of R.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) • [5](https://sites.math.rutgers.edu/~weibel/Kbook/Kbook.pdf)
2. The matrix ring Mmn(R) is a free R-module with basis the matrix units Eij for i = 1, ..., m and j = 1, ..., n.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
3. The polynomial ring R[X] is a free R-module with basis 1, X, X2, ....[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
Not every module is free. The Z-module Z/nZ is finitely generated, by the class of 1, but it is not free, since a free abelian group has no element of finite order other than 0.[2](https://math.hodapp.com/lectures/foundations/modules/modules.html) Over an integral domain, the existence of nonzero elements α ∈ R and x ∈ M with αx = 0 is an obstruction to M being free.[7](https://www.sciencedirect.com/topics/mathematics/free-module)
## Rank and the invariant basis property
If R is a division ring, every module is free, with a basis and a specific dimension or rank; this is the setting of ordinary linear algebra.[4](https://www.mathreference.com/mod%2Cbasis.html) For arbitrary rings the situation is weaker. Any two bases for a free module over a commutative ring have the same cardinality, proved by reducing modulo a maximal ideal: for a maximal ideal I of R, the quotient M/IM is a vector space over the field R/I, and a basis of M maps to a basis of this vector space, whose dimension is independent of the chosen basis.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
**Rings without invariant rank.** There are rings for which Rn ≅ Rn+t with t ≠ 0. A ring satisfies the invariant basis property (IBP, also called invariant basis number) if Rm and Rn are not isomorphic for m ≠ n, in which case the rank of a free module is an invariant independent of the choice of basis. Most rings encountered in practice satisfy it: commutative rings do, as do group rings Z[G], and any ring admitting a ring map to a field or division ring. For a finite-dimensional algebra R over a field F, the rank of a free module M is the invariant rank(M) = dimF(M)/dimF(R).[5](https://sites.math.rutgers.edu/~weibel/Kbook/Kbook.pdf) For infinite cardinals κ, a free module R(κ) on a basis of cardinality κ has every basis of cardinality κ, so it cannot be isomorphic to Rn for finite n.[5](https://sites.math.rutgers.edu/~weibel/Kbook/Kbook.pdf)
## Direct sums and submodules
The direct sum of free modules is free: taking the union of the basis elements over each component builds a basis for the sum. The converse fails. Z6 is the direct product of Z2 and Z3, yet the two summands are not free Z6-modules.[4](https://www.mathreference.com/mod%2Cbasis.html) For finite index sets the direct sum and direct product coincide, M1 ⊕ · · · ⊕ Mn = M1 × · · · × Mn, but this is not true for infinite sums and products.[3](https://ocw.mit.edu/courses/res-18-012-algebra-ii-student-notes-spring-2022/mit18_702s22_lect19.pdf) The direct product of infinitely many copies of Z is not a free Z-module.[8](https://math.virginia.edu/graduate/exams/algebra/alg-prep-3-1.html)
**Submodules over a PID.** Over a general ring, a submodule of a free module need not be free. Over a principal ideal domain (a ring in which every ideal is generated by one element, such as Z or a polynomial ring F[x] over a field), the situation is better: every submodule E of a free R-module F is itself free, and the rank of E is no greater than the rank of F. Consequently, every finitely generated torsion-free module over a PID is free.[9](https://people.math.sc.edu/mcnulty/algebra/grad/ringandmodules.pdf) • [10](https://math.ucla.edu/~sharifi/notes/algebra-ch08.html) This underlies the structure theorem for finitely generated modules over a PID, which decomposes any such module as a direct sum of a free module and cyclic torsion modules, generalizing both the classification of finitely generated abelian groups (the case R = Z) and the Jordan canonical form (the case R = F[x]).[2](https://math.hodapp.com/lectures/foundations/modules/modules.html)
## Homomorphisms and matrices
A homomorphism of R-modules is a homomorphism of abelian groups compatible with the R-action, φ(rm) = rφ(m); over a field this is exactly a linear map.[3](https://ocw.mit.edu/courses/res-18-012-algebra-ii-student-notes-spring-2022/mit18_702s22_lect19.pdf) Homomorphisms from a free module are easy to control: once a homomorphism is defined on a basis, it is defined on all of M, since values on basis elements extend uniquely by linearity. This is the module analog of a linear transformation on [Euclidean space](https://www.edgepedia.org/euclidean-space) being determined by its values on the standard basis vectors.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
For free modules M and N of finite ranks n and m with fixed ordered bases, every homomorphism M → N is represented by an m × n matrix with entries in R, with matrix multiplication implementing composition; homomorphisms between based free modules are naturally identified with matrices over R. The automorphisms of Rn form the group GLn(R) of invertible n × n matrices.[1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf) • [5](https://sites.math.rutgers.edu/~weibel/Kbook/Kbook.pdf)
**Why free modules matter.** Free modules play a central role in algebra because any module is the homomorphic image of some free module: given generators u1, ..., un of M, the map sending a1t1 + · · · + antn to a1u1 + · · · + anun is a surjective homomorphism from the free module on generators t1, ..., tn onto M. By the first isomorphism theorem, every module is a quotient of a free module, which is the starting point for constructing free resolutions in homological algebra.[11](https://mathworld.wolfram.com/FreeModule.html) • [1](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
## References
1. [Chapter 4, Module Fundamentals (Ash), University of Texas at Austin](https://web.ma.utexas.edu/users/voloch/Homework/ash-ch4.pdf)
2. [Modules — vector spaces over a ring, Math foundations lecture notes](https://math.hodapp.com/lectures/foundations/modules/modules.html)
3. [Lecture 19: Modules over a Ring, MIT RES.18-012 Algebra II Student Notes (Spring 2022)](https://ocw.mit.edu/courses/res-18-012-algebra-ii-student-notes-spring-2022/mit18_702s22_lect19.pdf)
4. [Modules, Free Module, Basis, Math Reference](https://www.mathreference.com/mod%2Cbasis.html)
5. [An Introduction to K-Theory (Weibel), Rutgers University](https://sites.math.rutgers.edu/~weibel/Kbook/Kbook.pdf)
6. [Abstract Algebra — Chapter 5: Advanced ring theory (Sharifi), UCLA](https://math.ucla.edu/~sharifi/notes/algebra-ch05.html)
7. [Free Module — an overview, ScienceDirect Topics](https://www.sciencedirect.com/topics/mathematics/free-module)
8. [Algebra General Exam Review, 3.1 Modules, University of Virginia](https://math.virginia.edu/graduate/exams/algebra/alg-prep-3-1.html)
9. [Ring and Modules lecture notes (McNulty), University of South Carolina](https://people.math.sc.edu/mcnulty/algebra/grad/ringandmodules.pdf)
10. [Abstract Algebra — Chapter 8: Topics in ring theory (Sharifi), UCLA](https://math.ucla.edu/~sharifi/notes/algebra-ch08.html)
11. [Free Module, Wolfram MathWorld](https://mathworld.wolfram.com/FreeModule.html)
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*Topic: Encyclopedia › Physical world and mathematics › Mathematics and statistics › Numbers and algebra › Algebraic structures › Module theory › Classes of modules*
*Initially written Sep 17, 2026 · Reviewed: Oct 15, 2026 · Edited: Oct 15, 2026 · Last review: Oct 15, 2026*
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