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Free module

In ring theory, a free module over a ring R with identity is an R-module M that admits a basis, which is a linearly independent subset of M that generates M through all possible R-linear combinations. Equivalently, M is isomorphic to a direct sum \bigoplus_{i \in I} R of copies of R, indexed by some set I whose cardinality determines the rank of the module.^[1]^[2] Free modules satisfy a universal property: for any set X and R-module M freely generated by X, any R-linear map from X to another R-module N extends uniquely to an R-linear homomorphism from M to N. The ring R itself is a free module of rank 1, generated by the identity element $1_R, and finite direct sums like R^n serve as prototypical examples of rank-n free modules. Over a field F, free F-modules coincide with vector spaces, where bases are the standard Hamel bases.^[2]^[3]^[1] Not all modules are free; for instance, the cyclic group \mathbb{Z}/n\mathbb{Z} is not free as a \mathbb{Z}-module for n > 1, since it lacks a basis. Every R-module is a quotient of some free module, highlighting their foundational role in module theory. In free modules with infinite bases, all bases have the same cardinality, ensuring a well-defined notion of rank even in infinite dimensions. These structures are central to homological algebra and representation theory, generalizing vector spaces to the setting of commutative and non-commutative rings.^[4]^[2]^[1]

Fundamentals

Definition

In ring theory, a module M over a ring R with identity is called a free R-module if there exists a subset B \subseteq M, called a basis, such that every element of M can be uniquely expressed as a finite linear combination \sum_{i=1}^n r_i b_i with r_i \in R and distinct b_i \in B (the sum having finite support, meaning only finitely many r_i are nonzero).^[1] The basis B is said to span M if every element arises as such a linear combination, and the elements of B are linearly independent over R if whenever \sum_{i=1}^n r_i b_i = 0 with distinct b_i \in B and r_i \in R, it follows that all r_i = 0. This uniqueness of expression follows directly from linear independence, ensuring that the representation is well-defined without ambiguity.^[1] Free modules generalize the notion of vector spaces, which are precisely the free modules over a field (a special case of a ring). When R = \mathbb{Z}, the integers, a free \mathbb{Z}-module is known as a free abelian group, where the basis elements generate the group additively with integer coefficients.^[1] Equivalently, a free R-module M with basis B is isomorphic to the direct sum of copies of R indexed by B, denoted M \cong \bigoplus_{b \in B} R.^[1]

Basis

In a free module M over a ring R, a basis B \subseteq M is defined as a generating set that is linearly independent over R. This means that every element of M can be expressed uniquely as a finite linear combination \sum_{b \in B} r_b b with coefficients r_b \in R, where only finitely many r_b are nonzero, and the relation \sum_{b \in B} r_b b = 0 implies all r_b = 0.^[5]^[6] The module M generated by such a basis is precisely the set of all such finite sums:

M = \left\{ \sum_{b \in B} r_b b \;\middle|\; r_b \in R, \; |\{b \in B : r_b \neq 0\}| < \infty \right\}.

^[6] A key property of the basis is that it generates M freely: the canonical evaluation homomorphism \phi: R^{(B)} \to M, which sends the standard basis elements of the free module on B to the elements of B, is an isomorphism of R-modules. This isomorphism ensures that M is structurally identical to the free module on the set B, preserving the linear independence and unique representations.^[6]^[5] Over commutative rings with identity, the cardinality of any basis of a free module is unique, defining the rank of the module. This uniqueness follows from the invariant basis number (IBN) property of such rings, which asserts that if R^m \cong R^n as free modules, then m = n. To sketch the proof, suppose B and B' are two bases of M; then M \cong R^{(B)} \cong R^{|B|} and similarly M \cong R^{|B'|}, so R^{|B|} \cong R^{|B'|}, implying |B| = |B'| by IBN.^[7]^[5] The existence of bases for free modules follows from their explicit construction as direct sums of copies of R, but in specific cases such as over principal ideal domains (PIDs), Zorn's lemma is applied to establish bases for submodules of free modules. For a submodule N of a free module M over a PID R, consider the partially ordered set of linearly independent subsets of N; every chain has an upper bound (their union), so Zorn's lemma yields a maximal linearly independent set in N, which extends to a basis of M and implies N is free with its own basis.^[8]^[5] This mirrors the use of Zorn's lemma for Hamel bases in vector spaces over fields.

Examples

Classical examples

A quintessential class of free modules arises over fields, where every module is free. Specifically, for a field k, the module k^n consists of n-tuples of elements from k, forming a free k-module of rank n with the standard basis \{e_1, \dots, e_n\}, where e_i has a 1 in the i-th position and 0 elsewhere.^[1] This structure generalizes finite-dimensional vector spaces, and any vector space over k is a free k-module, possibly of infinite rank.^[9] Over the integers \mathbb{Z}, free modules correspond to free abelian groups. The direct sum \mathbb{Z}^{(I)} = \bigoplus_{i \in I} \mathbb{Z} for an index set I is a free \mathbb{Z}-module with basis the standard generators \{e_i \mid i \in I\}, where each e_i is the tuple with 1 in the i-th position and 0 elsewhere.^[10] For finite I with |I| = n, this recovers \mathbb{Z}^n as a free \mathbb{Z}-module of rank n.^[11] Polynomial rings provide another familiar setting. For a commutative ring R, the polynomial ring R is a free R-module with basis \{1, x, x^2, \dots \}, which is countably infinite, illustrating a free module of infinite rank.^[11] Elements of R are uniquely expressed as finite R-linear combinations of these basis elements. Direct sums and products distinguish finite and infinite cases for freeness. The direct sum R^{(I)} = \bigoplus_{i \in I} R is always free over R with basis the canonical elements, regardless of whether I is finite or infinite.^[11] In contrast, the direct product R^I = \prod_{i \in I} R is free only if I is finite; for infinite I, it fails to be free, as it lacks a basis despite being generated by uncountably many elements in general cases like R = \mathbb{Z}.^[12] Over commutative rings, row and column vectors exemplify finite-rank free modules. The module of row vectors (R^n, +) with componentwise addition and right R-action by scalar multiplication is free of rank n with basis the standard row vectors e_1, \dots, e_n.^[13] Similarly, the column vector module is free of rank n, and for n=1, both reduce to the free module R of rank 1.^[4]

Pathological examples

Over non-commutative rings, free modules can exhibit bases of different cardinalities, contrasting with the invariant dimension property of vector spaces over fields. A canonical example is the endomorphism ring R = \mathrm{End}_k(V), where k is a field and V is a countably infinite-dimensional vector space over k; here, R is a free left R-module of rank 1 with basis \{1_R\}, the identity endomorphism. However, R is isomorphic to R^n as left R-modules for any positive integer n, since the natural inclusions and projections yield such isomorphisms, implying R also admits bases of cardinality n for each finite n. Infinite-rank free modules provide another departure from finite-dimensional vector space behavior, as they are not finitely generated and lack compactness properties. Consider the direct sum M = \bigoplus_{i \in I} R, where I is a countably infinite set and R is any ring with identity; M consists of all families (x_i)_{i \in I} in R^I with only finitely many nonzero entries, and it is free on the standard basis \{ e_i \mid i \in I \}, where e_i has 1 in the i-th position and 0 elsewhere. This module is free of rank |I|, but any generating set must have at least |I| elements, and submodules may fail to be free even if finitely generated. Free modules over rings with zero divisors can contain torsion elements, unlike free modules over integral domains, which are torsion-free. For instance, over the ring R = \mathbb{Z}/4\mathbb{Z}, the module R itself is free of rank 1 with basis \{ \overline{1} \}, as it is generated by \overline{1} and the basis is linearly independent (multiplication by any nonzero element yields a nonzero multiple). Yet, \overline{2} is a torsion element, since \overline{2} \cdot \overline{2} = \overline{0} and \overline{2} \neq \overline{0}, annihilated by the nonzero element \overline{2}. This highlights how ring zero divisors induce torsion in otherwise "basis-like" structures. The module of functions with finite support from an infinite set to R exemplifies an infinite-rank free module in a functional context. Let X be a countably infinite set; the R-module M = \{ f: X \to R \mid f(x) = 0 \text{ for all but finitely many } x \in X \}, with pointwise addition and scalar multiplication, is free on the basis \{ \delta_x \mid x \in X \}, where \delta_x(y) = \delta_{x,y} (the Kronecker delta, taking value 1 at x and 0 elsewhere). Every element of M is a unique finite R-linear combination of these basis functions, but M requires infinitely many generators, and its endomorphism ring includes shifts and projections that complicate direct analogs to finite bases.

Constructions

Direct sum construction

The standard construction of a free module over a ring R utilizes the direct sum of copies of R indexed by a set I. Specifically, the free R-module of rank |I|, denoted R^{(I)}, is the direct sum \bigoplus_{i \in I} R, whose elements are families (r_i)_{i \in I} with r_i \in R and only finitely many r_i nonzero (finite support).^[14]^[11] Addition in R^{(I)} is defined componentwise: (r_i) + (s_i) = (r_i + s_i), and scalar multiplication by t \in R is t \cdot (r_i) = (t r_i), preserving the finite support condition.^[11] The set \{e_i \mid i \in I\}, where each e_i is the family with 1 in the i-th position and 0 elsewhere, forms a basis for R^{(I)}, as every element can be uniquely expressed as a finite R-linear combination \sum r_i e_i.^[14]^[11] Any free R-module M with basis B is isomorphic to R^{(B)} via the R-module homomorphism that sends each basis element b \in B to the corresponding standard basis vector e_b \in R^{(B)}; this map is an isomorphism because it is bijective and preserves the free structure.^[14]^[15] In the finitely generated case, where I = \{1, \dots, n\} is finite, R^{(n)} \cong R^n can be realized concretely as the set of n \times 1 column vectors (or $1 \times n row vectors) with entries in R, equipped with matrix addition and scalar multiplication.^[11] The standard basis consists of the vectors with a single 1 and zeros elsewhere.^[11]

Formal linear combinations

One standard set-theoretic construction of the free module over a ring R with basis a set B is as the set M of all functions f: B \to R with finite support, meaning f(b) = 0 for all but finitely many b \in B.^[16] Addition of functions is defined pointwise by (f + g)(b) = f(b) + g(b) for all b \in B, and scalar multiplication by elements of R is given by (r \cdot f)(b) = r f(b) for r \in R and all b \in B.^[16] This endows M with the structure of an R-module.^[16] The functions e_b: B \to R defined by e_b(b') = 1 if b' = b and $0 otherwise (for each b \in B) form a basis for M.^[16] Every element f \in M can be uniquely expressed as the finite sum \sum_{b \in B} f(b) e_b, where the sum is over the finite support of f, ensuring linear independence of the basis \{e_b \mid b \in B\}.^[16] The map \phi: M \to \bigoplus_{b \in B} R given by \phi(f) = \sum_{b \in B} f(b) b (where the right-hand side denotes the direct sum construction with basis elements b) is an isomorphism of R-modules.^[16] This equivalence identifies the functional construction with the direct sum of copies of R indexed by B. When R = \mathbb{Z}, this yields the free abelian group on B, which can alternatively be constructed as the quotient of the free group on B by its commutator subgroup. This special case highlights a combinatorial presentation, but the functional approach generalizes uniformly to any ring R, including non-unital rings (rngs).^[16]

Properties

Universal property

The free module F over a ring R generated by a set B is characterized by the following universal property: there exists a canonical map \iota: B \to F such that for every R-module M and every function \phi: B \to M, there is a unique R-module homomorphism \psi: F \to M satisfying \psi \circ \iota = \phi.^[17] This property establishes F as the "freest" R-module generated by B, ensuring that any assignment of elements of M to the generators in B extends uniquely to an R-linear map from F.^[18] To prove this, recall that elements of F are uniquely expressed as finite sums \sum_{b \in B} r_b \iota(b) with r_b \in R and only finitely many nonzero. Define \psi on such sums by \psi\left( \sum r_b \iota(b) \right) = \sum r_b \phi(b); this is well-defined due to the linear independence of \{ \iota(b) \mid b \in B \}, and it is clearly R-linear. For uniqueness, suppose \psi' is another such homomorphism; then \psi' agrees with \psi on the basis images \iota(b), and thus on all of F by linearity and spanning.^[17]^[18] Categorically, the free module F on B is the coproduct \coprod_{b \in B} R in the category of R-modules, where each component R corresponds to a basis element via the canonical inclusions.^[19] A key consequence is the natural isomorphism of sets \Hom_R(F, M) \cong \Map(B, M), where \Map(B, M) denotes the set of all functions from B to M, given by \phi \mapsto \psi as above; the inverse sends \psi to \phi = \psi \circ \iota.^[20] When B is empty, the free module F is the zero module \{0\}, with the empty map \iota: \emptyset \to \{0\}; the universal property holds vacuously, as the only function \phi: \emptyset \to M is the empty function, and the unique homomorphism is the zero map. This establishes the zero module as the free module of rank zero.^[20]

Rank and uniqueness

The rank of a free module M over a ring R is defined as the cardinality of any basis B of M, denoted \operatorname{rk}(M) = |B|.^[21] Over commutative rings with identity, the rank is unique: any two bases of M have the same cardinality, and any isomorphism between free modules preserves the rank.^[21] This is a consequence of the invariant basis number (IBN) property, which holds for all commutative rings.^[22] For finitely generated free modules over principal ideal domains (PIDs), uniqueness follows from the structure theorem, which decomposes such modules into a free part of unique rank and a torsion part; the rank is the number of basis elements in the free summand, determined by the invariant factors (the free rank is the number of invariant factors that are units).^[23] The general case for commutative rings extends this via localization and Krull's principal ideal theorem, ensuring consistent basis sizes across localizations.^[22] For infinite ranks, the rank is a cardinal number, and bases always have the same cardinality; moreover, if M and N are free with at least one of \operatorname{rk}(M) or \operatorname{rk}(N) infinite, then \operatorname{rk}(M \oplus N) = \operatorname{rk}(M) + \operatorname{rk}(N), where + denotes cardinal addition.^[24] Over non-commutative rings, rank need not be unique: for example, if V is an infinite-dimensional vector space over a field F and R = \operatorname{End}_F(V), then the free left R-modules R and R^2 are isomorphic, despite having bases of different finite cardinalities.^[25] For a finitely generated free module M over a local ring (R, \mathfrak{m}), the rank equals the dimension of the vector space M / \mathfrak{m}M over the residue field R / \mathfrak{m}.^[26]

Generalizations and Relations

Relation to projective modules

Free modules over a ring R are projective because they can be expressed as direct summands of free modules (in particular, of themselves), and projective modules are precisely those that lift homomorphisms over epimorphisms. Specifically, if F is a free R-module with basis \{e_i\}_{i \in I}, then for any epimorphism \pi: N \to M and homomorphism f: F \to M, there exists a lift \tilde{f}: F \to N such that \pi \circ \tilde{f} = f, constructed by extending the map on the basis elements.^[27] However, the converse does not always hold: there exist projective modules that are not free. A standard example occurs over the ring R = \mathbb{Z}/6\mathbb{Z}, where the module \mathbb{Z}/3\mathbb{Z} (viewed as an R-module via the quotient map \mathbb{Z}/6\mathbb{Z} \to \mathbb{Z}/3\mathbb{Z}) is projective but not free. It is projective because \mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z} as R-modules by the Chinese Remainder Theorem, making \mathbb{Z}/3\mathbb{Z} a direct summand of the free module R. Yet it is not free, as any nonzero free R-module has cardinality a power of 6, while |\mathbb{Z}/3\mathbb{Z}| = 3.^[28] Similarly, \mathbb{Z}/2\mathbb{Z} over \mathbb{Z}/6\mathbb{Z} provides another such example.^[27] Projective modules are free under certain ring conditions. Over principal ideal domains (PIDs), every finitely generated projective module is free, as it is a direct summand of a finitely generated free module, and submodules of free modules over PIDs are free by the structure theorem for modules over PIDs.^[29] Over fields, all modules (which are vector spaces) are free, hence projective modules are free.^[30] Over local rings, Kaplansky's theorem states that every projective module is free.^[31] A module is free if and only if it is projective and possesses a basis, or equivalently, if it is projective and has a well-defined rank (constant on localizations at maximal ideals).^[27] Historically, Serre conjectured that every finitely generated projective module over a polynomial ring k[x_1, \dots, x_n] in finitely many variables over a field k is free; this was affirmatively resolved by Quillen and Suslin in the 1970s, with proofs relying on homotopical and algebraic K-theory methods, respectively.^[32]

Free resolutions and homological algebra

In homological algebra, a free resolution of a module M over a ring R is a projective resolution in which each projective module is free. Specifically, it consists of an exact sequence \cdots \to F_2 \to F_1 \to F_0 \to M \to 0, where each F_i is a free R-module. Every R-module admits such a resolution, constructed iteratively by surjecting free modules onto successive kernels. For instance, if I is a principal ideal generated by a regular element a \in R, then the cyclic module M = R/I has a free resolution $0 \to R \to R \to R/I \to 0, where the first map is multiplication by a and the second is the canonical quotient map.^[33]^[34] Free modules play a central role in computing derived functors like Tor, as they are flat modules. A module F is flat if tensoring with it preserves exact sequences, and free modules satisfy this property because the tensor product with a free module R^{(n)} is isomorphic to the direct sum of n copies of the tensor product with R. Consequently, for a free module F and any R-module N, the higher Tor groups vanish: \Tor_i^R(F, N) = 0 for i > 0. The Tor functor itself is defined using free resolutions: to compute \Tor_i^R(M, N), take a free resolution of M, delete M, tensor with N, and compute the homology at the i-th spot; this measures the deviation of M from being flat. Similarly, free resolutions enable computation of the Ext functor by applying Hom in place of tensoring.^[35]^[36] Minimal free resolutions refine this construction by choosing basis maps with entries in the maximal ideal (for local rings) or homogeneous of minimal degree (for graded modules), ensuring the ranks \beta_i of the free modules F_i are minimal invariants called Betti numbers. These \beta_i count the minimal number of generators of the i-th syzygy module of M. In the graded case over polynomial rings, the Hilbert syzygy theorem guarantees that every finitely generated graded module has a finite minimal free resolution of length at most the number of variables, bounding the projective dimension.^[34]^[37] A prominent example is the Koszul complex, which provides a free resolution for quotient modules by ideals generated by regular sequences. For a polynomial ring R = k[x_1, \dots, x_n] and elements x_1, \dots, x_r forming a regular sequence, the Koszul complex K(x_1, \dots, x_r) is an exact sequence of free modules resolving R/(x_1, \dots, x_r), with terms given by exterior powers of the free module on the generators. This complex is minimal when the sequence is regular and facilitates computations of Tor and Ext in algebraic geometry and commutative algebra.^[38]^[34]

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