nLab free module




Homological algebra

homological algebra

(also nonabelian homological algebra)



Basic definitions

Stable homotopy theory notions



diagram chasing

Schanuel's lemma

Homology theories




A free module over some ring RR is freely generated on a set of basis elements.

Under the interpretation of modules as generalized vector bundles a free module corresponds to a trivial bundle.



Let CC be a monoidal category, and Alg(C)Alg(C) the category of monoids in CC; and for AAlg(C)A \in Alg(C) let AAMod(C)(C) be the category of AA-modules in CC.

There is the evident forgetful functor U:AMod(C)CU : A Mod(C) \to C that sends each module (N,ρ)(N,\rho) to its underlying object NCN \in C.


The left adjoint CAMod(C)C \to A Mod(C) is the corresponding free construction. The modules in the image of this functor are free modules.

Over rings

Let RR be a ring. We discuss free modules over RR.


For RR \in Ring a ring and SS \in Set, the free RR-module on SS is isomorphic to the |S|{\vert S\vert}-fold direct sum of RR with itself

R (S) sSR. R^{(S)}\simeq \oplus_{s \in S} R \,.


As a monoidal functor

Let RR be a commutative ring, and let R{X}R\{X\} denote the free RR-module on a set XX.


The free RR-module functor is strong monoidal with respect to the Cartesian monoidal structure on sets, and the tensor product of RR-modules.

In other words, the free module construction turns set-theoretic products into tensor products. Thus, it preserves algebraic objects (such as monoid objects, Hopf monoid objects, etc.) and their homomorphisms. In particular, if MM is a monoid in the category of sets (and hence a bimonoid with the canonical comonoid structure) then R{M}R\{M\} is a bimonoid object in RModR \mathsf{Mod}, which is precisely a KK-bialgebra. A group GG in the category of sets is a Hopf monoid, and hence R{G}R\{G\} is a Hopf algebra — this is precisely the group algebra of GG.

Submodules of free modules

Let RR be a commutative ring.


Assuming the axiom of choice, the following are equivalent

  1. every submodule of a free RR-module is itself free;

  2. every ideal in RR is a free RR-module;

  3. RR is a principal ideal domain.


(See also Rotman, pages 650-651.) Condition 1. immediately implies condition 2., since ideals of RR are the same as submodules of RR seen as an RR-module. Now assume condition 2. holds, and suppose xRx \in R is any nonzero element. Let λ x\lambda_x denote multiplication by xx (as an RR-module map). We have a sequence of surjective RR-module maps

Rλ x(x) JRRR \stackrel{\lambda_x}{\to} (x) \cong \oplus_J R \stackrel{\nabla}{\to} R

(where \nabla is the codiagonal map); by the Yoneda lemma, the composite map RRR \to R is of the form λ r\lambda_r, where rRr \in R is the value of the composite at 1R1 \in R. Since λ r\lambda_r is surjective, we have λ r(s)=rs=1\lambda_r(s) = r s = 1 for some ss, so that rr is invertible. Hence λ r\lambda_r is invertible, and this implies λ x\lambda_x is monic. Therefore RR is a domain. From that, we infer that if ff and gg belong to a basis of an ideal II, then

0fgRfRg0 \neq f g \in R\cdot f \cap R \cdot g

whence ff and gg are not linearly independent, so f=gf = g and II as an RR-module is generated by a single element, i.e., RR is a principal ideal domain.

That condition 3. implies condition 1. is proved here.


Assuming the axiom of choice, over a ring RR which is a principal ideal domain, every module has a projective resolution of length 1.

See at projective resolution – Resolutions of length 1 for more.

Over a field: vector spaces


Assuming the axiom of choice, if R=kR = k is a field then every RR-module is free: it is kk-vector space and by the basis theorem every such has a basis.



Last revised on April 26, 2023 at 06:23:34. See the history of this page for a list of all contributions to it.