nLab free module

Contents

Context

Algebra

Homological algebra

homological algebra

(also nonabelian homological algebra)

Introduction

Context

Basic definitions

Stable homotopy theory notions

Constructions

Lemmas

diagram chasing

Schanuel's lemma

Homology theories

Theorems

Contents

Idea

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.

Definition

General

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.

Definition

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.

Proposition

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 \,.

Properties

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.

Proposition

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.

Proposition

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.

Proof

(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.

Corollary

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

Proposition

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.

References

Textbooks:

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