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free module

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nonabelian homological algebra

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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

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

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

References

  • Rotman Advanced Modern Algebra, pp. 650–651

Revised on February 8, 2013 12:03:06 by Urs Schreiber (89.204.138.214)