example of free module with bases of diffrent cardinality

Let k be a field and V be an infinite dimensional vector spaceMathworldPlanetmath over k. Let {ei}iI be its basis. Denote by R=End(V) the ring of endomorphisms of V with standard additionPlanetmathPlanetmath and composition as a multiplication.

Let J be any set such that |J||I|.

PropositionPlanetmathPlanetmath. R and jJR are isomorphic as a R-modules.

Proof. Let α:IJ×I be a bijection (it exists since |I||J| and I is infiniteMathworldPlanetmath) and denote by π1:J×IJ and π2:J×II the projectionsPlanetmathPlanetmath. Moreover let δ1=π1α and δ2=π2α.

Recall that jJR={f:JR} (with obvious R-module structureMathworldPlanetmath) and define a map ϕ:jJRR by defining the endomorphismPlanetmathPlanetmath ϕ(f)R for fjJR as follows:


We will show that ϕ is an isomorphismMathworldPlanetmathPlanetmathPlanetmath. It is easy to see that ϕ is a R-module homomorphismMathworldPlanetmath. Therefore it is enough to show that ϕ is injectivePlanetmathPlanetmath and surjectivePlanetmathPlanetmath.

1) Recall that ϕ is injective if and only if ker(ϕ)=0. So assume that ϕ(f)=0 for fjJR. Note that f=0 if and only if f(j)=0 for all jJ and this is if and only if f(j)(ei)=0 for all jJ and iI. So take any (j,i)J×I. Then (since α is bijective) there exists i0I such that α(i0)=(j,i). It follows that δ1(i0)=j and δ2(i0)=i. Thus we have


Since j and i were arbitrary, then f=0 which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath this part.

2) We wish to show that ϕ is onto, so take any hR. Define fjJR by the following formulaMathworldPlanetmathPlanetmath:


It is easy to see that ϕ(f)=h.

Corollary. For any two numbers n,m there exists a ring R and a free moduleMathworldPlanetmathPlanetmath M such that M has two bases with cardinality n,m respectively.

Proof. It follows from the proposition, that for R=End(V) we have


For finite setMathworldPlanetmath J module jJR is free with basis consisting |J| elements (productPlanetmathPlanetmathPlanetmath is the same as direct sumMathworldPlanetmathPlanetmathPlanetmath). Therefore (due to existence of previous isomorphisms) R-module R has two bases, one of cardinality n and second of cardinality m.

Title example of free module with bases of diffrent cardinality
Canonical name ExampleOfFreeModuleWithBasesOfDiffrentCardinality
Date of creation 2013-03-22 18:07:18
Last modified on 2013-03-22 18:07:18
Owner joking (16130)
Last modified by joking (16130)
Numerical id 13
Author joking (16130)
Entry type Example
Classification msc 16D40
Related topic IBN