standard duality on modules over algebras

Let k be a field and let A be an associative unital algebra. Throughout we will assume that all A-modules over k are unital. If M is a right A-module, then the space of all linear mappings


can be equipped with a left A-module structureMathworldPlanetmath as follows: for any fHomk(M,k) and aA put


Note that action direction need to be reversed, because


Analogously Homk(-,k) takes left A-modules to right A-modules. Also this action is compatible with functoriality of Homk(-,k), which means that it takes A-homomorphismsMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to A-homomorphisms. In particular we obtain a (contravariant) functorMathworldPlanetmath from categoryMathworldPlanetmath of left (right) A-modules to category of right (left) A-modules. Obviously Hom does not change the dimension of spaces, so we have well defined functors


which are restrictionsPlanetmathPlanetmath of Hom (here mod means finite dimensional modules left/right modules) and are known in literature as ,,standard dualities”.

PropositionPlanetmathPlanetmathPlanetmath. Both D’s are quasi inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath dualities of categories.

Proof. Let M be a finite dimensional A-module. We need to define a natural isomorphism between M and DD(M). Indeed, define


We will show that each τ is an isomorphismMathworldPlanetmathPlanetmath.

  1. 1.

    First we will show that τ is a monomorphismMathworldPlanetmathPlanetmathPlanetmath. Assume that τM(m)=0 for nonzero mM. This is if and only if α(m)=0 for every linear mapping α:Mk. But m is nonzero, so there is a basis of M (as linear space) which contains m. In particular there is a linear mapping f:Mk such that f(m)=1. ContradictionMathworldPlanetmathPlanetmath. Thus m=0, which completesPlanetmathPlanetmathPlanetmathPlanetmath this part.

  2. 2.

    τ is an epimorphismMathworldPlanetmath. Indeed, let F:D(M)k be a linear mapping. We need to show, that there is mM such that


    for any αD(M). Since M is finite dimensional, then let {e1,,en} be a k-basis of M. Of course {e1*,,en*} is a basis of D(M), where ei* is given by ei*(ej)=1 if i=j and ei*(ej)=0 otherwise. Define


    and put


    We leave it as a simple exercise, that τ(m)=F.

What remains is to prove, that τ is natural. Consider an A-homomorphism f:XY. We need to show that the following diagram commutes:


Indeed, if xX, then let F=τX(x). We have that


and evaluating this at αD(M) we have


In particular we obtain that


which means that


which completes the proof.

Title standard duality on modules over algebras
Canonical name StandardDualityOnModulesOverAlgebras
Date of creation 2013-12-11 15:25:39
Last modified on 2013-12-11 15:25:39
Owner joking (16130)
Last modified by joking (16130)
Numerical id 5
Author joking (16130)
Entry type Theorem
Classification msc 16S99
Classification msc 20C99
Classification msc 13B99