finite dimensional modules over algebra


Assume that k is a field, A is a k-algebra and M is a A-module over k. In particular M is a A-module and a vector spaceMathworldPlanetmath over k, thus we may speak about M being finitely generatedMathworldPlanetmathPlanetmath as A-module and finite dimensional as a vector space. These two concepts are related as follows:

PropositionPlanetmathPlanetmath. Assume that A and M are both unital and additionaly A is finite dimensional. Then M is finite dimensional vector space if and only if M is finitely generated A-module.

Proof. ,,⇒” Of course if M is finite dimensional, then there exists basis

{x1,…,xn}⊂M.

Thus every element of M can be (uniquely) expressed in the form

∑i=1nλi⋅xi

which is equal to

∑i=1n(λi⋅1)⋅xi

since M and A are unital. This completesPlanetmathPlanetmathPlanetmath this implicationMathworldPlanetmath, because λi⋅1∈A for all i.

,,⇐” Assume that M is finitely generated A-module. In particular there is a subset

{x1,…,xn}⊂M

such that every element of M is of the form

∑i=1nai⋅xi

with all ai∈A. Let m∈M be with the decomposition as above. Now A is finite dimensional, so there is a subset

{y1,…,yt}⊂A

which is a k-basis of A. In particular for each i we have

ai=∑j=1tλi⁢j⋅yj

with λi⁢j∈k. Thus we obtain

m=∑i=1nai⋅xi=∑i=1n(∑j=1tλi⁢j⋅yj)⋅xi=
=∑i=1n∑j=1tλi⁢j⋅(yj⋅xi)

which shows, that all yj⋅xi∈M together make a set of generatorsPlanetmathPlanetmathPlanetmath of M over k (note that yj and xi are independent on m). Since it is finite, then M is finite dimensional and the proof is complete. □

Title finite dimensional modules over algebra
Canonical name FiniteDimensionalModulesOverAlgebra
Date of creation 2013-03-22 19:16:35
Last modified on 2013-03-22 19:16:35
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Definition
Classification msc 16S99
Classification msc 20C99
Classification msc 13B99