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=1nλixi

which is equal to

i=1n(λi1)xi

since M and A are unital. This completesPlanetmathPlanetmathPlanetmath this implicationMathworldPlanetmath, because λi1A 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=1naixi

with all aiA. Let mM 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=1tλijyj

with λijk. Thus we obtain

m=i=1naixi=i=1n(j=1tλijyj)xi=
=i=1nj=1tλij(yjxi)

which shows, that all yjxiM 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