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 space over k, thus we may speak about M being finitely generated
as A-module and finite dimensional as a vector space. These two concepts are related as follows:
Proposition. 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
n∑i=1λi⋅xi |
which is equal to
n∑i=1(λi⋅1)⋅xi |
since M and A are unital. This completes this implication
, 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
n∑i=1ai⋅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=t∑j=1λij⋅yj |
with λij∈k. Thus we obtain
m=n∑i=1ai⋅xi=n∑i=1(t∑j=1λij⋅yj)⋅xi= |
=n∑i=1t∑j=1λij⋅(yj⋅xi) |
which shows, that all yj⋅xi∈M together make a set of generators 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 |
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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 |