Artinian Modules

This one bothered me to no end until I looked it up...It seemed so simple, yet I couldn't see how to do it, no matter how I banged my head on the chalkboard. So, without further ado:

Say a module M is Artinian if it satisfies the descending chain condition on submodules: every chain of submodules
MÊM₁ÊM₂Ê...
stabilises: for some n, M_n=M_n+m for all m.

Let R be a ring, M a left R-module, K a submodule of M such that K and M/K are both Artinian. Then M is also Artinian.

Proof:

We have a short exact sequence 0®K®M®M/K®0, where the first map, i, is inclusion, and the second, p, is the natural projection onto the quotient. Let
MÊM₁ÊM₂Ê...
be any descending chain of submodules in M. Then this chain induces two other chains,
KÊKÇM₁ÊKÇM₂Ê...
and
M/KÊp(M₁)Êp(M₂)Ê....

By assumption, each of these induced chains terminates: for some n, and all k,
KÇM_n=KÇM_n+k and p(M_n)=p(M_n+k).
If xÎM_n, then p(x)Îp(M_n)=p(M_n+k)
so $yÎM_n+k with p(y)=p(x):
x-yÎkerpÇM_n=KÇM_n=KÇM_n+k.
Therefore xÎM_n+k, or M_n=M_n+k. QED!

Posted by aloysius at July 28, 2003 02:44 PM | TrackBack |