Jordan-Holder Theorem for Modules .... ....

  • Context: MHB 
  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Modules Theorem
Click For Summary
SUMMARY

The discussion centers on understanding Proposition 4.2.16 from Paul E. Bland's book "Rings and Their Modules," specifically regarding the Jordan-Hölder theorem for modules. The key point is that if \( M_1 \neq N_1 \), then \( M_1 + N_1 = M \) because \( N_1 \) is a maximal submodule of \( M \). This conclusion is drawn from the fact that \( M_1 + N_1 \) must contain \( N_1 \) and thus encompasses the entire module \( M \).

PREREQUISITES
  • Understanding of Noetherian and Artinian modules
  • Familiarity with the Jordan-Hölder theorem
  • Knowledge of submodules and maximal submodules
  • Basic concepts of module theory as presented in "Rings and Their Modules" by Paul E. Bland
NEXT STEPS
  • Study the proof of the Jordan-Hölder theorem in detail
  • Explore examples of maximal submodules in various modules
  • Review Section 4.2 of "Rings and Their Modules" for deeper insights
  • Learn about the implications of Noetherian and Artinian properties in module theory
USEFUL FOR

Mathematicians, particularly those studying abstract algebra, module theory, and anyone looking to deepen their understanding of the Jordan-Hölder theorem and its applications in module theory.

Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand the proof of part of Proposition 4.2.16 (Jordan-Holder) ... ...

Proposition 4.2.16 reads as follows:
View attachment 8240
View attachment 8241

Near the middle of the above proof (top of page 116) we read the following:

"... ... If $$M_1 \neq N_1$$ then $$M_1 + N_1 = M$$ since $$N_1$$ is a maximal submodule of $$M$$. ... ... "

Can someone please explain exactly how $$N_1$$ being a maximal submodule of $$M$$ implies that $$M_1 + N_1 = M$$ ... ... ?

Peter
 
Physics news on Phys.org
Hi Peter,

Peter said:
Near the middle of the above proof (top of page 116) we read the following:

"... ... If $$M_1 \neq N_1$$ then $$M_1 + N_1 = M$$ since $$N_1$$ is a maximal submodule of $$M$$. ... ... "

Since $M_{1}\neq N_{1}$, $M_{1}+N_{1}$ is a submodule of $M$ containing the maximal submodule $N_{1}$ and so must be $M$.
 
GJA said:
Hi Peter,
Since $M_{1}\neq N_{1}$, $M_{1}+N_{1}$ is a submodule of $M$ containing the maximal submodule $N_{1}$ and so must be $M$.
Thanks for the help, GJA ...

Peter
 

Similar threads

Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
9
Views
2K