MHB Noetherian Modules .... Cohn Theorem 2.2 .... ....

[Math Amateur]

I am reading P.M. Cohn's book: Introduction to Ring Theory (Springer Undergraduate Mathematics Series) ... ...

I am currently focused on Section 2.2: Chain Conditions ... which deals with Artinian and Noetherian rings and modules ... ...

I need help with understanding an aspect of the proof of Theorem 2.2 ... ...Theorem 2.2 and its proof (including some preliminary relevant definitions) read as follows:
View attachment 8003
https://www.physicsforums.com/attachments/8004

At the end of the above proof by Cohn we read the following:

" ... ... If $$a_j \in N_{i_j} $$ and $$k = \text{ max} \{ i_1, \ ... \ ... \ , i_r \}$$, then equality holds in our chain from N_k onwards. ... ... "
Can someone please explain how/why $$a_j \in N_{i_j} $$ and $$k = \text{ max} \{ i_1, \ ... \ ... \ , i_r \}$$ implies that equality holds in our chain from $$N_k$$ onwards. ... ... ?Help will be appreciated ...

Peter

 

[castor28]

Hi Peter,

Each $a_j$ is an element of $N=\bigcup N_i$, and therefore an element of some $N_i$, say $N_{i_j}$. If $k = \max \{ i_1, \ldots, i_r \}$, then $N_k$ contains all the $a_j$; as these elements generate $N$ and $N_k\subset N$, we have $N_k=N$.

Now, for any $i\ge k$, we have $N_k=N \subset N_i\subset N$, and therefore $N_i=N_k=N$.

 

[Math Amateur]

Hi Peter,

Each $a_j$ is an element of $N=\bigcup N_i$, and therefore an element of some $N_i$, say $N_{i_j}$. If $k = \max \{ i_1, \ldots, i_r \}$, then $N_k$ contains all the $a_j$; as these elements generate $N$ and $N_k\subset N$, we have $N_k=N$.

Now, for any $i\ge k$, we have $N_k=N \subset N_i\subset N$, and therefore $N_i=N_k=N$.

Thanks castor28 ...

Think I follow that argument ...

Just reflecting further to make sure I fully understand ...

Thanks again,

Peter