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

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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:
Cohn - 1 - Theorem 2.2 ... PART 1 ... .png

Cohn - 2 - Theorem 2.2 ... PART 2 ... .png

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
 

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Math Amateur said:
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 222878
View attachment 222879
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

It suffices to show that ##N_{k+1} \subseteq N_k##.

Take ##n \in N_{k+1} \subseteq N = (a_1, \dots, a_r)##. Then ##n = \sum \lambda_i a_i## for elements ##\lambda_i##. Now, ##a_1, \dots, a_r## are contained in ##N_k##, because we have an increasing chain, and hence ##n\in N_k## as well, since modules are closed under linear combinations.
 
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Math_QED said:
It suffices to show that ##N_{k+1} \subseteq N_k##.

Take ##n \in N_{k+1} \subseteq N = (a_1, \dots, a_r)##. Then ##n = \sum \lambda_i a_i## for elements ##\lambda_i##. Now, ##a_1, \dots, a_r## are contained in ##N_k##, because we have an increasing chain, and hence ##n\in N_k## as well, since modules are closed under linear combinations.
Thanks for the help Math_QED ...

BUT ... just some clarifications ...

You write:

" ... ... Take ##n \in N_{k+1} \subseteq N = (a_1, \dots, a_r)##. ... ... "

This means ##N_{k+1} \subseteq N## and also ##N = (a_1, \dots, a_r)## ... is that correct ...

But then surely ##n## may equal ##\sum_{ i = 1}^t \lambda_i a_i## where ##t \lt r## ... and so ##a_1, \dots, a_r## may not all be contained in ##N_{k + 1} ## let alone ##N_k## ...

I am really puzzled as to exactly why ##(a_1, \dots, a_r)## are contained in ##N_k## ... ... what is the exact argument?

Can you clarify?

Peter
 
Math Amateur said:
Thanks for the help Math_QED ...

BUT ... just some clarifications ...

You write:

" ... ... Take ##n \in N_{k+1} \subseteq N = (a_1, \dots, a_r)##. ... ... "

This means ##N_{k+1} \subseteq N## and also ##N = (a_1, \dots, a_r)## ... is that correct ...

But then surely ##n## may equal ##\sum_{ i = 1}^t \lambda_i a_i## where ##t \lt r## ... and so ##a_1, \dots, a_r## may not all be contained in ##N_{k + 1} ## let alone ##N_k## ...

I am really puzzled as to exactly why ##(a_1, \dots, a_r)## are contained in ##N_k## ... ... what is the exact argument?

Can you clarify?

Peter

For your first question, yes that's exactly what it means, for your second question:

We can find coefficients such that ##n = \sum_{i=1}^r \lambda_i a_i##, because ##n \in (a_1 \dots, a_r)##. But ##a_1 \in N_{i_1}##

and ##k = \max\{i_1, \dots, i_r\}##, so ##k \geq i_1##. This implies that ##N_{i_1} \subseteq N_k##, because the chain is increasing, so ##a_1 \in N_k## and you can do the same thing for the other indices, obtaining that ##a_1, \dots, a_r \in N_k##

EDIT: Maybe you don't know what I mean with ##(a_1, \dots, a_r)##. This is the module generated by the elements ##a_1, \dots a_r##. I.e., the smallest submodule that contains these elements.
 
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