MHB Noetherian Modules and Finitely Generated Modules

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we find Theorem 2.2 on Noetherian modules. I need help with showing that "every module of M is finitely generated" implies that "M is Noetherian"

Theorem 2 reads as follows:View attachment 3166
https://www.physicsforums.com/attachments/3167In the above text [at the very end of the text - in the argument for $$(d) \Longrightarrow (a)$$] we read:

" … … 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 … … "

I do not follow the argument in the above text … indeed, I am having some trouble interpreting the exact meaning of $$a_j \in N_{i_j}$$ … …

Can someone please help me to understand the above argument and notation … my apologies to readers for not being able to make my question/confusion clearer …

Hope someone can help.

Peter

 

[Euge]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we find Theorem 2.2 on Noetherian modules. I need help with showing that "every module of M is finitely generated" implies that "M is Noetherian"

Theorem 2 reads as follows:View attachment 3166
https://www.physicsforums.com/attachments/3167In the above text [at the very end of the text - in the argument for $$(d) \Longrightarrow (a)$$] we read:

" … … 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 … … "

I do not follow the argument in the above text … indeed, I am having some trouble interpreting the exact meaning of $$a_j \in N_{i_j}$$ … …

Can someone please help me to understand the above argument and notation … my apologies to readers for not being able to make my question/confusion clearer …

Hope someone can help.

Peter

Hi Peter,

First, let's start with the ascending chain $N_1 \subseteq N_2 \subseteq \cdots$. We want to show that this chain stabilizes, i.e., there exists some $k$ such that $N_k = N_{k+1} = \cdots$. To do this, we consider the least upper bound of the chain, that is, $N := \cup N_i$. Recall that the union of a chain of an ascending chain of submodules is a submodule; in particular, $N$ is a submodule of $M$. Since it's assumed that every submodule of $M$ is finitely generated, we know that $N$ is finitely generated. That's why Cohn can choose a finite generating set $\{a_1,\ldots, a_r\}$ for $N$. But since $N$ is the union of the $N_i$, we know that each $a_j$ belongs to at least one of the $N_i$. In other words, for each $j\in \{1,2,\ldots, r\}$, there exists an index $i_j$ such that $a_j\in N_{i_j}$. By choosing $k$ to be the largest of the indices $i_1,\ldots, i_r$, you ensure that $N_k = N$, and hence $N_k = N_{k+1} = \cdots$. Make sense?

Just in case you didn't see how $N_k = N$, I'll put in more detail here. Since $N_{i_j} \subseteq N_k$ for all $j$ (by the ascending chain and the fact that $i_j \le k$ for all $j$) and $a_j\in N_{i_j}$ for all $j$, we have $a_j \in N_k$ for all $j$. Hence, the submodule generated by $a_1,\ldots a_r$ is a submodule of $N_k$. But since $N$ is generated by $a_1,\ldots, a_r$, $N \subseteq N_k$. On the other hand, $N_k \subseteq N$ by construction of $N$. Hence, $N_k = N$.

 

[Math Amateur]

Hi Peter,

First, let's start with the ascending chain $N_1 \subseteq N_2 \subseteq \cdots$. We want to show that this chain stabilizes, i.e., there exists some $k$ such that $N_k = N_{k+1} = \cdots$. To do this, we consider the least upper bound of the chain, that is, $N := \cup N_i$. Recall that the union of a chain of an ascending chain of submodules is a submodule; in particular, $N$ is a submodule of $M$. Since it's assumed that every submodule of $M$ is finitely generated, we know that $N$ is finitely generated. That's why Cohn can choose a finite generating set $\{a_1,\ldots, a_r\}$ for $N$. But since $N$ is the union of the $N_i$, we know that each $a_j$ belongs to at least one of the $N_i$. In other words, for each $j\in \{1,2,\ldots, r\}$, there exists an index $i_j$ such that $a_j\in N_{i_j}$. By choosing $k$ to be the largest of the indices $i_1,\ldots, i_r$, you ensure that $N_k = N$, and hence $N_k = N_{k+1} = \cdots$. Make sense?

Just in case you didn't see how $N_k = N$, I'll put in more detail here. Since $N_{i_j} \subseteq N_k$ for all $j$ (by the ascending chain and the fact that $i_j \le k$ for all $j$) and $a_j\in N_{i_j}$ for all $j$, we have $a_j \in N_k$ for all $j$. Hence, the submodule generated by $a_1,\ldots a_r$ is a submodule of $N_k$. But since $N$ is generated by $a_1,\ldots, a_r$, $N \subseteq N_k$. On the other hand, $N_k \subseteq N$ by construction of $N$. Hence, $N_k = N$.

Thanks for an extremely clear and explicit argument, Euge … thanks to your post I now follow the proof ...

Indeed, if you are not already teaching algebra at a University … then you should be … ...

Peter