Example from Bland - Right Noetherian but not Left Noetherian

[Math Amateur]

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 Example 4 on page 108 ... ...

Example 4 reads as follows:
View attachment 6114
In the above text, I do not understand the proof that \(\displaystyle R\) is not left Noetherian ... I hope someone can clearly explain the way the proof works ...

To try to clarify my lack of understanding of the proof ... Bland, in the above text writes ... ..." ... ... On the other hand, \(\displaystyle \mathbb{Q}\) is not Noetherian when viewed as a \(\displaystyle \mathbb{Z}\)-module.

Hence if \(\displaystyle S_1 \subseteq S_2 \subseteq S_3 \subseteq\) ... ... is a non-terminating chain of \(\displaystyle \mathbb{Z}\)-modules of \(\displaystyle \mathbb{Q}\), then \(\displaystyle \begin{pmatrix} 0 & S_1 \\ 0 & 0 \end{pmatrix} \subseteq \begin{pmatrix} 0 & S_2 \\ 0 & 0 \end{pmatrix} \subseteq \begin{pmatrix} 0 & S_3 \\ 0 & 0 \end{pmatrix} \subseteq\) ... ... is a non-terminating chain of left ideals of \(\displaystyle R\) ... ... "
My questions are as follows:What is the justification for regarding \(\displaystyle \mathbb{Q}\) as a \(\displaystyle \mathbb{Z}\)-module ... indeed, couldn't we have done this while trying to prove/justify that \(\displaystyle R\) was right Noetherian and ended up with a non-terminating chain of right ideals of \(\displaystyle R\) ... ? Why is \(\displaystyle \mathbb{Q}\) not Noetherian when viewed as a \(\displaystyle \mathbb{Z}\)-module ... ... ?Can someone please simply and clearly explain the proof of \(\displaystyle R\) not being left Noetherian ... ... ?
Hope someone can help ...

Peter

 

[GJA]

Hi Peter,

I'm a little short on time, but I hope I can point you in the right direction for now.

What is the justification for regarding \(\displaystyle \mathbb{Q}\) as a \(\displaystyle \mathbb{Z}\)-module ... indeed, couldn't we have done this while trying to prove/justify that \(\displaystyle R\) was right Noetherian and ended up with a non-terminating chain of right ideals of \(\displaystyle R\) ... ?

The idea for thinking of $\mathbb{Q}$ as a module over $\mathbb{Z}$ in the second case comes from noting that multiplying the matrices in the chain on the left by matrices from $R$, the only nonzero entry is $\mathbb{Q}$ as a $\mathbb{Z}$ module.

Applying the same reasoning to the first two chains is not the same because multiplication of these chains on the right by elements of $R$ does not behave like those in the second case.

Why is \(\displaystyle \mathbb{Q}\) not Noetherian when viewed as a \(\displaystyle \mathbb{Z}\)-module ... ... ?

The ascending chain of submodules of $\mathbb{Q}$ does not terminate: $(1/2)\subseteq (1/4)\subseteq\ldots\subseteq (1/2^{n})\subseteq\ldots$

 

[Math Amateur]

Hi Peter,

I'm a little short on time, but I hope I can point you in the right direction for now.
The idea for thinking of $\mathbb{Q}$ as a module over $\mathbb{Z}$ in the second case comes from noting that multiplying the matrices in the chain on the left by matrices from $R$, the only nonzero entry is $\mathbb{Q}$ as a $\mathbb{Z}$ module.

Applying the same reasoning to the first two chains is not the same because multiplication of these chains on the right by elements of $R$ does not behave like those in the second case.
The ascending chain of submodules of $\mathbb{Q}$ does not terminate: $(1/2)\subseteq (1/4)\subseteq\ldots\subseteq (1/2^{n})\subseteq\ldots$

Hi GJA ... thank you very much for the help ...

But ... can you be more explicit ... still somewhat perplexed ...

Will try to post a specific question soon ...

Peter

 

[Math Amateur]

Hi GJA ... thank you very much for the help ...

But ... can you be more explicit ... still somewhat perplexed ...

Will try to post a specific question soon ...

Peter

Thanks Again for the help, GJA ...

... BUT ... still somewhat uneasy and perplexed by this example ...One of the points I am uneasy about is that if we "view" \(\displaystyle \mathbb{Q}\) as a \(\displaystyle \mathbb{Z}\)-module it is not Noetherian

... but if we "view" \(\displaystyle \mathbb{Q}\) as a field then it is Noetherian ... ?

... can you explain how this makes sense ...Maybe when thinking of \(\displaystyle \mathbb{Q}\) maybe I should be thinking of \(\displaystyle \mathbb{Z} \cdot \mathbb{Q}\) ... ... (is that right?) ... ... ... as the elements of the \(\displaystyle \mathbb{Q}\) as a \(\displaystyle \mathbb{Z}\)-module would presumably be of the form \(\displaystyle n \cdot \frac{a}{b}\) where \(\displaystyle n \in \mathbb{Z}\) and \(\displaystyle \frac{a}{b} \in \mathbb{Q} \) ... but ... then (?) on the other hand (?) ... ...

... \(\displaystyle n \cdot \frac{a}{b}\) just gives us another "fraction" ... that is, another member of \(\displaystyle \mathbb{Q}\)

... so ... ... there goes my idea that \(\displaystyle \mathbb{Z} \cdot \mathbb{Q}\) would at least be something different from \(\displaystyle \mathbb{Q}\) ... so the difference between Noetherian and non-Noetherian would not just depend weirdly on the view we take of \(\displaystyle \mathbb{Q}\) ...

... but maybe you do have to have a term or structure \(\displaystyle \mathbb{Z} \cdot \mathbb{Q}\) appearing in the problem you are working on, before you can declare that you are going to "consider" or "view" \(\displaystyle \mathbb{Q}\) as a \(\displaystyle \mathbb{Z}\)-moduleHope you can clarify this issue for me ...Peter

 

[GJA]

Hi Peter,

This is a good question

One of the points I am uneasy about is that if we "view" \(\displaystyle \mathbb{Q}\) as a \(\displaystyle \mathbb{Z}\)-module it is not Noetherian

... but if we "view" \(\displaystyle \mathbb{Q}\) as a field then it is Noetherian ... ?

... can you explain how this makes sense ...

and comes down to how one wants to think about/"view" things -- i.e. what part of the (in this case algebraic) structure of the object do you want to consider and what do you want to leave out. When the author says they are thinking of $\mathbb{Q}$ as a $\mathbb{Z}$-module, it means they are forgetting that $\mathbb{Q}$ has more structure (as a field) for the time being.

$\mathbb{Q}$ is Noetherian as ring (= field in this case) because every non-zero element of $\mathbb{Q}$ is a unit (see https://en.wikipedia.org/wiki/Field_(mathematics)) and so there are no non-trivial ascending chains. To help convince yourself of this, take any nonzero element from $\mathbb{Q}$ and compute the ideal it generates.

When viewed as a $\mathbb{Z}$-module, the previous reasoning doesn't apply because you are only dealing with $\mathbb{Q}$'s additive structure -- the notion of unit (see https://en.wikipedia.org/wiki/Unit_(ring_theory)) is not even defined in the case of a module. $\mathbb{Q}$ is not Noetherian as a $\mathbb{Z}$-module because there exist ascending chains of submodules that do not terminate (see https://en.wikipedia.org/wiki/Noetherian_module). The example I gave before was of the submodules generated by $1/2^{n}$, though you could pick many other examples.

Maybe when thinking of \(\displaystyle \mathbb{Q}\) maybe I should be thinking of \(\displaystyle \mathbb{Z} \cdot \mathbb{Q}\) ... ... (is that right?) ... ...

If what you mean by $\mathbb{Z}\cdot\mathbb{Q}$ is that you are considering $\mathbb{Q}$ as an additive group over the integers, then, yes, this is what $\mathbb{Q}$ being a $\mathbb{Z}$-module means. If you meant something else here, then I would need clarification.

But I would not only think of $\mathbb{Q}$ as a $\mathbb{Z}$-module all the time. Sometimes it is necessary to use the full field structure of $\mathbb{Q}$ -- how you think about something is largely dictated by the problem you're solving (or, in this case, how the author tells us is best to think about it to solve the problem).

... as the elements of the \(\displaystyle \mathbb{Q}\) as a \(\displaystyle \mathbb{Z}\)-module would presumably be of the form \(\displaystyle n \cdot \frac{a}{b}\) where \(\displaystyle n \in \mathbb{Z}\) and \(\displaystyle \frac{a}{b} \in \mathbb{Q} \) ... but ... then (?) on the other hand (?) ... ...

... \(\displaystyle n \cdot \frac{a}{b}\) just gives us another "fraction" ... that is, another member of \(\displaystyle \mathbb{Q}\)

... so ... ... there goes my idea that \(\displaystyle \mathbb{Z} \cdot \mathbb{Q}\) would at least be something different from \(\displaystyle \mathbb{Q}\)

The definition of a module $M$ over a ring $R$ insists that multiplication of the form $rm\in M,$ where $r\in R$ and $m\in M$, so $\mathbb{Z}\cdot\mathbb{Q}$ should at worst return a subset of $\mathbb{Q}$.

... so the difference between Noetherian and non-Noetherian would not just depend weirdly on the view we take of \(\displaystyle \mathbb{Q}\) ...

The "not" in this quote is what makes it incorrect. Depending on how we view $\mathbb{Q}$ (e.g. as a ring=field in this case, or as a $\mathbb{Z}$-module) will tell us whether it is Noetherian of that type or not, as I hope to have helped make a bit clearer above.One quick comment to help try and bring things back to the original post: The author is cleverly relating the non-Noetherian MODULE structure of $\mathbb{Q}$ over $\mathbb{Z}$ to the non-left-Noetherian RING structure of $R$ via how the matrix multiplication works out. The chain of left ideals (a RING concept) will never terminate because the left multiplication structure for the ascending chain of ideals in the RING in this case is essentially the same as the MODULE "multiplication" of $\mathbb{Q}$ over $\mathbb{Z}$.

 

[Math Amateur]

Hi Peter,

This is a good question
and comes down to how one wants to think about/"view" things -- i.e. what part of the (in this case algebraic) structure of the object do you want to consider and what do you want to leave out. When the author says they are thinking of $\mathbb{Q}$ as a $\mathbb{Z}$-module, it means they are forgetting that $\mathbb{Q}$ has more structure (as a field) for the time being.

$\mathbb{Q}$ is Noetherian as ring (= field in this case) because every non-zero element of $\mathbb{Q}$ is a unit (see https://en.wikipedia.org/wiki/Field_(mathematics)) and so there are no non-trivial ascending chains. To help convince yourself of this, take any nonzero element from $\mathbb{Q}$ and compute the ideal it generates.

When viewed as a $\mathbb{Z}$-module, the previous reasoning doesn't apply because you are only dealing with $\mathbb{Q}$'s additive structure -- the notion of unit (see https://en.wikipedia.org/wiki/Unit_(ring_theory)) is not even defined in the case of a module. $\mathbb{Q}$ is not Noetherian as a $\mathbb{Z}$-module because there exist ascending chains of submodules that do not terminate (see https://en.wikipedia.org/wiki/Noetherian_module). The example I gave before was of the submodules generated by $1/2^{n}$, though you could pick many other examples.
If what you mean by $\mathbb{Z}\cdot\mathbb{Q}$ is that you are considering $\mathbb{Q}$ as an additive group over the integers, then, yes, this is what $\mathbb{Q}$ being a $\mathbb{Z}$-module means. If you meant something else here, then I would need clarification.

But I would not only think of $\mathbb{Q}$ as a $\mathbb{Z}$-module all the time. Sometimes it is necessary to use the full field structure of $\mathbb{Q}$ -- how you think about something is largely dictated by the problem you're solving (or, in this case, how the author tells us is best to think about it to solve the problem).

The definition of a module $M$ over a ring $R$ insists that multiplication of the form $rm\in M,$ where $r\in R$ and $m\in M$, so $\mathbb{Z}\cdot\mathbb{Q}$ should at worst return a subset of $\mathbb{Q}$.

The "not" in this quote is what makes it incorrect. Depending on how we view $\mathbb{Q}$ (e.g. as a ring=field in this case, or as a $\mathbb{Z}$-module) will tell us whether it is Noetherian of that type or not, as I hope to have helped make a bit clearer above.One quick comment to help try and bring things back to the original post: The author is cleverly relating the non-Noetherian MODULE structure of $\mathbb{Q}$ over $\mathbb{Z}$ to the non-left-Noetherian RING structure of $R$ via how the matrix multiplication works out. The chain of left ideals (a RING concept) will never terminate because the left multiplication structure for the ascending chain of ideals in the RING in this case is essentially the same as the MODULE "multiplication" of $\mathbb{Q}$ over $\mathbb{Z}$.

Well GJA ... you have certainly given me a lot to think about ...

Still reflecting on what you said ...

Thanks for such a helpful and thought-provoking post ...

Peter