Example - Bland - Right Noetherian but not Left Noetherian

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In summary: S_i## of the form ##\begin{bmatrix}0 & S_i \\ 0 & 0 \end{bmatrix}##... in ##\mathbb{Q}##... Now all we are allowed to do, is multiply them with elements of ##R## from the left. Let's do it:$$ \begin{bmatrix}\mathbb{Z} & \mathbb{Q} \\ 0 &\mathbb{Q}\end{bmatrix}\,\cdot\,\begin{bmatrix}0 & S_i \\ 0 & 0 \end{bmatrix}=\begin{bmatrix}
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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:
?temp_hash=8f533acc4f31b449a5f54e3b598df8df.png

In the above text, I do not understand the proof that ##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, ##\mathbb{Q}## is not Noetherian when viewed as a ##\mathbb{Z}##-module.

Hence if ##S_1 \subseteq S_2 \subseteq S_3 \subseteq## ... ... is a non-terminating chain of ##\mathbb{Z}##-modules of ##\mathbb{Q}##, then##\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 ##R## ... ... "

My questions are as follows:What is the justification for regarding ##\mathbb{Q}## as a ##\mathbb{Z}##-module ... indeed, couldn't we have done this while trying to prove/justify that ##R## was right Noetherian and ended up with a non-terminating chain of right ideals of ##R## ...Why is ##\mathbb{Q}## not Noetherian when viewed as a ##\mathbb{Z}##-module ... ... ?Can someone please simply and clearly explain the proof of ##R## not being left Noetherian ... ... ?
Hope someone can help ...

Peter
 

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  • #2
Math Amateur said:
What is the justification for regarding ##\mathbb{Q}## as a ##\mathbb{Z}##-module
It is easy to show that ##\mathbb{Q}## can be regarded as a ##\mathbb{Z}##-module. All that is needed is to show that it satisfies the four module axioms.
Why is ##\mathbb{Q}## not Noetherian when viewed as a ##\mathbb{Z}##-module ... ... ?
Because there exist non-terminating ascending chains of ideals, for instance the chain whose ##k##th element is ##2^{-k}\mathbb Z##.
 
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  • #3
Math Amateur said:
What is the justification for regarding ##\mathbb{Q}## as a ##\mathbb{Z}##-module
Hi Peter.

We consider ##R##-left-modules ##S_i## of the form ##\begin{bmatrix}0 & S_i \\ 0 & 0 \end{bmatrix}##
We simply choose them as an arbitrary chain ##S_1 \subseteq S_2 \subseteq S_3 \subseteq \dots## in ##\mathbb{Q}##

Now all we are allowed to do, is multiply them with elements of ##R## from the left. Let's do it:
$$ \begin{bmatrix}\mathbb{Z} & \mathbb{Q} \\ 0 &\mathbb{Q}\end{bmatrix}\,\cdot\,\begin{bmatrix}0 & S_i \\ 0 & 0 \end{bmatrix}=\begin{bmatrix}0 & \mathbb{Z} \cdot S_i \\ 0 & 0 \end{bmatrix} $$
Therefore the ascending chain condition of the ##S_i## as left ##\mathbb{Z}##-modules comes into play.
... indeed, couldn't we have done this while trying to prove/justify that ##R## was right Noetherian and ended up with a non-terminating chain of right ideals of ##R## ...
Do the multiplication from the right and see what happens.
Why is ##\mathbb{Q}## not Noetherian when viewed as a ##\mathbb{Z}##-module ... ... ?
The short answer is Proposition 5.33. in Joseph J. Rotman's book: Advanced Modern Algebra.
The long one is to define ##S_i := \mathbb{Z} \cdot \frac{1}{2^i} ## for ##i \in \mathbb{N}##.
 
  • #4
Thanks for the help fresh_42 and Andrew ... help is much appreciated ...

... BUT ... still somewhat uneasy and perplexed by this example ...

One of the points I am uneasy about is that if we "view" ##\mathbb{Q}## as a ##\mathbb{Z}##-module it is not Noetherian ... but if we "view" ##\mathbb{Q}## as a field then it is Noetherian ... ? ... can you explain how this makes sense ...

Maybe when thinking of ##\mathbb{Q}## maybe I should be thinking of ##\mathbb{Z} \cdot \mathbb{Q}## as the elements of the ##\mathbb{Q}## as a ##\mathbb{Z}##-module would presumably be of the form ##n \cdot \frac{a}{b}## where ##n \in \mathbb{Z}## and ##\frac{a}{b} \in \mathbb{Q}## ... but ... then (?) on the other hand (?) ... ... ##n \cdot \frac{a}{b}## just gives us another "fraction" ... that is another member of ##\mathbb{Q}## ... ... there goes my idea that ##\mathbb{Z} \cdot \mathbb{Q}## would at least be something different from ##\mathbb{Q}## ... so the difference between Noetherian and non-Noetherian would not just depend weirdly on the view we take of ##\mathbb{Q}## ...

Hope you can clarify this issue for me ...

Peter
 
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  • #5
Math Amateur said:
Thanks for the help fresh_42 and Andrew ... help is much appreciated ...

... BUT ... still somewhat uneasy and perplexed by this example ...

One of the points I am uneasy about is that if we "view" ##\mathbb(Q)## as a ##\mathbb(Z)##-module it is not Noetherian ... but if we "view" ##\mathbb(Q)## as a field then it is Noetherian ... ? ... can you explain how this makes sense ...

Maybe when thinking of ##\mathbb(Q)## maybe I should be thinking of ##\mathbb(Z) \cdot \mathbb(Q)## as the elements of the ##\mathbb(Q)## as a ##\mathbb(Z)##-module would presumably be of the form ##n \cdot \frac{a}{b}## ... but ... then (?) on the other hand (?) ... ... ##n \cdot \frac{a}{b}## just gives us another "fraction" ... that is another member of ##\mathbb(Q)## ... ...

Hope you can clarify this issue for me ...

Peter
We consider ##\begin{bmatrix}0 & S_i \\ 0 & 0\end{bmatrix}## as ##R = \begin{bmatrix} \mathbb{Z} & \mathbb{Q} \\ 0 & \mathbb{Q}\end{bmatrix}## left modules.

But multiplication from left means we have ##\mathbb{Z} \cdot S_i \subseteq S_i \subseteq \mathbb{Q}##. So the ##S_i \subseteq \mathbb{Q}## are left ##\mathbb{Z}##-modules by their construction. The requirement ##s_i \cdot r \in \begin{bmatrix}0 & S_i \\ 0 & 0\end{bmatrix} \; (s_i \in \begin{bmatrix}0 & S_i \\ 0 & 0\end{bmatrix} \;,\; r\in R)## forces the fact that the ##S_i## have to be left ##\mathbb{Z}##-modules.

And yes, ##n \cdot \frac{a}{b} \in \mathbb{Q}##. So? That's the ##\mathbb{Z}##-module property of the ##S_i## which is inherited by the ##R##-module property of the ##S_i##-matrices ##\begin{bmatrix}0 & S_i \\ 0 & 0\end{bmatrix}##.

The point is that we get an ascending chain of ##R##-modules that does not end:
$$ \begin{bmatrix} 0 & \mathbb{Z} \cdot 2^{-1} \\ 0 & 0 \end{bmatrix} \subsetneq \begin{bmatrix} 0 & \mathbb{Z} \cdot 2^{-2} \\ 0 & 0 \end{bmatrix} \subsetneq \begin{bmatrix} 0 & \mathbb{Z} \cdot 2^{-3} \\ 0 & 0 \end{bmatrix} \subsetneq \ldots \subsetneq \begin{bmatrix} \mathbb{Z} & \mathbb{Q} \\ 0 & \mathbb{Q} \end{bmatrix} =: R $$
We define them in this way. All in ##R \, ##! And one such chain is sufficient, but any other prime will do as well.

Edit: If we replace all ##\mathbb{Z}## by ##\mathbb{Q}## then the example breaks down as all inclusions become equalities. But then we would talk about another ring.
 
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  • #6
Thanks so much for the further assistance ... most helpful!

I think I'm closer to understanding things now ... but still reflecting ...

Thanks again ...

Peter
 
  • #7
Correction to post #5: It has to be ##r\,\cdot\,s_i \in \begin{bmatrix}0 & S_i \\ 0 & 0\end{bmatrix}## since we consider left modules here. It doesn't work for right modules.
Sorry.
 
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  • #8
Any commutative ring ##R## can be considered as either a ##R##-module ##R_R## or a ##\mathbb Z##-module ##R_{\mathbb Z}##. In the former, the scalar product ##rm## where ##r## is in ##R## as ring and ##m## is in ##R## as module, is the module element corresponding to the ring element ##rm## (calculated using ring multiplication). In the latter, the scalar product ##nm## with ##n\in\mathbb Z## means ##m+m+...+m## with ##n## terms.

##R_{\mathbb Z}## in general has a richer set of submodules than ##R_R##. In fact if ##R## is a field then the only submodules of ##R_R## are {0} and ##R_R##. Can you figure out why?
 
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  • #9
andrewkirk said:
Any commutative ring ##R## can be considered as either a ##R##-module ##R_R## or a ##\mathbb Z##-module ##R_{\mathbb Z}##. In the former, the scalar product ##rm## where ##r## is in ##R## as ring and ##m## is in ##R## as module, is the module element corresponding to the ring element ##rm## (calculated using ring multiplication). In the latter, the scalar product ##nm## with ##n\in\mathbb Z## means ##m+m+...+m## with ##n## terms.

##R_{\mathbb Z}## in general has a richer set of submodules than ##R_R##. In fact if ##R## is a field then the only submodules of ##R_R## are {0} and ##R_R##. Can you figure out why?
Andrew ... you ask ... " ... ... In fact if ##R## is a field then the only submodules of ##R_R## are {0} and ##R_R##. Can you figure out why? "

Basically that is because if an ideal contains a unit then it is necessarily equal to the whole ring ... and all elements of a field, except 0, are units ...

Peter
 
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  • #10
Math Amateur said:
Andrew ... you ask ... " ... ... In fact if ##R## is a field then the only submodules of ##R_R## are {0} and ##R_R##. Can you figure out why? "

Basically that is because if an ideal contains a unit then it is necessarily equal to the whole ring ... and all elements of a field, except 0, are units ...

Peter
This is also the reason, why the example above, regarded as right modules leads to a different result.
The multiplication ##s_i\cdot r## becomes ##S_i \cdot \mathbb{Q}##.
 
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FAQ: Example - Bland - Right Noetherian but not Left Noetherian

1. What does it mean for a ring to be right Noetherian but not left Noetherian?

A ring is considered right Noetherian if every ascending chain of right ideals eventually stabilizes. Similarly, a ring is left Noetherian if every ascending chain of left ideals eventually stabilizes. A ring that is right Noetherian but not left Noetherian means that there exists at least one ascending chain of left ideals that does not stabilize.

2. Can you give an example of a ring that is right Noetherian but not left Noetherian?

One example of a ring that is right Noetherian but not left Noetherian is the ring of upper triangular matrices with coefficients in a field. This ring has a finite descending chain of left ideals, but an infinite ascending chain of left ideals.

3. Why is it important to distinguish between right Noetherian and left Noetherian rings?

The distinction between right Noetherian and left Noetherian rings is important because it affects the structure and properties of the ring. For example, a right Noetherian ring may have a finite basis for its right ideals, while a left Noetherian ring may have a finite basis for its left ideals. This can have implications for the solvability of systems of equations and the existence of certain types of submodules.

4. Can a ring be both right Noetherian and left Noetherian?

Yes, a ring can be both right Noetherian and left Noetherian. In fact, most commonly studied rings, such as polynomial rings and finite-dimensional algebras, are both right and left Noetherian.

5. How does the concept of Noetherianness relate to other properties of rings?

Noetherianness is related to other properties of rings in that it is often used as a condition in theorems and proofs involving other properties. For example, a ring that is both Noetherian and Artinian is called a Noetherian ring, and it has many desirable properties such as being finitely generated and having a unique decomposition into primary ideals. Noetherianness is also closely related to the concept of homological dimension, which measures the complexity of a ring's modules.

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