MHB Composition Series and Noetherian and Artinian Modules ....

Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
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 the proof of part of Proposition 4.2.14 ... ...

Proposition 4.2.14 reads as follows:

https://www.physicsforums.com/attachments/8237
https://www.physicsforums.com/attachments/8235
In the above proof by Bland we read the following:

"... ... Since $$M / M_1$$ is a simple R-module, $$M / M_1$$ is artinian and noetherian ... ... Can someone please explain why $$M / M_1$$ being a simple R-module implies that $$M / M_1$$ is artinian and noetherian ... ... ?Peter
 
Last edited:
Physics news on Phys.org
Peter said:
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 the proof of part of Proposition 4.2.14 ... ...

Proposition 4.2.14 reads as follows:
In the above proof by Bland we read the following:

"... ... Since $$M / M_1$$ is a simple R-module, $$M / M_1$$ is artinian and noetherian ... ... Can someone please explain why $$M / M_1$$ being a simple R-module implies that $$M / M_1$$ is artinian and noetherian ... ... ?Peter
It now occurs to me that the answer to my question is quite straightforward ... indeed ...$$M / M_1$$ is simple $$\Longrightarrow$$ only submodules of $$M / M_1$$ are $$\{ 0 \}$$ and $$M / M_1$$$$\Longrightarrow$$ only descending and ascending chains of submodules are finite ... that is terminate in a finite number of elements$$\Longrightarrow$$ $$M / M_1$$ is artinian and noetherian ...
Is that correct ... ?

Peter
 
Yes, it is correct. It means that every simple module is fingen.

It is not true for rings, though.
 
steenis said:
Yes, it is correct. It means that every simple module is fingen.

It is not true for rings, though.
Thanks steenis ...

Appreciate your help ...

Peter
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K