MHB Endomorphism Rings - Bland Example 7 - page 10

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Example Rings
Click For Summary
The discussion centers on understanding the notation used in Paul E. Bland's example of endomorphism rings in his book "Rings and Their Modules." Specifically, there is confusion regarding the subscript $$\mathbb{Z}$$ in the notation $$\text{End}_{\mathbb{Z}}(G)$$, where G is an abelian group. It is clarified that the notation $$\text{End}_{R}(M)$$ refers to the endomorphisms of an R-module M, and since any abelian group can be viewed as a $$\mathbb{Z}$$-module, the subscript is relevant. This highlights the connection between abelian groups and modules over the integers. Understanding this relationship is crucial for grasping the concepts presented in Bland's work.
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Paul E. Bland's book, "Rings and Their Modules".

In Chapter 1: Basic Properties of Rings and Modules, Bland gives endomorphism rings as a basic example of a ring.

The example (Example 7) reads as follows:https://www.physicsforums.com/attachments/3572

I do not feel that I fully understand Bland's notation in this example.

Bland talks about the ring $$\text{End}_{\mathbb{Z} } (G)$$ where G is an abelian group ... ... BUT ... ... why do we have $$\mathbb{Z}$$ as a subscript in this definition? $$\mathbb{Z}$$ seems to have no relevance to this definition.

Hope someone can help ...

Peter
 
Physics news on Phys.org
Hi Peter,

Usually, the notation $$End_{R}(M)$$ denotes the endomorphism of an R-module M.

Any abelian group can be thought on as a $$\Bbb{Z}$$ module.
 
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 2 ·
Replies
2
Views
1K
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K