MHB Endomorphism Rings .... Bland, Example 7, Section 1.1 .... ....

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 1.1 Rings and need some help to fully understand the proof of part of Example 7 on page 10 ... ...

Example 7 on page 10 reads as follows:View attachment 8197In the above example from Bland we read the following:

" ... ... $$\text{ End}_\mathbb{Z} (G) $$ denotes the set of all group homomorphisms $$f \ : \ G \to G$$ ... ... "Can someone explain exactly why $$\mathbb{Z}$$ is in the symbol/notation $$\text{ End}_\mathbb{Z} (G) $$ for the set of all group homomorphisms $$f \ : \ G \to G$$ ... ... ?Peter
 
Physics news on Phys.org
The $”R”$ in $Hom_R(M,N)$ denotes that we are working with $R$-modules and $R$-maps.
$Hom_{\mathbb{Z}}(G,H)$ denotes that we are dealing with $\mathbb{Z}$-modules and $\mathbb{Z}$-maps. Now $\mathbb{Z}$-modules are additive abelian groups and $\mathbb{Z}$-maps are homomorphisms between additive abelian groups.

$End_{\mathbb{Z}}(G)$ is short for $Hom_{\mathbb{Z}}(G,G)$ and the $”\mathbb{Z}”$ denotes that the underlying ring is $\mathbb{Z}$.
 
steenis said:
The $”R”$ in $Hom_R(M,N)$ denotes that we are working with $R$-modules and $R$-maps.
$Hom_{\mathbb{Z}}(G,H)$ denotes that we are dealing with $\mathbb{Z}$-modules and $\mathbb{Z}$-maps. Now $\mathbb{Z}$-modules are additive abelian groups and $\mathbb{Z}$-maps are homomorphisms between additive abelian groups.

$End_{\mathbb{Z}}(G)$ is short for $Hom_{\mathbb{Z}}(G,G)$ and the $”\mathbb{Z}”$ denotes that the underlying ring is $\mathbb{Z}$.
Thanks for the help, steenis ...

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 ·
Replies
1
Views
1K
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
742
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K