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

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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
 
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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
 
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This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
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