# Finding the injective hull

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## Main Question or Discussion Point

Let's suppose that I have an element $e$ of order $p$ in the group of complex numbers whose elements all have order $p^n$ for some $n\in\mathbb{N}$ (henceforth called $K$), and the module generated by $(e)$ is irreducible.

How do I show that the injective hull of the module generated by $(e)$ is in fact, equal to $K$?

Attempted Work. I was told that this submodule generated by $(e)$ is isomorphic to $\mathbb{Z}[p^{-1}]$. I don't know how to proceed from there... but I think it involves showing that two maps are surjective.

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fresh_42
Mentor
Let's suppose that I have an element $e$ of order $p$ in the group of complex numbers whose elements all have order $p^n$ for some $n\in\mathbb{N}$ (henceforth called $K$), and the module generated by $(e)$ is irreducible.

How do I show that the injective hull of the module generated by $(e)$ is in fact, equal to $K$?

Attempted Work. I was told that this submodule generated by $(e)$ is isomorphic to $\mathbb{Z}[p^{-1}]$. I don't know how to proceed from there... but I think it involves showing that two maps are surjective.
What is the module operation? Module above what?

mathwonk
Homework Helper
do you know what a divisible group is? and how it relates to injectivity? it seems obvious in your example that to make your subgroup divisible, you would need to include everything in K.

https://en.wikipedia.org/wiki/Divisible_group

do you know what a divisible group is? and how it relates to injectivity?
I think there is a theorem that for a principle ideal domain X, X is divisible iff. X is injective. Tell me, is it iff. or if?

it seems obvious in your example that to make your subgroup divisible, you would need to include everything in K.

https://en.wikipedia.org/wiki/Divisible_group
It isn't obvious to me. Could you explain why?

mathwonk
Homework Helper
divisibility involves the presence of roots, and in your example everything is a root of your basic elements. i.e. everything in K becomes an element of your small subgroup after raising to a high power. note that your group is written multiplicatively and the definition for abelian groups is usually stated additively so that means when they say a group is divisible if na belonging to the goup forces a to also belong, means that if a^n belongs then a also belongs.
try to relax over this stuff. just try to actually learn what some of these things are, if you don't finish everything in a certain time, that matters less than actually learning some of it so you don't have to repeat that part.

mathwonk