# Finding the injective hull

• A
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.

fresh_42
Mentor
2021 Award
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