Direct sum of divisible abelian groups is divisible.

Question

I'm trying to solve this in the following way: 1.Divisible abelian group is injective $\Bbb Z$-module 2.Direct product of injective modules is injective (Here I need to show direct sum of injective modules is injective but I can't. As i know it is only true for Noetherian Rings.)\ 3.An injective $\Bbb Z$-module is a divisible abelian group. Please help me to solve this problem.

Answer 1

Have you tried the most direct way? Given divisible abelian groups $A_i$ how does an element $a$ in the sum $\bigoplus_i A_i$ look like? And if you know that $a_i/n$ exists for each $a_i \in A_i$ and $0\neq n\in \mathbb Z$, how does a choice for $a/n$ look like?

Answer 2

Hint:

An abelian group $A_i$ is divisible if for each $n$, the homomorphism $$A_i\xrightarrow{\enspace\times n\enspace}A_i\qquad\text{(in additive notation)}$$ is surjective. Just prove a direct sum of surjections is a surjection.