Every torsion free divisible abelian group D is direct sum of the copies of the $\mathbb{Q}$

257 Views Asked by At

Prove that every torsion free divisible abelian group $D$ is direct sum of the copies of the $\mathbb{Q}$.

If $a\in D$ then there exists unique $b \in D$ and $n\neq 0 \in \mathbb{Z} $ such that a=nb. Now, D is torsional free means that every element except identity is not of finite order. b can be written as b= 1/n a but I am unable to think which result should I use now?

Can you outline a complete proof?

1

There are 1 best solutions below

2
On BEST ANSWER

Here are some hints, if you can answer these, you should have your proof.

  1. What is another name for modules over the ring $\mathbb{Z}$?
  2. What is another name for modules over the ring $\mathbb{Q}$?
  3. What structural theorems do we know about $\mathbb{Q}$-modules?