Is $\Bbb Q$ a decomposable module over $\Bbb Z$ or not?

Question

Is $\Bbb Q$ a decomposable module over $\Bbb Z$ or not?

My attempt:

let, $p_1,p_2,\dots,p_k,\dots$ be an enumeration of primes in $\Bbb N$. Then, can't we write $\Bbb Q = \Bbb Z \oplus Z(\frac{1}{p_1})\oplus \Bbb Z(\frac{1}{p_2})\oplus \dots \oplus \Bbb Z(\frac{1}{p_k}) \oplus \dots $ ?

If the above expression is true, for some fixed $l \in \Bbb N$, we consider $A=\Bbb Z \oplus Z(\frac{1}{p_1})\oplus \Bbb Z(\frac{1}{p_2})\oplus \dots \oplus \Bbb Z(\frac{1}{p_l})$ and $B=Z(\frac{1}{p_{l+1}}) \oplus \dots $ .

Then we are able to write $\Bbb Q$ as a direct sum of two non-zero $\Bbb Z$-modules.

Please give a correction if I'm mistaken. Thanks in Advance for help!

Answer 1

If the rationals were decomposable, then you could write the rationals as the direct product of nontrivial modules $\mathbb{Q}=A\oplus B$. But as the sum is direct, any $0\neq v\in A, 0\neq w\in B$ are linearly independent over the integers. However, this is a contradiction, as two rational numbers are always linearly dependent over the integers as $$ (q a)\cdot \frac{p}{q} +(-pb) \frac{a}{b}=0.$$

This also shows you your mistake, none of your summands is direct.

Answer 2

No, the element $1$ lies in both $\mathbb{Z}$ and $\mathbb{Z}[1/p]$, so the sum is not direct.

Answer 3

Suppose $\mathbb{Q}$ is decomposable as $\mathbb{Q}=X\oplus Y$. Then $X$ is

1. divisible, because it is a homomorphic image of $\mathbb{Q}$;
2. torsionfree, because it is a subgroup of $\mathbb{Q}$.

Similarly for $Y$. Therefore $X$ and $Y$ are vector spaces over $\mathbb{Q}$ and $\mathbb{Q}=X\oplus Y$ is a decomposition as direct sum of subspaces. Hence either $X$ or $Y$ must have dimension $0$.