TITLE: How can we show that $\mathbb Q$ is not a free $\mathbb Z$-module?
QUESTION [35 upvotes]: I am really confused from the definition.
How do we know that $\mathbb Q$ is not a free $\mathbb Z$-module?
In class people use it as a trivial fact, but I don't seem to understand.

REPLY [2 votes]: It follows from the definition of free modules.
Let us suppose to the contradictory that $\mathbb{Q}$ is a free $\mathbb{Z}$ module, so by definition of free modules, for a given injective  map $\alpha: X \rightarrow \mathbb{Q}$ and for any map $f : X \rightarrow \mathbb{Z}$, there exist a unique $\mathbb{Z}$-homomorphism $g: \mathbb{Q} \rightarrow \mathbb{Z}$ such that $f=g\alpha$. Every $\mathbb{Z}$ module homomophism is a group homomorphism and we know that there is only trivial group homomorphism from $\mathbb{Q}$ to $\mathbb{Z}$. Since we can define a lot of distinct maps from $X$ to $\mathbb{Z}$ and we don't have any homomorphism from $\mathbb{Q}$ to $\mathbb{Z}$ corresponding to non-zero maps $f:X \rightarrow \mathbb{Z}$, thus $\mathbb{Q}$ is not a free module over $\mathbb{Z}$.