As the title says;
Is there always a free $\mathbb{Q}$-module on a set $A$?
I believe so, since you can always form the free $\mathbb{Z}$-module on a set $A$, $F(A)$ and then identify $\mathbb{Z}$ as a subset of $\mathbb{Q}$.
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5How does identifying $\mathbb{Z}$ as a subset of $\mathbb{Q}$ do anything? In any case, yes, there is always such a free module, and $\mathbb{Q}$ can be replaced with any ring. – Qiaochu Yuan Sep 30 '23 at 15:19
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You're right that the free $\Bbb Q$-module will always have a submodule which is a free $\Bbb Z$-module on $A$. Technically you could go from the free $\Bbb Z$-module to the free $\Bbb Q$-module, but that's confusing and takes way more effort than you need! Since you're happy the free $\Bbb Z$-module exists, just look at the proof of that and replace "$\Bbb Z$" with "$\Bbb Q$". In general in algebra, "the free object on $A$" will usually exist. This is not actually much harder to prove that existence of free $\Bbb Z$-modules. – Izaak van Dongen Sep 30 '23 at 16:14