Is $\mathbb{Q}^n$ a vector space over $\mathbb{Z}$ or over $\mathbb{Q}$?
$\mathbb{Q}^n$ is clearly not a vector space over $\mathbb{R}$, because scalar multiplication of some $q \in \mathbb{Q}$ by $\pi$ renders $\pi q$, which is irrational.
Is $\mathbb{Q}^n$ a vector space over $\mathbb{Z}$ or over $\mathbb{Q}$?
$\mathbb{Q}^n$ is clearly not a vector space over $\mathbb{R}$, because scalar multiplication of some $q \in \mathbb{Q}$ by $\pi$ renders $\pi q$, which is irrational.
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$\mathbb{Q}^n$ can't be a vector space over $\mathbb{Z}$ since $\mathbb{Z}$ isn't a field.
To see that $\mathbb{Q}^n$ is a vector space over $\mathbb{Q}$ you could first prove that every field $\mathbb{F}$ is a vector space over itself and then since the product of vector spaces is a vector space $\mathbb{F}^n$ is a vector space over $\mathbb{F}$ and so $\mathbb{Q}^n$ is a vector space over $\mathbb{Q}$ since $\mathbb{Q}$ is a field.
You can't have a vector space over $\Bbb Z$. By definition, a vector space is required to be over a field. If you take away the field requirement, what you're left with is calles a module. And yes, $\Bbb Q^n$ is definitely a $\Bbb Z$-module.
That being said, there is a list of $10$ requirements (or thereabouts, it varies slightly) for a space like $\Bbb Q^n$ to be a vector space over a field like $\Bbb Q$. All of them should be easily verifiable. So yes, $\Bbb Q^n$ is a vector space over $\Bbb Q$.