Is $\mathbb{Z}(p^{\infty})$ a vector space over some field $\mathbb{F}$?

I don't know how to write in good English, so I will follow Hungerford's word from his book Algebra.

The following relation on the additive group $\mathbb{Q}$ of rational numbers is a congruence relation $a\sim b \leftrightarrow a-b\in \mathbb{Z}$. Denote for $\mathbb{Q}/\mathbb{Z}$ the set of all those equivalence classes. Let $p$ be a prime and $$\mathbb{Z}(p^{\infty})=\{\overline{a/b}\in \mathbb{Q}/\mathbb{Z}\;|\; a,b \in \mathbb{Z}\text{ and }b=p^{i}\text{ for some } i\geq 0\}.$$

I am trying to define a vector space structure on $\mathbb{Z}(p^{\infty})$ over a field $\mathbb{F}$, but I think it is impossible. For example, if $\mathbb{F}$ is uncountable, it can't be done. But I don't know to answer this question when $\mathbb{F}=\mathbb{Q}$ or $\mathrm{char}(\mathbb{F})=q$, where $q$ is a prime number.

-

A vector space over $\mathbf{F}$ must satisfy that $\alpha\mathbf{v}=\mathbf{0}$ if and only if $\alpha=0$ or $\mathbf{v}=\mathbf{0}$.

In particular, if $\mathbf{F}$ is of characteristic $0$, then the underlying abelian group structure of $V$ must be torsion-free, since writing $n$ as a sum of $n$ copies of $1_{\mathbf{F}}$, we get $$\underbrace{\mathbf{v}+\cdots+\mathbf{v}}_{n\text{ summands}} = (\underbrace{1_\mathbf{F}+\cdots+1_{\mathbf{F}}}_{n\text{ summands}})\mathbf{v} = \mathbf{0}\implies \mathbf{v}=\mathbf{0}.$$

Similarly, a vector space over a field of characteristic $p\gt 0$ must have underlying abelian group structure that is $p$-torsion ($p\mathbf{v}=\mathbf{0}$ for all $\mathbf{v}\in V$). This does not hold for the Prüfer group, so the Prüfer group cannot be given a vector space structure over any field.

-

Let me write $V=\mathbb Z(p^\infty)$ for brevity.

Notice that the element $x=\overline{1/p}$ of $V$ is non-zero and is such that $p\cdot x=0$. It follows from this that if $V$ is to be a vector space, it has to be a vector space over a field of characteristic $p$. In particular, $p\cdot y=0$ for all $y\in V$.

But now $z=\overline{1/p^2}\in V$ is an element such that $p\cdot v\neq0$.

-

There is no such field. The group has the property that every element has a finite order, so it cannot be a vector space over a field of characteristic $0$; but the group has elements of order other than $1$ and $q$, so cannot be a vector space over a field of characteristic $q$ either. (After all, if $q = 0$, then $q v = 0$ as well.)

-