Xjyuk

This question is inspired by the question Are these vector spaces? but is free-standing.

Suppose $mathbbF$ is a field, and that $X$ is the multiplicative group of $mathbbF$.

Let us write this abelian group $X$ additively. To be precise the set is $mathbbFsetminus0$, the zero element is $hat0:=1$, the addition operation is $x hat+y:=xy$, and the negative operation is $widehat-x:=x^-1$.

Let $V$ be the abelian group of $n$-tuples $X^n$ with the usual co-ordinatewise operations.

Question : Can we define an $mathbbF$-scalar multiplication on $V$ so that $V$ becomes an $mathbbF$-vector space?

By looking at $-1$ in $mathbbF$, which satisfies $(-1)hat+(-1)=(-1)^2=1=hat0$ we see that if there is to be any chance of turning $V$ into a vector space then the field $mathbbF$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $mathbbF$ has no finite subfields $mathbbF_2^k$ except $mathbbF_2$. Beyond that I cannot go.

The answer is no except for the trivial case $mathbb F =mathbb F_2$ (in which $V=0$)

Indeed, assume $mathbb F$ is such a field and take $xin mathbb F^*$. In the structure of $V$, we have $2cdot x = 0$, which translates to $x^2= 1$ in $mathbb F$ (for each coordinate).

This implies that any element of $mathbb F$ is a root of $X^2-1$ which has at most to roots : $|mathbb F| leq 2$

Not an answer, but hopefully a useful observation. Your question (if I have understood it correctly) can be reframed as follows.

We are given an abelian group $X = Bbb Fsetminus 0$, which we present as a $Bbb Z$-module where $x+y := xy$ and $nx := x^n$. Is it possible to take $V = X^n$ and extend multiplication by $Bbb Z$ into multiplication by $Bbb F$ in such a way that $V$ becomes a $Bbb F$-module (i.e. a vector space over $Bbb F$)?

As Max's comment below observes: any $Bbb F$-space structure $V$ exists corresponds to an $Bbb F$-linear map $V otimes_Bbb Z Bbb F to V$. Notably, $V otimes_Bbb Z Bbb F$ is the usual extension of scalars.