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

Suppose $\mathbb{F}$ is a field, and that $X$ is the multiplicative group of $\mathbb{F}$.

Let us write this abelian group $X$ additively. To be precise the set is $\mathbb{F}\setminus\{0\}$, the zero element is $\hat{0}:=1$, 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 $\mathbb{F}$-scalar multiplication on $V$ so that $V$ becomes an $\mathbb{F}$-vector space?

By looking at $-1$ in $\mathbb{F}$, which satisfies $(-1)\hat{+}(-1)=(-1)^2=1=\hat{0}$ we see that if there is to be any chance of turning $V$ into a vector space then the field $\mathbb{F}$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $\mathbb{F}$ has no finite subfields $\mathbb{F}_{2^k}$ except $\mathbb{F}_2$. 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 $x\in \mathbb F^*$. In the structure of $V$, we have $2\cdot x = 0$, 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 : $|\mathbb F| \leq 2$

As suggested in the comments, let us try to see what happens if we have two fields a field $K$, a field $\mathbb F$ and we want to see when $K^*$ has a $\mathbb F$ vector space structure. Suppose $K$ has characteristic $\neq 2$. Then $(-1)^2 = 1$ implies that either $K^* = 0$ as a vector space (and so $K=\mathbb F_2$, conversely, this can be given a vector space structure over any field) or $2=0$ in $\mathbb F$, so $\mathbb F$ has characteristic $2$. But then $x^2=1$ for all $x\in K^*$ which implies that $K^*$ has at most two elements, in particular $K=\mathbb F_2$ or $\mathbb F_3$. Conversely, $\mathbb F_3^*$ is a $\mathbb F_2$-vector space.

Now if $K$ has characteristic $2$: as Jyrki Lahtonen points out, at least all the Mersenne primes give fields over which it works. I haven't figured out the whole thing yet. If $K$ has characteristic $2$ (and is not $\mathbb F_2$) then $\mathbb F$ can't, therefore $2$ is invertible : in particular $K$ must be a perfect field as every element has a square root. If $\mathbb F$ is of positive characteristic, then in particular $p=0$ for some $p$ so that $K$ must be finite, and in particular $K^*$ is cyclic : it must be $1$-dimensional over the prime subfield of $\mathbb F$ which must therefore equal $\mathbb F$, say it is $\mathbb F_p$ and then $p$ is a Mersenne prime, and conversely any Mersenne prime yields an example.

It remains to see what happens if $\mathbb F$ has characteristic $0$. In this case any element of $K^*$ is uniquely divisible. It follows that $K$ can't contain any nontrivial finite field : if $x$ is algebraic over $\mathbb F_2$, then it has exactly $3$ cube roots, all in the same finite subfield, instead of just the one.

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  • $\begingroup$ Fantastic! Any clue what $V \otimes \Bbb F_{2^k}$ looks like? Is it ever non-zero? $\endgroup$ – Omnomnomnom Aug 30 '19 at 17:11
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    $\begingroup$ @omnomnomnom : yes, in fact if $\mathbb F$ is e.g. $\mathbb F_2 (X) $ then its multiplicative group is free, so $V\otimes \mathbb F$ is a direct sum of copies of $\mathbb F$ $\endgroup$ – Maxime Ramzi Aug 30 '19 at 17:17
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    $\begingroup$ @Omnomnomnom But for $\mathbb{F}_{2^k}$ and $V=\mathbb{F}_{2^k}\setminus{0}$ under multiplication, it is zero: every element of $V$ satisfies $v^{2^k}=v$, hence the generators are all zero: $v\otimes\alpha = v^{2^k}\otimes\alpha = v\otimes(2^k\alpha) = v\otimes 0 = \mathbf{0}$. $\endgroup$ – Arturo Magidin Aug 30 '19 at 17:20
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    $\begingroup$ @jyrkilahtonen : I think that's the case : the notation $\mathbb F$ is used for both $\endgroup$ – Maxime Ramzi Aug 30 '19 at 17:26
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    $\begingroup$ @Jyrkilahtonen : yes you are correct I hadn't thought it through. Actually my argument rules out characteristic $0$ for the field that becomes the vector space because of $(-1)^2 = 1$ $\endgroup$ – Maxime Ramzi Aug 30 '19 at 20:19

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 $X = \Bbb F\setminus \{0\}$, 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.

  • $\begingroup$ Why does $V\otimes \mathbb F = 0$ ? if it is the case then there is no $\mathbb F$-vector space structure on $V$ as such a thing would be a map $V\otimes \mathbb F\to V$ $\endgroup$ – Maxime Ramzi Aug 30 '19 at 16:46
  • $\begingroup$ @Max I see no justification for my statement, now that I think about it. I'll remove it. $\endgroup$ – Omnomnomnom Aug 30 '19 at 16:48

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