# Turning an Abelian Group into a Vector Space

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.

• Fantastic! Any clue what $V \otimes \Bbb F_{2^k}$ looks like? Is it ever non-zero? – Omnomnomnom Aug 30 '19 at 17:11
• @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 $\mathbb F$ – Maxime Ramzi Aug 30 '19 at 17:17
• @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: $v\otimes\alpha = v^{2^k}\otimes\alpha = v\otimes(2^k\alpha) = v\otimes 0 = \mathbf{0}$. – Arturo Magidin Aug 30 '19 at 17:20
• @jyrkilahtonen : I think that's the case : the notation $\mathbb F$ is used for both – Maxime Ramzi Aug 30 '19 at 17:26
• @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 $(-1)^2 = 1$ – 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.

• 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 $V\otimes \mathbb F\to V$ – Maxime Ramzi Aug 30 '19 at 16:46
• @Max I see no justification for my statement, now that I think about it. I'll remove it. – Omnomnomnom Aug 30 '19 at 16:48