Are there always multiple scalar multiplications over non-prime fields?

Question

Let $F$ a non-prime field and $V$ an abelian group and denote by transposition a scalar multiplication operation $F \times V \to V$ making $V$ into a vector space.

Does there always exist another scalar multiplication $\cdot: F \times V \to V$ that is distinct from the original one that still makes $V$ into a vector space?

Some initial results I have:

It is easy enough to show that if $F$ is prime, scalar multiplication is uniquely determined. On the other hand it is easy enough to construct examples in which there are multiple scalar multiplications, e.g. if $V$ is a complex vector space, we can make another scalar multiplication by $c \cdot v = \overline{c}v$.

More generally if $K \subsetneq F$ and $F = K(\alpha)$ where $\alpha$ is algebraic and has a Galois conjugate $\beta$, we can construct an automorphism $\tau : (F, \alpha) \to (F, \beta)$ that preserves $K$ and use that to define scalar multiplication $r \cdot v = \tau(r) v$.

Cases I haven't figured are e.g. $\mathbb{Q}(\sqrt[3]{2})$, where we don't have any Galois conjugates in our field.

Note: This question is not answered here.

Answer 1

Yes, as long as $V$ is nonzero. We don't need to construct an automorphism of $F$; if $\varphi : V \to V$ is any automorphism of $V$ (as an abelian group!) and $\rho : F \to \text{End}(V)$ is our original scalar multiplication operation then the pointwise conjugate $\varphi \rho \varphi^{-1}$ is a scalar multiplication distinct from the original unless $\rho(f) = \varphi \rho(f) \varphi^{-1}$ for all $f \in F$, and we'll always be able to arrange for this not to be true.

Proof. By hypothesis $F$ isn't a prime field, so it has dimension at least $2$ over its prime subfield $k$, hence $V$ is also a vector space of dimension at least $2$ over the prime subfield $k$. The endomorphism ring $\text{End}(V)$ is then a (possibly infinite-dimensional) matrix algebra over $k$, and in particular is not only noncommutative but has center $k$ (exercise). Hence the action of $F$ on $V$ by scalar multiplication has image not lying in the center, meaning there is some $\varphi \in \text{Aut}(V)$ which does not commute with it. $\Box$

Example. Let $F = \mathbb{Q}(\sqrt[3]{2})$ acting on itself; abstractly $\sqrt[3]{2}$ acts by some $3 \times 3$ matrix over $\mathbb{Q}$ with eigenvalues $\sqrt[3]{2}, \omega \sqrt[3]{2}, \omega^2 \sqrt[3]{2}$ and the above argument just conjugates it to some other such matrix in $M_3(\mathbb{Q})$.