4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ The linked question answers it in case $F$ has a nontrivial automorphism. Thus it remains to see what happens if $F$ is not prime but has trivial automorphism group. $\endgroup$ Aug 30, 2020 at 15:37
  • $\begingroup$ Hence it would suffice to show that every pair of scalar multiplications induces a non-trivial automorphism $F \to F$, then take an extension with a trivial automorphism group to get a counterexample. To that end it would be useful to show that the for any fixed set, the span under any given scalar multiplication is the same. I'm not sure if this is true, though it is interesting in its own right. $\endgroup$ Aug 30, 2020 at 15:58

1 Answer 1

3
$\begingroup$

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 $M_3(\mathbb{Q})$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.