How many scalar multiplications are possible on a $\mathbb{Q}$-Vector Spaces?

Question

Let $V$ be a $\mathbb{Q}$ Vector-Space. Then from the definition of Vector-Spaces, we have $1.x=x$ for all $x \in V$. From this one can show $n.x=x+x+...+x(\text{n times})$ for all $n \in \mathbb{Z}$.

So any scalar multiplication defined on a $\mathbb{Q}$-Vector Space, when restricted to $\mathbb{Z}$ becomes identical. Now the question is, is there only one possible way to define scalar multiplication on a $\mathbb{Q}$-Vector Space?

This can be generalized to ask

How many scalar multiplications are possible on $\mathbb{R}$-Vector Space and $\mathbb{C}$-vector space?

Answer 1

Hint: If $q = m/n \in \mathbb Q$, then $n\cdot(q\cdot v)=m\cdot v$.

For the general case, if $V$ is a vector space over a field $F$ and $\phi$ is an automorphism of $F$, then $\lambda \circ v := \phi(\lambda) v$ defines a scalar multiplication on $V$. If $\phi$ is not the identity, then this scalar multiplication is not the original one.

The argument for $\mathbb Q$ boils down to the fact that $\mathbb Q$ has only one automorphism. Nor does $\mathbb R$ (see here). On the other hand, complex conjugation is a nontrivial automorphism of $\mathbb C$. This gives an indication of how to handle the general case.

Answer 2

$\newcommand{\Z}{\mathbb Z}$ $\newcommand{\Q}{\mathbb Q}$ $\newcommand{\R}{\mathbb R}$ $\newcommand{\C}{\mathbb C}$
Yes, the scalar multiplication on a $\Q$-vector space is unique.
We only need to show that for every fixed non-zero vector $v\in V$, the scalar multiplication on the subspace $\Q v$ is unique. And, $\Q v$ is isomorphic to $\Q$ as additional groups, so we only need to prove the statement for $V=\Q$ and $v$ is a rational number.
Let $f:\Q\to \Q$ be a function such that $f(q)=q*v$ for every $q\in\Q$. ($xv$ is the common multiplication on $\Q$ and $x*v$ is the scalar multiplication on the vector space.) Then, $f(1)=v$ and $f(kq)=k*f(q)=kf(q)$ for $k\in\Z$. So $f(k)=kv$ for integer $k$. Therefore, if $q=\dfrac mn\in\Q$, then $nf(q)=n*f(q)=f(m)=mv$, so $f(q)=\dfrac{mv}{n}$ is uniquely determined.