TITLE: Vector spaces: Is (the) scalar multiplication unique?
QUESTION [6 upvotes]: Notation
Consider an arbitrary vector space $(V, \oplus, \odot)$ over a field $F$ with

vector addition $\oplus : V \times V \to V$ and
scalar multiplication $\odot : F \times V \to V$,

both satisfying all the axioms defining a vector space.
Background
Let us fix the field $F$ and the set $V$. It is obvious that the vector addition does not have to be unique. Any binary operation $\oplus$ that makes $(V, \oplus)$ an Abelian group, would actually work. But I am not so sure if this is also true for (the) scalar multiplication.
Question
Let us fix the field $F$ and the Abelian group $(V, \oplus)$. Is the action of $F$ on $(V, \oplus)$, i.e., the scalar multiplication $\odot$ satisfying the axioms of a vector space, a unique operation?

REPLY [9 votes]: No. On any complex vector space $(V,+,\cdot)$ you can introduce a new scalar multiplication $*$ given by $z * v = \overline{z} \cdot v$ for all $z \in \mathbb{C}$ and $v \in V$.
More generally: If $(V,+,\cdot)$ is an $F$-vector space and $\phi \colon F \to F$ a field automorphism then $z * v = \phi(z) \cdot v$ defines a new scalar multiplication $*$.
PS: The scalar multiplication is unique if $F$ is a prime field, i.e. if $F = \mathbb{Q}$ or $F = \mathbb{F}_p$ with $p > 0$ prime. This follows because the action of $1 \in F$ is uniquely determined by the axioms of the scalar multiplicaton and each element in these fields is a multiple of $1$ (if $F = \mathbb{F}_p$) or can be written as a quotient of multiples of $1$ (if $F = \mathbb{Q}$).