Confusion on the scalar multiplication identity of the vector space.

Question

For the axiom of vector space, the book says "There exists $1\in\mathbb F$, such that $1\cdot x=x, \forall x\in V$".

I am confused here. Does "$1$" means the multiplication identity in $\mathbb F$, or it depends on how we define $c\cdot x=?$ in a particular problem? To be more concrete, suppose $V=\{(a,b)|a,b\in\mathbb R \}$, and choose $\mathbb F=\mathbb R$, if we define

$$c\cdot (a,b)=(\frac c2a,b)$$

Surely this example is NOT a vector space, but it is sufficient to clarify the question. Under this definition, the scalar multiplication identity "$1$" is

$$"1"=2$$ Correct?

Answer 1

Just to elaborate slightly.

Let $a$ be the identity of $\Bbb F$ and let $b$ be the scalar-multiplication identity for $V$. Let $z$ denote the zero vector of $V$. As in Noah's post let $\times$ denote field multiplication and $\cdot$ denote scalar multiplication.

Then: $$a\cdot v=b\cdot(a\cdot v)=(b\times a)\cdot v=b\cdot v=v$$Therefore $a\cdot v=v$ for all $v$. Therefore $(a-b)\cdot v=z$ for all $v$. If $a-b\neq 0\in\Bbb F$ then because $\Bbb F$ is a field, there is $c$ with $c\times(a-b)=a$. Hence: $$v=a\cdot v=(c\times(a-b))\cdot v=c\cdot((a-b)\cdot v)=c\cdot z=z$$For all $v\in V$. If $V$ is not trivial i.e. contains a nonzero vector $v$, it follows from the contradiction of "$v=z,\,\forall v$" that no such $c$ can exist - it follows that $a-b$ must be zero.

Therefore, if $V$ is not the zero space, $a=b$ is forced; the scalar multiplication identity and the field multiplication identity must coincide.

Answer 2

In fact, this will be forced by the other vector space axioms. In your example, note that (writing "$\times$" for the scalar-scalar multiplication and "$\cdot$" for the scalar-vector multiplication for clarity) we have $$(2\times 2)\cdot (a,b)\not=2\cdot (2\cdot (a,b)).$$ But one of the vector space axioms is $$(\lambda_1\times \lambda_2)\cdot v=\lambda_1\cdot(\lambda_2\cdot v)$$ for all scalars $\lambda_1,\lambda_2$ and all vectors $v$. So whether or not we explicitly require the "scalar-vector" identity to be the "scalar-scalar" identity, they will turn out to coincide.