Logical dependence between vector space axioms

Question

My question is not very long, but I'd like to explain where it comes from. Consider the classical definition of vector spaces:

$E$ is said to be a vector space over a field $F$ when:

A) E is a commutative group.
B) There is an scalar multiplication satisfying $1.x=x$ and $(\alpha\beta).x=\alpha.(\beta.x)$
C)
1. $\alpha.(x+y)= \alpha.x + \alpha.y$ (scalar multiplication distributivity)
2. $(\alpha+\beta).x= \alpha.x + \beta.x $ (vector multiplication distributivity)

This is the definition that can be found in Wikipedia, or in Halmos' Finite Dimensional Vector Spaces. While reading the latter, I was rather surprised by the following disclaimer:

"These axioms are not claimed to be logically independent; they are merely a convenient characterization of the objects we wish to study."

Some rather trivial examples (setting stuff like $\alpha.x= \alpha^{2}x$) show that:

A) and B) are not logically connected
Neither C)1. or C)2. are implied by A) together with B)
A), B), C)1. together do not imply C)2.

Now the tricky question is: do A), B), C)2. together imply C).1 ??

If $F=\mathbb{Q}$ the answer is yes. In a nutshell this is because $n.x = x+x+...x$ so that indeed $n.(x+y)=n.x +n.y$, and \begin{equation*} x+y=m.(m^{-1}.x+m^{-1}.y)) \end{equation*} so that $m^{-1}.(x+y)=m^{-1}.x +m^{-1}.y$. Putting both together with scalar associativity gives C)1.

What happens for general $F$ ? I have not managed to find a counter example, and have no idea how to extend the proof above.

Thanks for your help

I think you'll have trouble showing it for irrational scalars — David P, Sep 06 '14 at 08:40

score 1 · Accepted Answer · answered Sep 07 '14 at 12:26

1

The axioms are not logically independent, in the sense that you do not need to assume that $E$ is commutative; you can prove that from the other axioms.

answered Sep 07 '14 at 12:26

Gerry Myerson

179,216

But this has come up before, e.g., http://math.stackexchange.com/questions/840637/proving-commutativity-of-addition-for-vector-spaces – Gerry Myerson Sep 07 '14 at 12:29
Ok thanks a lot ! – Sergio Sep 08 '14 at 09:36

score 0 · Answer 2 · answered Sep 07 '14 at 12:10

Ok I'm going to answer my own question, in case someone has interest in the answer.

Take $F= \mathbb{Q}+\sqrt{2}\mathbb{Q}$, $E=\mathbb{R}^{2}$. Let $\alpha.x$ be $\alpha x$ except if $x=(a,0)$ in which case put $q+p\sqrt{2}.x=qx-p\sqrt{2}x$. All conditions except C).1) are satisfied, and C).1) does not hold with, for example, $x=(1,0), y=(0,1)$.

Logical dependence between vector space axioms

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