I'm writing some notes on Linear Algebra and am trying to show the independence of an "optimal list" for vector space axioms. I am using the following definition: A vector space is a structure $(V,0,+,\cdot)$, where $V$ is a set, $0$ is an element of $V$ and $+\colon V\times V\to V$ and $\cdot\colon\mathbb{R}\times V\to\mathbb{V}$ are binary operations such that
- $v+(u+w)=(v+u)+w$
- $v+0=v$
- For all $v$ there exists $w$ such that $v+w=0$.
- $\alpha(\beta v)=(\alpha\beta)v$
- $\alpha(v+w)=\alpha v+\alpha w$.
- $(\alpha+\beta)v=\alpha v+\beta v$
- $1v=v$
Question: Is there a simple example of a structure which satisfies all axioms except only 4?
A few remarks:
- I used just the "right-identity/right-inverses" version of the group axioms in 1.-3., which is well-known to be sufficient for having a group (see here)
- I added "$0$" as a constant because axiom 3. depends on the given $0$ which satisfies 2. This is a way to avoid the usual ambiguities with the way axiom 3. is commonly stated (see this question and the accepted answer)
- It is well known that commutativity is implied by the other axioms (ignoring 4., even), so I ommited it.
- I am considering only real vector spaces. (So the example in this answer, for example, does not apply)
For each axiom in the list except 5., I have already managed to find a structure which satisfies all other axioms except it. However, my example for axiom 4. is quite complicated: Let $V=\mathbb{R}$ as a $\mathbb{Q}$-vector space, and $L_{\mathbb{Q}}(\mathbb{R})$ be the $\mathbb{Q}$-vector space of $\mathbb{Q}$-endomorphisms of $\mathbb{R}$.
Take any $\mathbb{Q}$-linear map $\Phi\colon\mathbb{R}\to L_\mathbb{Q}(\mathbb{R})$ which takes $1$ to the identity and $\sqrt{2}$ to the zero map (which can be done by difining $\Phi$ on a basis of $\mathbb{R}$ over $\mathbb{Q}$ which contains $1$ and $\sqrt{2}$), and let $\lambda x = \Phi(\lambda)(x)$ for $\lambda\in\mathbb{R}$ and $x\in V$.
Then axioms 1.-3. are immediate for $V$; axiom 5. follows from $\Phi(\alpha)$ being $\mathbb{Q}$-linear (additive) for each $\alpha$, 6. follows from $\Phi$ itself being $\mathbb{Q}$-linear, and 7. from $\Phi(1)$ being the identity. Axiom 4. is not true: Taking $\alpha=\beta=\sqrt{2}$ and $v=1$, we have $\alpha(\beta v)=0$ but $(\alpha\beta)v=2v=2$.
However, this example depends on the axiom of choice (for having a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space) and some linear algebraic facts as well. If we allowed for vector spaces over other fields, then this would give us an example which does not depend on choice simply by taking $\mathbb{Q}[\sqrt{2}]$ instead of $\mathbb{R}$, but this is not what I want.
Is there any other simpler example of a structure which satisfies all axioms except 4.? The distributivity axioms actually allow us to prove that such a given structure will be a divisible group, and that in fact the product will be associative for integers (although not necessarily for rationals). It's also not clear to me whether the subjacent group will be torsion-free.
