Vector Spaces: Redundant Axiom? Question
Why are the axioms for vector space independent?
More precisely $1x=x$ seems redundant...
(I take the axioms from: Wikipedia)
Explanation
One has for zero vector:
$$\lambda0+\lambda0=\lambda(0+0)=\lambda0\implies\lambda0=0$$
And for zero scalar:
$$0x+0x=(0+0)x=0x\implies0x=0$$
In familiar form:
$$\lambda x=0\implies\lambda=0\lor x=0$$
Threrefore one calculates:
$$1(1x+x^{-1})=1(1x)+1(x^{-1})=(11)x+1(x^{-1})=1x+1(x^{-1})=1(x+x^{-1})=10=0$$
Hence for nontrivial field:
$$1\neq0\implies1x+x^{-1}=0\implies1x=x$$
But where is the flaw in that check??
 A: The axioms of the list you cite are not independent, but $1x=x$ is not the problem.
Commutativity of addition follows from the other axioms; let $x,y\in V$ and set
$$
z=(1+1)(x+y)
$$
Then
$$
z=1(x+y)+1(x+y)=x+y+x+y
$$
(parentheses can be omitted because of associativity). On the other hand
$$
z=(1+1)x+(1+1)y=1x+1x+1y+1y=x+x+y+y
$$
Therefore, being $V,+$ a group, we can do
$$
(-x)+x+y+x+y+(-y)=(-x)+x+x+y+y+(-y)
$$
which gives
$$
y+x=x+y
$$

An correct objection would be that, removing the commutativity of addition axiom, only right zero and right opposites are assumed. However a general result about monoids applies.

Let $M$ be a (multiplicative) semigroup, with right identity $e$. If every element has a right $e$-inverse, then every element has a left $e$-inverse and the right identity $e$ is also a left identity.

The assumption is that $ae=a$, for all $a\in M$, and that, for all $a\in M$, there exists $b\in M$ such that $ab=e$.
Let $a\in M$ and $b\in M$ such that $ab=e$. Then $bab=be=b$, so, if $c\in M$ and $bc=e$, we have $babc=bc$, hence $ba=e$. Moreover, $ea=aba=a$. Thus $e$ is also a left identity and $b$ a left inverse of $a$.
A: The axiom system you quote does not have
$$\lambda x=0\implies \lambda=0 \lor x=0 $$
as an axiom.
If we drop the axiom $1v=v$, the following becomes an example of a "vector space" over $\mathbb R$:


*

*$V=(\mathbb R,+)$, $F=(\mathbb R,+,\cdot)$

*for $\lambda \in F$ and $v\in V$ let $\lambda v=0$.


We do not want this to happen.
A: A mistake: You only showed that $\lambda=0$ or $x=0$ $\implies$ $\lambda x=0$.
You did not show the reverse implication: $\lambda x=0 \implies \lambda=0$ or $x=0$. A proof of that implication uses the axiom $1x=x$.
A standard counterexample of a structure that satisfies all the other axioms save $1x=x$ is the following:


*

*$V=\Bbb{F}^2$

*$(x_1,x_2)+(y_1,y_2)=(x_1+y_1,x_2+y_2)$, i.e. the usual componentwise vector addition

*$a*(x_1,x_2)=(ax_1,0)$.


The system $(V,+,*)$ satisfies all the other axioms but $1x=x$. Note that that reverse implication does not hold in this system:
$$1*(0,1)=(0,0)=0_V.$$
A: In order to prove
$$
\lambda x=0\implies \lambda=0\vee x=0\tag1
$$
you need to use the axiom $1x=x$. 
Here is how you prove (1): if $\lambda\neq0$, then $\lambda x=0$ implies
$$
\lambda^{-1}(\lambda x)=\lambda^{-1}(0)
$$
$$
(\lambda^{-1}\lambda)x=
0
$$
$$
1x
=0
$$
$$x=0$$
so $\lambda\neq0$ implies $x=0$, or equivalently, $\lambda=0\vee x=0$.
Thus, your proof of the axiom $1x=x$ being redundant goes in circles.
