Is every axiom in the definition of a vector space necessary?
Solution 1:
I think they are redundant after all! Here's a proof that axiom 1 is redundant. Let $a,b\in V$, and consider $(1+1)\cdot (a+b)$. By axiom 7 and 8, this is equal to $(a+b)+(a+b)$; on the other hand by axiom 6 it is $(1+1)\cdot a + (1+1)\cdot b$, or $(a+a)+(b+b)$ by axiom 7 and 8. We can then use axioms 2, 3, 4 to show that \begin{align*} a^{-1} + (a+b) + (a+b) + b^{-1} &= a^{-1} + (a+a) + (b+b) + b^{-1}\\ b + a &= a + b \end{align*} and $V$ is Abelian.
Necessity of some of the other axioms:
4: Take $V=[0,\infty)$ under multiplication, and $K=\mathbb{R}$, with $z\cdot x \mapsto \begin{cases} x^z, & x\neq 0\\0, &x=0.\end{cases}$
5: Consider $K=\mathbb{C}$, $V=\mathbb{R}$ with $z\cdot x = \Re(z)x$.
6: Necessary once you toss out commutativity. Take $K=F_3$, and $V$ the Heisenberg group over $F_3$, with $z\cdot x = x^z$. Since all elements of $V$ have order dividing 3, axiom 7 is satisfied, but $$\left(\left[\begin{array}{ccc}1 & 0 &0\\0 & 1 & 1\\0 & 0& 1\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 0\\0 & 1 & 0\\0 & 0 &1\end{array}\right]\right)^2 \neq \left[\begin{array}{ccc}1 & 0 &0\\0 & 1 & 1\\0 & 0& 1\end{array}\right]^2\left[\begin{array}{ccc}1 & 1 & 0\\0 & 1 & 0\\0 & 0 &1\end{array}\right]^2.$$
7: Take $K=\mathbb{C}$, $V=\mathbb{R}$, and $z\cdot x = |z|x$.
8: See comment by Jyrki below.
Solution 2:
Selected progress on the definition of Vector Space:
At 1971, Bryant proved that the commutativity of $\oplus$ can be deduced by other axioms$^{(1)}$.
At 1973, Rigby and Wiegold proved that only 6 axioms are needed$^{(2)}$.
In fact, the set of axioms of a vector space may reduce to only 6 as described:
Definition. A vector space over a field $K$ consists of a set $V$ and two binary operations $\oplus: V\times V→V$ and $\odot: K×V→V$ satisfying the following axioms:
$(a\oplus b)\oplus c=a\oplus (b\oplus c), \forall a,b,c\in V$,
$\lambda\odot(a\oplus b)=(\lambda \odot a)\oplus(\lambda \odot b),\forall a,b\in V, \forall \lambda\in K,$
$(\lambda +\mu)\odot a=(\lambda \odot a)\oplus(\mu \odot a),\forall a\in V,\forall\lambda,\mu\in K,$
$(\lambda\times\mu)\odot a=\lambda\odot (\mu \odot a), \forall a\in V,\forall \lambda,\mu\in K,$
$0\odot a=0\odot b,\forall a,b\in V,$
$1\odot a=a, \forall a \in V.$
We have the follow theorems, in which, the Theorem 1. implies the existence of additive inverse and Theorem 2. implies the commutativity axiom.
Theorem 1. If $V$ satisfies axioms, $1, 3, 5, 6,$ then it have additive inverse.
Hints:$$ a =1\odot a=(1+0)\odot a=(1\odot a)\oplus(0\odot a)=a\oplus z$$ $$z=0\odot a=(1+(-1))\odot a=(1\odot a)\oplus(-1)a=a\oplus((-1)\odot a)$$
Denote the element $(-1)\odot a$ as the additive inverse of $a$.
Theorem 2. If $V$ satisfies axioms, $1,2,3,5,6,$ then it is a commutative additive under $\oplus$.
On counterexamples
The example corresponding to additive inverse axiom described in the answer (@user7530) doesn't satisfy the axioms $2, 3$, too. According to Theorem 1., it certainly failed for additive inverse axiom. Therefore, this example cannot prove the necessary of this axiom.
Last but not least
An important rule that intrinsically holds is VERY often to overlook! I call it Rule 0. as follows:
Rule 0. $\lambda\odot a\in V, \forall a\in V, \forall \lambda\in K. $ (Closure under scalar multiplication)
Consider the following counterexample for being a vector space:
$V=\{(a_1,...,a_n):a_i\in \mathbb{R}, i=1,...,n\}, K=\mathbb{C}$ with the operations of coordinatewise addition and multiplication.
It seems satisfy all the axioms listed above. But pay attention to it that $\exists \lambda\in K,$ s.t. $(\lambda\odot a)\notin V$. Now that so, how can we apply axioms $2,3,$ or $4$?
Reference
(1) Bryant, V. (1971). Reducing Classical Axioms. The Mathematical Gazette, 55(391), 38-40. doi:10.2307/3613304
(2) Rigby, J., & Wiegold, J. (1973). Independent Axioms for Vector Spaces. The Mathematical Gazette, 57(399), 56-62. doi:10.2307/3615171
Solution 3:
You won't be able to entirely remove axiom 3, since otherwise $V=\emptyset$ would (vacuously) satisfy the other axioms. However, you can remove axiom 4 if you replace axiom 3 with this slightly stronger version (which I will call axiom 3*):
(Axiom 3*) There exists an element $\mathbb{0'} \in V$ such that for all $x \in V$, $0_K \cdot x = \mathbb{0'}$.
(Here, I use the notation $\mathbb{0'}$ to denote that this is a nonstandard definition of $\mathbb{0}$).
Axiom 3* implies axiom 3 and 4
That this element is an additive identity follows from axioms 6 and 8: we have $$\mathbb{0'} + x = 0_k \cdot x + 1_k \cdot x = (0_k + 1_k)\cdot x = 1_k \cdot x = x.$$
Also, for every $x \in V$, we have $$x + (-1_K)\cdot x = (1_K)\cdot x + (-1_k)\cdot x = (1_K + -1_K)\cdot x = 0_k \cdot x = \mathbb{0'}$$ so each $x \in V$ has as an inverse $(-1_K)\cdot x$.
Axioms 3 and 4 imply axiom 3*
We have that
$$(0_K)\cdot x + x = (0_k + 1_K) \cdot x = x$$
so, denoting the inverse of $x$ by $-x$,
$$(0_K)\cdot x + x + (-x) = x+(-x)$$ $$(0_K)\cdot x + \mathbb{0} = \mathbb{0}$$ $$(0_k) \cdot x = \mathbb{0}$$
Solution 4:
I believe $6$ is indeed indispensible, and may be equivalent to $1$, here's my reasoning:
what $6$ actually says is we have an action of $(F,+)$ upon $(V,+)$, that is, the map $v \mapsto a\cdot v$ (let's call this map $\phi_a$) induces a group homomorphism (the operation being $+$):
$F \to V$ via $a \mapsto \phi_a(v)$ for any fixed $v \in V$.
Indeed, we can relax the vector space axioms to allow $R$ to be a commutative ring with unity, and obtain an $R$-bimodule. Now there is a unique homomorphism $\psi:\Bbb Z \to R$ sending $1 \mapsto 1_R$, and this allows us to define on any $R$-bimodule $M$, a $\Bbb Z$-action by:
$n\cdot m = \psi(n)\cdot m$.
Now the intuitive way to try to impose a $\Bbb Z$-action on a group $G$, is to try to set:
$n\cdot g = g^n$.
However, $g \mapsto g^n$ is an element of $\text{End}(G)$ for all $n \in \Bbb Z$ if and only if $G$ is abelian (the "if" part is obvious, the "only if" can be proved using $n = 2$, which is essentially user7530's argument).