Vector spaces without additive inverses I was writing out the axioms of a vector space, in preparation for teaching next week, and I started wondering: Do I actually need to impose that vectors have additive inverses?
To be precise: Let $(F,+,\times,0,1)$ be a field. Let $V$ have a binary operation $+ : V \times V \to V$, another binary operation $\cdot : F \times V \to V$ and an element $\vec{0}$, obeying

*

*$(V,+, \vec{0})$ is a commutative semigroup.


*$(a+b) \cdot \vec{v} = a \cdot \vec{v} + b \cdot \vec{v}$ and $a \cdot (\vec{v}+\vec{w}) = a \cdot \vec{v} + a \cdot \vec{w}$


*$a \cdot (b \cdot \vec{v}) = (a \times b) \cdot \vec{v}$.
Can we deduce that $(-1) \cdot \vec{v}$ is an additive inverse of $\vec{v}$? Of course, we can immediately write $\vec{v}+(-1) \cdot \vec{v} = (1+(-1)) \cdot \vec{v} = 0 \cdot \vec{v}$, so the question is whether we can deduce that $0 \cdot \vec{v} = \vec{0}$ without using that additive inverses exist.
 A: The answer is no, unless I made a mistake. The set $\{\vec{0},\vec{z},\vec{u}\}$ over the field $\mathbb{F}_2 = \{0,1\}$ with the operations




$+$
$\vec{0}$
$\vec{z}$
$\vec{u}$




$\vec{0}$
$\vec{0}$
$\vec{z}$
$\vec{u}$


$\vec{z}$
$\vec{z}$
$\vec{z}$
$\vec{u}$


$\vec{u}$
$\vec{u}$
$\vec{u}$
$\vec{z}$




and




$\cdot$
$\vec{0}$
$\vec{z}$
$\vec{u}$




$0$
$\vec{0}$
$\vec{z}$
$\vec{z}$


$1$
$\vec{0}$
$\vec{z}$
$\vec{u}$




satisfies every axiom in the question, and there is nothing that forces $\vec{z}$ to be equal to $\vec{0}$.
A: In my class, we proved the fact that 0v= 0_v using the fact that an additive inverse exits.
I don't know whether this answers your question, but since you are using this fact to deduce the existence of an additive inverse, I would say this is the reason why you need to have the axiom.
here's the proof:
Let x be a scalar and v a vector in V. Consider the quantity
(x+0)v
Now by the distributive property this is xv+0v
If we do the parentheses first we have xv, and these must be equal.
xv+0v=xv
Now since xv is still a vector, there exists a unique additive inverse denoted -xv. Using the associative and communtative property of +, we have
0v=0_v.  (0_v is the zero vector) for any v in V
