Vector spaces without additive inverses

Question

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.

Answer 1

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 $\vec{0}$.

Answer 2

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