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
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$(V,+, \vec{0})$ is a commutative semigroup.
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$(a+b) \cdot \vec{v} = a \cdot \vec{v} + b \cdot \vec{v}$ and $a \cdot (\vec{v}+\vec{w}) = a \cdot \vec{v} + a \cdot \vec{w}$
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$a \cdot (b \cdot \vec{v}) = (a \times b) \cdot \vec{v}$.
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.
Best Answer
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
and
satisfies every axiom in the question, and there is nothing that forces $\vec{z}$ to be equal to $\vec{0}$.