# A unitarity axiom in the vector space axioms

Consider the standard unitarity axiom $1\cdot \boldsymbol x=\boldsymbol x$ in the set of axioms for vector space $V\ni \boldsymbol x$ over a field. Let the latter be denoted by $F$ and its elements are $\alpha,\beta$ etc. Look at another axiom of $V$. Namely, $$\alpha\cdot(\beta\cdot\boldsymbol x)=(\alpha\times\beta)\cdot\boldsymbol x.$$ Plug-in here the unity $\beta=1$ of the field $F$. One gets $$\alpha\cdot(1\cdot\boldsymbol x)=(\alpha\times1)\cdot\boldsymbol x$$ and, hence, $$\alpha\cdot(1\cdot\boldsymbol x)=\alpha\cdot\boldsymbol x.$$ May I do cancel the left multiplier $\alpha$ here and get $$1\cdot\boldsymbol x=\boldsymbol x\quad?\qquad\qquad(*)$$ This looks plausible since this $\alpha$ and $\boldsymbol x$ are free elements of their domains. But $(*)$ is an independent axiom as noted in the first sentence of the post. Is (*) an axiom or theorem? Where is a contradiction?

Canceling $\alpha$ in $\alpha\cdot(1\cdot\boldsymbol x)=\alpha\cdot\boldsymbol x$ is actually multiplying both sides by $\alpha^{-1}$: $$\alpha^{-1}(\alpha\cdot(1\cdot\boldsymbol x))=\alpha^{-1}(\alpha\cdot\boldsymbol x)$$ Now you need to use associativity of scalar multiplication: $$(\alpha^{-1}\alpha)\cdot(1\cdot\boldsymbol x)=(\alpha^{-1}\alpha)\cdot\boldsymbol x$$ and then $$1\cdot(1\cdot\boldsymbol x)= 1\cdot\boldsymbol x$$ which begs the question.
Perhaps it is easier to see this by trying to prove that $\alpha\cdot \boldsymbol x=\alpha\cdot\boldsymbol y$ implies $\boldsymbol x=\boldsymbol y$. You'll end up with $1\cdot\boldsymbol x=1\cdot\boldsymbol y$.