Friedberg's definition of a Vector space I'm reading Linear Algebra by Friedberg, Insel, and Spence. They define a vector space in chapter 1. Among the points they use to characterize vector spaces is

$a(bx)=b(ax)$ [where $a$ and $b$ are arbitrary real scalars and $x$ is in the vector space]

How does this impose a constraint upon $x$? Is it not a trivial consequence of commutativity of real multiplication? I would really appreciate an example of a collection of entities that satisfies all vector space properties except this one.
 A: Let's put aside the debate in the comments about a probable typo in your PDF. For a counter-example, take $F = \Bbb{Q}[X]/(f)$, where $f$ is an irreducible polynomial of degree $4$, so that $F$ is a (commutative) field that is also a $4$-dimensional vector space over $\Bbb{Q}$. Take $V$ to be the ring $M_2(\Bbb{Q})$ of $2\times 2$-matrices with entries in $\Bbb{Q}$ and pick a vector space isomorphism $\iota : F \to V$ that maps $1$ to $\pmatrix{1 & 0 \\ 0 & 1}$. Then the operation $(a, x) \mapsto \iota(a) x : F \times V \to V$, satisfies all the vector space axioms except the axiom $a(bx) = (ab)x$ (and also does not satisfy $a(bx) = b(ax)$).
Motivation: a vector space is essentially a homomorphism of rings from a field $F$ to the automorphism ring $\mathrm{Aut}(V, V)$ of an abelian group $V$. The axiom $a(bx) = (ab)x$ is the one that requires that homomorphism to preserve multiplication. We can thwart it by identifying $F$ with $\mathrm{Aut}(V, V)$ via a mapping that preserves the vector space structures but not multiplication (while still arranging for $1x = x$ to hold).
A: This should be a typo. Let $F=V=\mathbb C$. Define a scalar multiplication $\otimes:F\times V\to V$ as $a\otimes v=\operatorname{Re}(a)v$, the usual multiplication of the real part of $a$ with $v$. Clearly $\otimes$ is distributive, $1\otimes v=v$ and
$$
a\otimes(b\otimes v)=\operatorname{Re}(a)\operatorname{Re}(b)v=\operatorname{Re}(b)\operatorname{Re}(a)v=b\otimes(a\otimes v),\tag{1}
$$
but both sides of $(1)$ are not equal to $(ab)\otimes v=\operatorname{Re}(ab)v$ in general.
