I suspect that your 9 axioms for a set and two operations, which I denote by $+$ (addition) and $\cdot$ (multiplication), to be a field are the following:
- Closure [$a,b \in F \implies a\cdot b, a+b \in F$]
- Associativity [$(a\cdot b) \cdot c = a \cdot (b \cdot c), (a + b) + c = a + (b + c) \quad \forall \;a,b,c \in F$]
- Commutativity [$a \cdot b = b \cdot a, \; a + b = b + a$]
- Additive identity [$\exists 0\in F \;s.t. a + 0 = a \quad \forall a \in F$]
- Multiplicative identity [$\exists 1 \in F \; s.t. a \cdot 1 = a \quad \forall 0 \neq a \in F$]
- Additive inverse [$\forall a \in F, \exists b =: -a \; s.t. a + (-a) = (-a) + a = 0$]
- Multiplicative inverse [$\forall a\neq 0 \in F, \exists b =: \frac{1}{a} \; s.t. a \cdot \frac{1}{a} = \frac{1}{a} \cdot a = 1$]
- Distributivity [$a(b+c) = ab + ac, \; (a+b)c = ac + bc$]
- Non-triviality [$1 \neq 0$]
Or at least the 9 axioms will be more or less equivalent (maybe not including (9.) - that's not so important). The idea is that additive inverses are denoted by negative numbers. Thus the additive inverse of $a$ is $-a$. And we know the field is closed under addition, thus $b + (-a) \in F$ is guaranteed. This is completely independent of your ordering.