# Are those properties sufficient for defining a field?

Consider the following properties:

1. $a(x+y) = ax + ay$
2. $x + y = y + x$
3. $ax = xa$
4. $x + 0 = x$
5. $x \cdot 1 = x$
6. for every $x\ne 0$ there's a $y$ such that $xy=1$.

Are those enough for defining a field?

• As a minimum they would need some quantifiers. But I doubt you can derive the associativity from these. – Tobias Kildetoft Mar 20 '17 at 12:55
• Another example of a structure that satisfies these axioms, but is not a field, is the set $\{0, 1\}$, with the $+$ operator being the "maximum" operator and the $\cdot$ operator being the "minimum" operator. – Tanner Swett Mar 20 '17 at 19:35
• Without $1 \not = 0$ you could have the zero ring with a single element – Henry Mar 20 '17 at 23:06
• You also need $0 \neq 1$. The set with one item: {0} satisfies these axioms, too (with 0=1). – ypercubeᵀᴹ Mar 20 '17 at 23:07

No. You still need additive inverses and the two associative properties. As a quick counterexample, consider the set of all nonnegative integers reals $S=\{x\in\mathbb{R}\mid x\ge0\}$. They satisfy all six properties. But it's not a field because there are no additive inverses.