# 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

## 2 Answers

These axioms do not require the existence of the opposite (for addition). The set of the not negative rational numbers with the usual operations satisfies these axioms and is not a field.

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.

• The nonnegative integers do not have inverses as required here. – Tobias Kildetoft Mar 20 '17 at 13:04
• @TobiasKildetoft: Oops, sorry... I still need to wake up fully. :-) Of course, they don't. But reals or rationals would do. Thanks! – zipirovich Mar 20 '17 at 13:06