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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?

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

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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.

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

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