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In a field, there usually are certain axioms that are defined. Specifically, the two linear maps of addition and multiplication are defined. Is it possible to just define a field as having ADDITION associativity, commutativity, distributivity, identity, and inverses, and ONLY multiplication identity and inverses, and be able to prove associativity of multiplication? Meaning, if we have $a,b,c \in \mathbb{F}$:

1) $(a+b) + c = a +(b+c)$

2) $a+b = b+a$

3) $a(b+c) = ab+ac$

4) $a+0=a=0+a$

5) $a+(-a) = 0 = (-a) + a$

6) $a \cdot 1 = a = 1 \cdot a$

7) $aa^{-1} = 1 = a^{-1}a$ if $a \neq 0$.

can we use these to prove: $(ab)c = a(bc)$? Thanks.

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    $\begingroup$ You usually also assume commutativity of multiplication, which you have left out. $\endgroup$ – Eric Wofsey Sep 4 '16 at 4:27
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    $\begingroup$ There exist semifields. $\endgroup$ – Mariano Suárez-Álvarez Sep 4 '16 at 4:30
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    $\begingroup$ Multiplication of sedonions is neither commutative nor associative, for example. $\endgroup$ – hardmath Sep 4 '16 at 4:33
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    $\begingroup$ "The two linear maps of addition and multiplication"? What linear? $\endgroup$ – DonAntonio Sep 4 '16 at 8:22
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No, you cannot prove multiplication is associative from your axioms. A famous counterexample is the octonions, which satisfy all your axioms but are not associative. Briefly, octonions are $8$-tuples of real numbers equipped with coordinatewise addition and a certain complicated multiplication law (similar to quaternions) which is $\mathbb{R}$-bilinear and such that every element has a two-sided inverse, but which is not associative.

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    $\begingroup$ It is sort of implied and not directly stated in the paper, and the algebra to check it might be complicated, but it looks like there is a six-dimensional example of nonassociativity: arxiv.org/abs/math/0411428 . $\endgroup$ – zyx Sep 4 '16 at 5:52

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