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.

  • 1
    $\begingroup$ You usually also assume commutativity of multiplication, which you have left out. $\endgroup$ Sep 4, 2016 at 4:27
  • 2
    $\begingroup$ There exist semifields. $\endgroup$ Sep 4, 2016 at 4:30
  • 3
    $\begingroup$ Multiplication of sedonions is neither commutative nor associative, for example. $\endgroup$
    – hardmath
    Sep 4, 2016 at 4:33
  • 1
    $\begingroup$ "The two linear maps of addition and multiplication"? What linear? $\endgroup$
    – DonAntonio
    Sep 4, 2016 at 8:22

1 Answer 1


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.

  • 1
    $\begingroup$ 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, 2016 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.