I am thinking about construction of a structure that "obeys" all of the axioms for a field except an axiom of distributivity of multiplication over addition, and I am not sure is that possible at all?

I mean, it ought to be possible, because, it seems that an axiom of distributivity is not necessarily redundant in a sense that it is implied by other axioms of a field.

But I am really really not sure.

Do you have somewhere already an example of such a structure, it could even be some familiar field with addition and multiplication specially defined to suit the purpose.


Without the distributive axiom, there is no connection between the addition and multiplication operations at all (other than that they happen to be defined on the same set). So, you could take any set $S$ and consider any abelian group structure on $S$ at all as addition and any abelian group structure on $S\setminus\{0\}$ at all as multiplication (and $0$ times anything is $0$). Or, the multiplicative group structure could include $0$, with $0$ having an inverse, as long as $0$ is not the multiplicative identity.

For an explicit example, for instance, let $S=\{0,1\}$, with addition being the usual addition mod $2$, and multiplication being addition mod $2$ with the roles of $0$ and $1$ swapped (so $0\cdot 0=1$, $0\cdot 1=0$, $1\cdot 1=1$). This then satisfies all the field axioms except distributivity.

For another example, you could let $S$ be a set with $6$ elements, with addition having the structure of $\mathbb{Z}/6$ and multiplication of nonzero elements having the structure of $\mathbb{Z}/5$ (and $0$ times anything is $0$). This cannot be a field since there is no field with $6$ elements, but it satisfies all the axioms except distributivity.


There are several families of algebraic structures that are "almost" fields.

  1. Skew fields (aka division rings) satisfy all of the conditions of a field, except that the multiplication is not required to be commutative. If we have a finite structure, there are no skew fields that are not in fact fields.
  2. Near fields satisfy all of the conditions of a skew field, except that we require only one of the distributive laws.
  3. Semifields satisfy all of the conditions of a skew field, except for associativity of multiplication.
  4. Quasifields (as mentioned by Berci) are essentially the weakest of the batch: they are not required to have associativity of multiplication, and are only required to have one of the distributive laws.

These are all interesting to consider as finite structures, because we do have nontrivial examples of finite near/semi/quasifields. There are even finite semifields with commutative multiplication that are not examples of finite fields.

The smallest example of a near field has order 9, and is due to Hall. We take an irreducible polynomial of $\mathbb{F}_{3}$, say $f(x) = x^{2}+x+2$. You use this to define a new multiplication on $\mathbb{F}_{3}^{2}$ by $$(x_{1},y_{1})\cdot(x_{2},0) = (x_{1}x_{2}, y_{1}x_{2}),$$ and $$(x_{1},y_{1})\cdot(x_{2},y_{2}) = (x_{1}x_{2}-y_{1}y_{2}^{-1}f(x_{2}), x_{1}y_{2}-y_{1}x_{2}+2y_{1}).$$ (The 2 as coefficient in the last term comes from the polynomial $f(x)$, if $f(x) = x^{2}+ax+b$, that coefficient is $-a$.)

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    $\begingroup$ Note that the multiplicative group of the given nearfield of 9 elements is the quaternion group, and we can represent it by additive basis $1,i$ over $\Bbb F_3$ and setting $j=i-1$ and $k=i+1$. $\endgroup$ – Berci Jan 11 '20 at 14:01

Quasifields are not commutative and they obey distributivity only on one side. They arise in the study of (finite) projective planes.

In fact, there are many irregular projective planes that induce irregular field-like structure.
Briefly, remove a line from the given projective plane and regard it the 'line at infinity', then fix a line, on which easy geometric constructions - using parallels - lead to an 'addition' and a 'multiplication', which in general may satisfy only some relaxed versions of a few field axioms.


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