TITLE: Non-distributive fields?
QUESTION [10 upvotes]: 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.

REPLY [14 votes]: There are several families of algebraic structures that are "almost" fields.

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.
Near fields satisfy all of the conditions of a skew field, except that we require only one of the distributive laws.
Semifields satisfy all of the conditions of a skew field, except for associativity of multiplication.
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 $-a$.)