METHODS OF CONSTRUCTING HYPERFIELDS

In this paper we introduce a class of hyperfields which contains non quotient hyperfields. Thus we give a negative answer to the question of whether every hyperfield K is isomorphic to a quotient of a field K by some subgroup G of its multiplica-


INTRODUCTION
A hyperfield is a triplet (H, +, -) where H is a non-empty set, + is a hyper- composition (i.e. a mapping whose domain is H H and whose range is the power set of H), and an internal composition of H (i.e. for every x,y H, x-y H).These two operations satisfy the following axioms: I. Multiplicative axiom H is an almost-group with regard to the multiplication.An almost-group is a semigroup S which is the union GU{0} where G is a group and 0 a bilaterally absorbing element of S By we shall denote its neutral element and by 0 its ab- sorbing element.

II. Additive axioms
(i) x+y=y+x (ii) (x + y) + z x + (y + z) (iii) For every x H there exists one and only one x' H such that 0 e x + x'.
x' is written -x and called the opposite of x; moreover we shall write x y instead of x + (-y) (iv) z E x + y implies that y e z x III.Distributive  (a) By usual convensions if "-" is an internal composition of a set E and X,Y are subsets of E, X-Y is the set of all x y such that (x,y) XY If "-" is a hypercomposition in E then X Y signifies the union U(x,y xxyX y In both cases x Y and X y will have the same meaning as {x} Y and X {y} re- spectively. (b) By virtue of axioms ll.(iii) and ll.(iv) one has x + 0 x for each x H Indeed; 0 x x so x x + O. Let's suppose that some y different than x be- longs to x + 0. Then 0 x y which is absurd because of ll.(iii).So x + 0 x.Also, because + is a hypercomposition, the distributive axiom gives: (a + b)(c + d) ac + ad + bc + bd. (properties of hyperfields can be found in [3], [5] (c) If we replace the multiplicative axiom I by I'.H is a multiplicative semigroup having a bilaterally absorbing element O.
we obtain a more general structure (H,+,-) which is called hyperring. (see [4], [6]) The concept of the hyperfield has been introduced by M. Krasner in his study [I].
The hyperfield that occurs there was obtained by considering a quotient of the type K where K is a valued field and an equivalence relation defined from the 0 field's valuation.This hyperfield was called the residual of K(mod 2. THE QUOTIENT HYPERFIELD. Let F be a field and G a subgroup of F's multiplicative group.Then the multiplicative classes modulo G in F form a partition of F We shall denote by F F the set of the classes of this partition.A multiplication can be introduced in F by defining the product x y of two classes x,y to be the class which results from their setwise product.It can also be proved (see [2]) that if we add x,y as subsets of F the set that results is a union of classes modulo G This enables us to define the sum x + y of x,y to be the set of all classes z F which are contained in the setwise sum of x,y F endowed with these two operations becomes a hyperfield ([2]) called quotient hyperfield of F by G.
If F is a ring and G a normal subgroup of F (: such that (G,-) is a group, and xG Gx for all x in R then F F will be a hyperring called the quotient hyperring of F by G The question that Krasner sets ( [2])is whether there are other hyperfields be- sides the quotient hyperfields.Indeed it is quite important to find out how rich the class of all hyperfields is, for if there were only quotient hyperfields then a good part of hyperfield theory could have been deduced directly from corresponding results in field theory.
At first one can notice that the residual hyperfields are always quotient hyper- fields Next we shall prove that every subhyperfield of a quotient hyperring is a R quotient of a field by a multiplicative group.Indeed let be a hyperring and F R G a subhyperfield of Then F LEMMA i.If uG is the unit of , then uG is a mu]tiplicative group.
PROOF.Let x uy be an element of uG Since uG Gu we have: xuG (u)uG u(u')G u2G cuG.uG uG So the class xuG is a subset of uG thus xuG uG In the same way we can show that uGx uG and therefore uG is indeed a group.Now if we denote by [[FG]] the subring of R which is generated by the set FG we observe that [ The unit of uG is also the unit of I_FG]] PROOF.Let e be the unit of uG If s [[FG]] then there exist s' [ [FG]] and y G such that s s'(uy) exists an element t' [[FG]] such that tG-t'G uG, and therefore there exist y,7' G such that (t) (t'y') e where e is both the unit of G and the unit

of [[FG]]
Thus yt'y' is the inverse element of t and therefore [[FG]] is a field.
We shall construct a class of hyperfields which contains hyperfields that are not isomorphic to quotient ones.For this purpose we consider a commutative multiplica- tire almost-group (H,.) in which we introduce a hypercomposition + defined as follows: x + y {x,y} if x,y z0 and x y x + x H\{x} if x 0 and x + 0 0 + x x for every x e H Then: PROPOSITION 2. The triplet (H,+,.) is a hyperfield.
PROOF.At first one can notice that the opposite of x is x itself since (x + y) + z (x + x) + z =[ H{x} + z (z + z)U(w + z)= {z} U{w,z} H with w z Now as far as the axiom ll.iv, is concerned we have: (i) if y x then x + y x + x H{x} This means that for every z H\{x} we must have x z + x which is in fact true.
(ii) If y x 0 then x + y {x,y}, thus z x or z y If z x then x z + y {z,y} and if z y then x E z + y Hy} So this axiom is valid as well.
Finally it is straight forward to verify the distributive axiom.
THEOREM.The class of the hyperfields constructed as above contains elements that do not belong to the class of quotient hyperfields.
PROOF.We choose an almost group (H,.) with card H 3 and such that x for every x E H\{0} Next we consider the hyperfield (H, +, .)which is con- structed according to the above method and we suppose that this hyperfield is iso- F F morphic to a quotient hyperfield(..)Then for the following must be valid: ( Because of (i) C must contain all the squares of F. But if F is not of charac- teristic 2 then every element of F can be written as a difference of two squares: ---------) +(-1) This contradicts (iii).
Now if the characteristic of F is 2 then the sum of two squares is always a square; thus G G + G, which contradicts again (iii).So H cannot be isomorphic to any quotient hyperfield.

REMARKS. (a)
This theorem gives something more, that is, H cannot be isomor- phic even to a subhyperfield of a quotient hyperring.
(b) Two other classes of hyperfields completely different can be found in [7].
The problem of those hyperfields' isomorphism to quotient hyperfields is also dis- cussed there but no final answer is given.