Structure of unital 3-fields

We investigate fields in which addition requires three summands. These ternary fields are shown to be isomorphic to the set of invertible elements in a local ring R having Z/2Z as a residual field. One of the important technical ingredients is to intrinsically characterize the maximal ideal of R. We include a number illustrative examples and prove that the structure of a finite 3-field is not connected to any binary field.


INTRODUCTION
Most of us seem to be biologically biased towards thinking that it always requires two in order to generate a third. In mathematics or physics, however, this idea does not seem to rest on a sound foundation: The theory of symmetric spaces, for example, is nicely described in terms of Lie or Jordan triple systems (CHU [2012], UPMEIER [1985]; see e.g. BOHLE AND WERNER [2015] for a recent development), and in physics, higher Lie algebras have come into focus in NAMBU [1973] (for later development see e.g. KERNER [2000], DE AZCARRAGA AND IZQUIERDO [2010]) and were e.g. applied to the theory of M2-branes in BAGGER AND LAMBERT [2008]. Ternary Hopf algebras were introduced and investigated in DUPLIJ [2001].
In this paper we investigate 3-fields, a structure in which the binary operations of the classical theory are replaced by ternary ones. There is a marked difference between addition and multiplication here. Whereas the multiplicative structure of higher arity in rings seems to easier make contact with binary algebra (see e.g. LEESON AND BUTSON [1980] or ELGENDY AND BREMNER [2012]), ternary addition apparently produces phenomena of a more unusual kind and has been, to the knowledge of the authors, treated less thoroughly. We therefore keep multiplication in the fields binary for the moment and stick to ternary addition. Technically, this fact is hidden behind the expression unital, as in ternary group theory one can very well dispose of a unit, and, even more strikingly, the truly ternary case is characterized by the absence of one.
It turns out that there are a number (actually, one that might turn out to be too large) of interesting examples of 3-fields, finite ones, a certain subset of the 2-adic numbers, a class of finite skew-3-fields, based on the quaternion group, or a number of group 3-algebras which actually turn out to be 3-fields.
Here is what we will do in the following: The first section collects some basic theory (based on the pioneering papers by Dörnte and Post DÖRNTE [1929], POST [1940], section 2 introduces the main technical tool that permits a connection to binary algebra, in the third section we deal with ideals which probably feature the most uncommon definition in this paper, section 4 is brief on 3-vector spaces and 3-algebras, just enough in order to be well equipped for a first attack on the classification of finite 3-fields in the final section. Among other things, we will prove here that the number of elements of a finite 3-field is a power of two, that their structure is governed by certain polynomials, with coefficients from the unit disk B 2 of the 2-adic number field Q 2 and (in the case of a single generator, mapping ∂B 2 into the interior of B 2 . Furthermore, each such field carries a structure totally different from classical fields, because essentially none of the finite ternary unital 3-fields embeds into a binary field, when the latter is supposed to carry its canonical ternary structure.

BASICS AND EXAMPLES
The present topic has some precursors. Besides the ones mentioned in the previous section, so called multioperator linear operations in a vector space were considered in abstract form in the 60's by KUROSH [1969]. Our construction is connected with the notion of (n, m)-rings introduced inČUPONA [1965] and further studied in CELAKOSKI [1977], CROMBEZ AND TIMM [1972]. The Post theorem for (n, m)-rings was formulated in CROMBEZ [1973].
(1.6) Definition 1.1. A set X with two operations ν and µ satisfying distributivity and for which X, ν is a (commutative) ternary group and X, µ is a ternary semigroup is called a (3, 3)ring, or for shortness, a 3-ring.
Definition 1.2. If ternary multiplication µ on R is commutative, i.e. if µ = µ • σ, where σ is any permutation from S 3 , then we call R a commutative 3-ring.
A neutral element for R, ν , also called a ternary zero 0, is defined through for all x ∈ R.
If such an element exists, R is called a 3-ring with zero. Distributivity then leads to the following relation with respect to ring multiplication µ (0, x, y) = µ (x, 0, y) = µ (x, y, 0) = 0. (1.8) Another important notion is the one of a querelement DÖRNTE [1929], denoted byx for the addition ν and byx for the multiplication µ. They are supposed to satisfy (1.10) for all x ∈ R The existence of querelements for all x ∈ R is equivalent to unique solvability (1.6).
(3) Whenever R has both, 1 and 0, then it is a binary ring.
Proof. Define in the first case a • b = µ (a, 1, b), and a + b = ν (a, 0, b) in the second case, and check that all axioms are fulfilled in both cases.
Definition 1.5. We call R a proper (3, 3)-ring, iff none of µ or ν are derived, and R is a proper unital 3-ring, iff its multiplication µ is derived.
In LEESON AND BUTSON [1980], a 3-field was defined as a commutative 3-ring R for which R * , µ is a group. Definition 1.6. A unital 3-field F is called proper, iff µ is nonderived, i.e. iff there is no zero element in F.
In the following, all 3-fields will be proper and unital.
Example 1.7. When equipped with ternary addition and multiplication inherited from the complex number field, the set iR becomes a (2, 3)-field.
Example 1.8. For a class of proper (3, 3)-fields, start with a unital 3-field F. Fix a unital 3-subfield F 1 as well as an element t ∈ F \ F 1 so that t 2 ∈ F 1 . Then tF 1 is a proper (3, 3)-field. Example 1.9. A finite unital 3-field is given by (1.11) The fact that each element has a multiplicative inverse follows from the fact that gcd (a, 2 n ) = 1, for all a ∈ (Z 2 n Z) odd .
Recall that a cancellative and commutative 3-ring R is called a 3-integral domain CROMBEZ AND TIMM [1972].
Example 1.10 (3-field of fractions CROMBEZ AND TIMM [1972]). For any proper 3integral domain the 3-field of fractions is a proper 3-field. For instance, starting with we arrive at the proper 3-field (1.12) Trying to find a completion of Q odd which itself is a proper 3-field one has to avoid a zero element in the process. The easiest way to do this seems to be to exploit the relationship of Q odd with the field of dyadic numbers, Q 2 . Recall the definition of the absolute value |·| 2 . If p q = 2 r p 0 q 0 , where neither of the integers p 0 and q 0 is divisible by 2, we have (1.13) Completion of Q w.r.t. |·| 2 results in the field Q 2 , the elements of which can be formally written as (1.14) and |x| 2 = 2 n0 . Then |x| 2 = 1, iff (1.15) Example 1.11. The set Q odd 2 = Z × 2 = {x ∈ Q 2 | |x| 2 = 1} is a unital 3-field w.r.t multiplication and ternary addition inherited from Q 2 . This field is the completion of Q odd w.r.t. |·| 2 . Note that this 3-field is compact. Furthermore, similar to the binary case, Q odd 2 is an inverse limit Q odd 2 = lim ←− (Z 2 n Z) odd .

PAIRS
Let us introduce one of the most important constructions of this paper, an additive operator pair of a unital 3-ring R.
Definition 2.1. Let R be a unital 3-ring, and for a, b ∈ R denote by q a,b the additive operator pair Note that the above definition yields an equivalence relation on the Cartesian product R × R. This kind of construction has been used in POST [1940] in order to "reduce arity": For any fixed c ∈ R we can use operator pairs to introduce the binary retract addition ⊕ c by Since the choice of c is arbitrary, this construction is not functorial, and we will follow a different path here. First, we convert the set of pairs into a (binary) ring. In order to reduce the technical effort to a minimum we use pairs in their standard forms: For each pair we have q a,b = q a+b−1,1 , and whenever q s,1 = q t,1 then s = t. With this notation we (well-)define binary addition + q and the binary product × q for pairs in the following way (2.5) These operations are well-defined and we furthermore have For the proof it is very convenient to use pairs in their standard forms. We leave the details to the reader.
as well as (2.7) In the same vein, Denote by F 3 the category of unital 3-fields. For each morphism φ : F (1) → F (2) define mappings Qφ : QF (1) → QF (2) and Uφ : UF (1) → UF (2) , by It is easily seen that Qφ and Uφ are (unital) morphisms. It follows that U is a functor between the category F 3 and the category of binary unital rings R 2 .
In the following, we write i F for the embedding F→U (F).

Theorem 2.4. A unital 3-ring R carries a derived structure iff it is a retract of U(R).
Proof. If R is a binary unital ring, the retract mapping is given by Ψ(q a,b ) = a + b on Q(R) and the identity otherwise. Conversely, if such a map exists, we define a binary product on R through a + b = Ψ(q a,b ).
A direct consequence of this is Theorem 2.5 (Universal functor theorem). The functor U is universal in the following sense. Suppose R is a unital binary ring, F is a unital 3-field, and ϕ : F→R is a morphism of unital 3-rings, where R is supposed to carry the derived ternary structure. Then, there exists a morphismφ : U (F) →R of binary rings, such thatφ • i F = ϕ.
We investigate next, how unital 3-fields can be analyzed in terms of binary algebraic structures.
(1) Let R be a (unital) local binary ring with (unique) maximal ideal J so that R J ∼ =Z 2Z. Then R \ J is a unital 3-field.
(2) For any unital 3-field F, there exists a local binary ring R with residual field Z 2Z such that F ∼ =R \ J , where J is the maximal ideal of R and R \ J carries the derived ternary structure inherited from R.
Similarly, the product a 1 a 2 ∈ R \ J , and distributivity is satisfied. It remains to show that each a ∈ R \ J has a multiplicative inverse. Suppose a has no inverse, by Krull's theorem it is contained in a maximal ideal different from J , thus contradicting the locality of R. ( It remains to show that U (F) Q (F) = Z 2Z. Take r ∈ U (F). If r ∈ F, then r + Q (F) = 1 + Q (F), because r +r + 1 = 1, and therefore r ∼ 1. If r ∈ Q (F), then, of course, r + Q (F) = 0 + Q (F), i.e. r ∼ 0. So there are only two equivalence classes and hence U (F) Q (F) = Z 2Z.
Remark 2.7. It is not difficult to see that the functor U actually establishes an equivalence of the categories of unital local rings with residual field Z 2Z and the category of unital 3-fields.
Example 2.8. In the case of Q odd 2 , its local ring is the valuation ring Here is another application: Theorem 2.9. For any unital 3-field F the following are equivalent.
(1) There exists an embedding of F into a binary field K, where the latter is supposed to carry its derived ternary structure.
(3) For each y = 1 the equation has the only solution x = 1.
Proof. It is easily checked that conditions (2) and (3) are equivalent. Suppose that Q(F) is an integral domain, and denote by K its field of quotients. Then is an embedding, and so (a) follows. Conversely, whenever there exists an embedding F → K, Q(F) is injectively mapped into Q(F), which is an integral domain.

IDEALS
Because of the absence of zero in a proper 3-ring, the usual correspondence between ideals and kernels of morphisms is no longer available. Instead, we apply the results of the previous section.
Let us consider a morphism of unital 3-rings φ : R 1 → R 2 . Then ker U (φ) is an ideal of U (R 1 ), and the underlying equivalence relation on R 1 is given by (3.1) Note that ker U (φ) is contained in Q (R 1 ), and so q must be an additive pair. The above is a motivation for Definition 3.1. An ideal for a unital 3-ring R is any (binary) ideal of Q (R). We furthermore denote the quotient an ideal of Q(R) defines on R by R J .
Proposition 3.2. Suppose R and S are unital 3-rings, and φ : R → S is a morphism. Then the quotient R ker U (φ) is a unital 3-ring, and R ker U (φ) ≃ Im φ.
Remark 3.3. We prefer the expression "an ideal for a unital 3-ring" over "an ideal of...", as the former is not a subset of R.
The following theorem is an analogue to the fact that for a binary ring the quotient by an ideal is a field, iff the ideal is maximal.
Theorem 3.4. For a unital 3-ring R and an ideal I, the quotient R I is a unital 3-field, iff for any proper ideal J of U (R) for which J ⊇ I it follows that J ∩ R = ∅.
Proof. Suppose that F = R I is a unital 3-field but J ∩ R = ∅ for some proper ideal J containing I. Let π : U (R) → U (R) I be the quotient map. It then follows that π (J ∩ R) = K, hence π (J ) = U(K), and so J = U(R).
If, on the other hand, for any proper ideal J ⊇ I we have J ∩ R = ∅, we choose r ∈ R I as well as r 0 ∈ R with π (r 0 ) = r. If r were not invertible, then the ideal J 0 generated by r 0 and I would be proper, contain I and intersect R.
Let F be a unital 3-field, then each ideal in U (F) is "evenly maximal", and so for each J of U (F), F J again is a field. This is quite different from the binary case.
Example 3.6. The proper ideals for Q odd = r ∈ Q | ∃p, q ∈ Z odd , r = p q are of the form J n = r ∈ U Q odd | ∃q ∈ Z odd , ∃u ∈ Z, r = 2 n u q = 2 n , n ∈ N. (3.2) Obviously, all the J n are ideals for Q odd . Conversely, let J be an ideal, and n 0 = min n ∈ N | ∃p, q ∈ Z odd , 2 n p q ∈ J . (3.3) Because any r ∈ J is of the form 2 n u q , u ∈ Z, q ∈ Z odd , n ≥ n 0 we must have J n0 ⊇ J .
Fix an element 2 n0 p 0 q 0 ∈ J , p 0 , q 0 ∈ Z odd . Then 2 n0 u q ∈ J for all u ∈ Z, q ∈ Z odd and hence J ⊇ J n0 .
We apply this observation to prime fields. Let us consider a unital 3-field F with unit 1 and define F prim to be the 3-subfield generated by 1, i.e. F prim = 1 .
Definition 3.7. The characteristic of a unital 3-field is χ (F) = F prim .
Theorem 3.8. If F prim is finite, then there is n ∈ N 0 so that F prim ∼ = (Z 2 n Z) odd .
Proof. Define a morphism ψ : Q odd → F prim by ψ (p q) = pq −1 which is welldefined and surjective (since imψ is a 3-subfield containing 1), and so we must prove that Q odd J n = Z odd 2 n , for n ∈ N 0 . Since the case ker ψ = {0} is trivial, we suppose n > 1. Then division with reminder by 2 n yields a morphism Z odd → Z odd 2 n , which extends to the quotient 3-field Q odd . It is easily checked that the kernel of this extension is the ideal J n .

3-VECTOR SPACES AND UNITAL 3-ALGEBRAS
Let us define a ternary analogue of the concept of a vector space.
Definition 4.1. A 3-vector space consists of a commutative 3-group of vectors, V , a unital 3-field F as well as an action of F on V . Furthermore, 1v = v for all v ∈ V , and the (ternary analog of) usual distributivity relations are supposed to hold. Linear mappings between 3-vector spaces are defined in the obvious way.
It is easily seen that for any 3-vector space V over F we have a canonical action of U (F) on U (V ).

Definition 4.2.
A subset E ⊆ V of a 3-vector space over a unital 3-field F is called a generating system, iff any element of V can be represented as n i=1 λ i a i with λ i ∈ U (F), a i ∈ E, and n i=1 λ i ∈ F. A is called a basis, iff this representation is unique. If A is any subset of V we denote by lin A the 3-vector subspace of V generated by A.

Remark 4.3. It is important to observe that any linear combination
Example 4.4. A 3-vector space V over the unital 3-field F is given by It has a basis consisting of elements e i = (δ ij ) j ∈ (F n ) free . Note that F n is a 3-vector space as well, which however does not possess a basis if n is different from 1. If n = 1, F free = F, and any element of F is a basis.
Proposition 4.5. Every 3-vector space over a unital 3-field has a free resolution V free .
Proof. We pick a generating set A = {a 1 , . . . , a n }, and let V free

Corollary 4.6. The number of elements of a 3-vector space over the finite 3-field F, generated by n elements is
For the proof, note that |U (F)| = 2 |F|, using, e.g. the standard form of pairs.
Thus, a free resolution is given by Definition 4.8. Let A be a 3-vector space over the unital 3-field F. We call A a unital (commutative) 3-algebra, iff there exists a binary multiplication (•) on A so that (A, +, •) is a (commutative) unital 3-ring.
In the following, 3-algebras will be mostly commutative.
Example 4.9. Let G be a binary group and F a unital 3-field. The group algebra of G over F is defined by together with the convolution product (φ * ψ) (g) = g1g2=g φ (g 1 ) ψ (g 2 ). It can be shown that φ * ψ ∈ FG. These 3-algebras quite often seem to be 3-fields. For example, in the case where G equals the additive group Z nZ, we have FG = F (n), as defined below.
Example 4.10. We define the Toeplitz field of order n over F, T (n, F), as the set of all matrices Note that the inverse of each t is of the same form, and hence the Toeplitz fields are commutative 3-subfields of the triangular 3-fields from Example 5.16. The number of elements in this field is |T (n, F)| = |F| |U (F)| n−1 = 2 n−1 |F| .
(4.5) Actually, these 3-fields are isomorphic to the members of the next class of examples.
Example 4.11. Fix a unital 3-field F as well as a natural number n, and define it is easily checked that this is a unital 3-algebra, generated by the single polynomial x − 1.
That each element in this 3-algebra has an inverse is a consequence of the isomorphism established below.
In fact, that F (n) ∼ = T (n, F) can be seen as follows. The 3-vector space F (n) has basis and if we consider an element P = f + . . , n − 1 as a linear map on F (n), then its matrix representation w.r.t. E is given by (4.8) Since the product of F (n) turns out to be the matrix product of these matrices, the claim has been proven.
Definition 4.12. Fix a unital 3-field F and let (4.9) We call this space the polynomial algebra in n variables over the 3-field F. Note that this space actually is a unital 3-algebra when we use the usual product of polynomials.
Note that (4.10) and Theorem 4.13 (Universality of polynomial algebras). The polynomial algebra F [x 1 , . . . , x n ] is universal in the class of unital 3-algebras over F, generated by n elements.
The polynomial algebra Q odd [x 1 , . . . , x n ] is universal in the class of all unital 3algebras over any of the prime fields, generated by n elements.
The proof closely follows the standard line of reasoning: If A is an algebra generated by a 1 , . . . , a n , define Ψ : F [x 1 , . . . , x n ] → A by Ψ |α|≤N f α x α1 1 , . . . , x αn n = |α|≤N f α a α1 1 , . . . , a αn n and use section 2 to see that A ∼ = F [x 1 , . . . , x n ] / ker Ψ. The statement about Q odd [x 1 , . . . , x n ] follows by applying the first part of this theorem for the respective prime field and then by combining the quotient mapping with the one from Example 3.6.

FINITE FIELDS
In this final section we collect some results and examples concerning finite unital 3fields. We will mainly be concerned with the case χ (F) = 1 and put F 0 = {1}. As a general idea, the theory looks quite different from the classical (binary) one. One of the major differences lies in the fact that the minimal number of generators may be strictly larger than one.
Our first result still looks familiar, though.
Theorem 5.1 (Cardinality of finite fields). For each finite unital 3-field the number of elements is a power of 2.
Proof. Clearly, each finite unital 3-field F is a 3-vector space over Prim F. By Corollary 4.6, the number of elements in F is 2 n−1 |Prim F| n |ker φ F | = 2 n−1 χ (F) n |ker φ F | . According to Theorem 3.8, χ (F) is a power of 2, which |ker φ F | must divide, and the result follows.
For any polynomial P = ν a ν x ν in Q[x] we let P 2 = max ν |a ν | 2 . Then P 2 ≤ 1 iff P ∈ U(Q odd )[x], and P 2 = 1 iff, moreover, P has at least one coefficient in Q odd . Then, for any P, Q ∈ Q[x], we have P Q 2 = P 2 Q 2 . This follows from he fact that the product of two polynomials in Z[x], having both at least one odd coefficient, possesses itself at least one odd coefficient.
Lemma 5.2. The irreducible polynomials for the ring U(Q odd )[x] are: the constant polynomial 2 and those polynomials P which are irreducible in Q[x], and for which P 2 = 1. Definition 5.3. Let F be any prime field. We call a polynomial P with coefficients from in U(F) completely even iff (1) P is even, and, (2) up to units, P does not admit a factorization in which one of the factors is odd.
Theorem 5.4. Suppose F is a prime field and that P 0 is any polynomial in F[x] even . Then F[x] P 0 is a unital 3-field iff P 0 is completely even.
Proof. Clearly, whenever P 0 has a factorization P 0 = QP , with Q a non-invertible odd polynomial, Q is an ideal larger than P 0 , strictly smaller than F[x] even , and intersecting F[x] odd . For the converse, suppose P 0 is completely even. For the purpose of this proof, we will also assume that P 0 does not contain any invertible odd factor. We first look at the case in which |F| = ∞. For an ideal I ∈ U[x] we write I Q for the ideal I generates in Q[x], i.e. I Q = 2 −n P | P ∈ I, n ∈ N .
(5.1) Since (U[x]P 0 ) Q is the principal ideal generated by P 0 in Q[x], we find for any Ideal I of (and different from) U[x], larger than U[x]P 0 , a factorization P 0 = P Q so that I Q = Q[x]P . Since for no n ∈ N, 2 −n is a factor of P 0 , we actually may suppose that P, Q ∈ U [x] and still have I Q = Q[x]P . It follows that P and Q are even and so Now suppose |F| = 2 n and denote by the canonical quotient map, reducing the coefficients of elements in Q odd [x] to coefficients in (Z 2 n Z) odd . We again will suppose that P 0 does not contain any invertible odd factor. Fix a polynomial P 1 ∈ U[x] even , of the same degree as P 0 , such that π n (P 1 ) = P 0 . Then P 1 has to be completely even as well, and so by the first part, Q odd [x] P 1 is a unital 3-field. The result then follows from the fact that for |F| = 2 n Example 5.5. The polynomial P = x n − 1 is completely even, iff n is a power of 2. In fact, if n = 2 k n 0 , with n 0 > 1 and odd, then where n0 m=1 x n0−m is odd. Corollary 5.6. Each finite, singly generated unital 3-field F is the quotient of a singly generated unital 3-field F with prime field Q odd .
Theorem 5.7. Suppose F is a finite unital 3-field with χ(F) = 1, and denote by F 0 = {1} its prime field. If F is generated by a single element, then it is isomorphic to Proof. By the above result, and since U(F 0 ) = Z 2Z is a (binary) field, there must be a completely even polynomial P in powers of (x − 1) we have that Q is even iff its constant term is even, and it is odd iff its constant term is. Now, a polynomial P for which there are n 0 < . . . < n k with clearly is not completely even, and we are done.
is an epimorphism with kernel (x − 1) n .
We look into some further examples and, especially, determine their automorphism groups. To this end we will use Lemma 5.8. There is a 1-1 correspondence between the automorphisms of F 0 (n) and polynomials P = j f j x j , f j ∈ U(F 0 (n)), f j ∈ F 0 , for which there exists another such polynomial Q with P • Q = x.
Example 5.9. Let us start with a classical field extension. For χ (F) = 1 we formally adjoin a square root of 3 and obtain F √ 3 . More explicitly, (5.9) and it turns out that this 3-field is isomorphic to F (2). Note that this field has a trivial automorphism group. a= x (5.16) This is the dihedral group of order 8.
Note that all these automorphism groups would be the "Galois" groups for the respective extensions of {1}. Also, F (n) is a quotient of F (m) whenever n < m.
Theorem 5.13. A finite unital 3-field F admits an embedding into a binary field K iff F = {1}.
Proof. If χ(F) > 1, then Q(Prim F) is not an integral domain, and so Prim F (and much less F) can be embedded into a binary field. In case χ(F) = 1, the statement follows since Q(F 0 (n)) in this case is not an integral domain, either.
We conclude with some remarks on finite unital 3-fields in characteristic 1. We begin with two examples Fix a unital 3-field F, and put This extension of F is characterized by the fact that it displays the fewest possible relations a field of k generators possibly can have (this will be made precise below). It can be shown that F 0 (n 1 , . . . , n k ) =    1 + (1,...,1)≤α≤(n1−1,...,n k −1) where (1 − x) α = (1 − x 1 ) α1 · · · (1 − x n ) αn . Much more relations are necessary in order to present the Cartesian product F 0 (n 1 ) × . . . × F 0 (n k ). Denote by ξ i the generator of the 3-field F 0 (n i ) and by x i the element of F 0 (n 1 ) × . . . × F 0 (n k ) which has the unit element in each entry except at the place i where it is ξ i . Then for each element (f 1 , . . . , f n ) ∈ F 0 (n 1 ) × . . . × F 0 (n k ) there are ε ij = 0, 1 so that Consequently, Theorem 5.14. Let F be a finite field with χ(F) = 1, generated as a unital 3-field by n elements. Let, as before, (1) There exist natural numbers k 1 , . . . , k n such that F is a quotient of F 0 (k 1 , . . . , k n ). ( where the polynomials P 1 , . . . , P N are neither divisible by an odd polynomial nor by any of the (x k − 1) n k .
Our final examples show that there are finite unital 3-fields which are noncommutative.
Example 5.16 (Triangular 3-fields). Let F be a unital 3-field an put . . . , n} , (5.25) where It is easily seen that, D (n, F) is a noncommutative unital 3-algebra (if n > 1). Note that each A n (f i , b ij ) is invertible, and that its inverse has the form  whereb ij ∈ U (F). (Observe that the standard procedure for the inversion of a matrix yields expressions which are well-defined within F.) So it follows that D (n, F) actually is a noncommutative unital 3-field, which is finite in case F is.
Example 5.17 (Quaternion 3-fields). We start by selecting a unital 3-field F, let where i µ , µ = 0, 1, 2, 3, denotes a set of basis elements. We suppose that the quaternion relations i 2 1 = i 2 2 = i 2 3 = i 1 i 2 i 3 = −1, i 0 = 1 hold and extend this multiplication to the whole of HF. This product is well-defined, since the sum of the coefficients c k ∈ U(F) in which clearly is an element of F. So HF is a unital 3-algebra. In addition, if q = a 0 +a 1 i 1 + a 2 i 2 +a 3 i 3 , we letq = a 0 −a 1 i 1 −a 2 i 2 −a 3 i 3 and observe that qq = (a 2 0 +a 2 1 +a 2 2 +a 2 3 )i 0 ∈ HF. Then, for each element q of HF, an inverse is given by q −1 =q qq = a 0 − a 1 i 1 − a 2 i 2 − a 3 i 3 a 2 0 + a 2 1 + a 2 2 + a 2 3 ∈ HF (5.30) and thus, HF is a non-commutative 3-field.