Related Links (Outside this Site)

(2006-03-16)  
Is a "field" a type of skew field like "tea" is a type of leaf tea ?

At the very least, the analogy  (attributed tobourbakist Roger Godement) is amusing.  It deserves much better consideration thanwhat it receivedwhen somebody  (Robinh)  kindly posted it within theWikipedia articledefining a division ring,  only to see the remarkhastily dismissed as"patentnonsense".

Well, no matter how you slice it, a qualifier which widens the scope of whatever it qualifies results in a confusing expression,unless it's recognized as an idiom... Scientific locutions which go against common usage aren't helpfuland a concerned scientist like Godement (who isn't a native speaker of English) has every right to be disturbed when the lingua franca  of Science is butchered this way.

A number of noted authors have used  (albeit fleetingly) the term skew field as synonymous with theinclusiveconcept of division ring  (commutative or not).

We argue that the term skew field  should only designatea noncommutative  division ring (the only popular example consists of Hamilton's quaternions). To many of us, a division ring  is either a field  or a skew field. However, because this is not universally accepted, it's best to use thelocution  "skew field"  only inspecialcontexts where noncommutativity is otherwise clearly stated...

(2006-03-16)  
Commutativeringsin which all nonzero elements are invertible.

Although it's just based on the ancient rules of ordinary arithmetic,the field  concept emerged as such only whenthe Norwegian Niels Henrik Abel (1802-1829)and the Frenchman Evariste Galois (1811-1832)where led to consider finite fields in their independent investigationsconcerning the impossibility of solving "by radicals" a generalpolynomialof degree greater than 4. This had been  the  central problem of Algebra ever since RenaissanceItalian algebraists gave solutions by radicals for equations of the third and fourth degree.

Fields  became a primary focus of investigation in their own right with thejoint work ofLeopold Kronecker (1823-1891) andRichard Dedekind (1831-1916). Ernst Steinitz  (1871-1928) published an axiomatic definition of fields in 1910:

A field  is a commutative ring  in whichevery element but "zero"  (the neutral element for addition) has a multiplicative inverse (areciprocal). This means that the properties listed below holdfor "addition" and "multiplication", which are otherwise only assumed to be well defined  internal operations (this is to say that a sum or a product of two elements of thefield is also an element of the field).

(*)   Both sides of the distributivity law are shown,so that the table remains correct for a division ring with just the deletion of the (highlighted) entry concerning multiplicativecommutativity.  Two-sided versions of the other multiplicativeproperties can be derived from their one-sided counterparts without  assumming commutativity (seeelsewhere on this site for a proof).

The terms commutative  and distributive (French: commutatif & distributif) were both introduced in a memoir ofJoseph Servois(1768-1847) published in the Annales de Gergonne (5:4, October 1, 1814).

Associativity  was so named by W.R. Hamilton in 1843, shortly after he realized that the multiplication ofoctonions does not have this property...

Commutativity of Addition :

In 1905,Leonard Dickson (1874-1954)pointed out that commutativity of addition need not be postulated ifthe commutativity of multiplication is  (which isn't always so,especially in texts of French origin).  This is an easy theorem whichcan be proved by expanding the equal quantities  (1+x)(1+y) and  (1+y)(1+x)  using the other  field axioms, including  xy = yx.

Addition must be commutative even if multiplication isn't :

Actually, in a unital ring, distributivity by itself impliesthat any associative addition is commutative, even when multiplication is not. Proof:

x + y   =   (-x+x) + x + y + (y+(-y))   =   -x + (1+1) x + (1+1) y + (-y)
=   -x + (1+1) (x+y) + (-y)   =   -x + (x+y) + (x+y) + + (-y)
=   -x + x + y + x + y + (-y)   =   (-x + x) + y + x + (y + (-y))   =   y + x 

At the heart of that proof are the two possible expansions of  (1+1) (x+y) using distributivity on either side  (as both are explicitely allowed by thering axioms which are included in the above field axioms).

(2006-02-06)  
Smallest field containing a given ring A  (withoutzero-divisors).

In a ring,  a zero-divisor  x  is a nonzero element whose product with somenonzero element  y  is zero.  In a subring of a field,  that never happensbecause any nonzero  y  has an inverse in the field. So,  if  xy = 0  ,  then:

x   =   x ( y y^-1 )   =   ( x y ) y^-1   =   0 y^-1   =   0

However,  if a ring A  has no zero-divisors, then we may always find a field K with a subring isomorphic to A.  The smallest such field is called the quotient field  of A  and it can be constructed as follows:

We define an equivalence relation within the Cartesian product A x A*  (i.e,  all ordered pairs ofelements from A  where the second one is nonzero) by stating that  (a,b) and  (c,d)  areequivalent  when:

a d   =  b c

The equivalence-class of  (a,b)  is then called the quotient  ofa  and b. (When all is said and done we'll denote it a/b.)

All such quotients form the ring's quotient field K (also called field of fractions) on which addition and multiplication are induced by the following operations between pairs, which can be shown to respect the above equivalence relation (i.e., the class of the result depends only on the classes of the operands):

(a,b) + (c,d)   =   (ad+bc,bd)
(a,b) (c,d)   =   (ac,bd)

The first key observation is that the resulting pairs are indeed also in A x A* (the second element of either result is never zero because there are no zero-divisors in A).

Next we can show that any element  x  of A  is uniquely associated with the class consisting of allpairs  (bx,b)  where b  is a nonzero element of A. Indeed,  all such pairs are equivalent  and the class associated with  x  can't beassociated with another element  y  of A  (:  otherwise  x-y would be a divisor of zero). It's also easy to verify that this one-to-one mapping is an homomorphism (i.e., it respects both operations).  So,  we may as well identify an element of A with its associated class and consider that A  is just a subring of K  (just like we routinely consider that integers are part of the rational numbers).

Likewise,  (b,b)  is the neutral element for multiplication, which we may call  1  (whether or such a neutral element was already present in A).

The tedious  verification of all field properties is just routine.

Some common examples of quotient fields :

Ring A	Quotient Field K
Integers	Rationals
p-adic Integers  Zp	p-adic Numbers  Qp
(Formal) Polynomials	(Formal) Rational Functions
(Formal)Power Series	(Formal)Laurent Series

Here  (and elsewhere)  the qualifier formal denotes the algebraic definition of an object independently ofwhatever applications it may have.  For example,  a formalpolynomial is nonzero whenever some of its coefficients are nonzero, although its value may be zero everywhere in a finite field. Likewise, formal power series are well-defined irrespective of convergence.

(2006-03-26)  
In the next section, we'll show that it's actually a  (commutative) field.

We call  integral ring (anneau intègre  in French) aring  (commutative or not) where the product of two nonzero elements is never zero. Below is a proof that every nonzero element in a finite integral ring is invertible:

First, we establish the existence of a neutral element 1  (unity)  for multiplication: Consider the successive powers of a nonzero  element  y :

y¹ = y,    y² = yy,    y³ = yyy,    y⁴,    y⁵,     ...

As there are only finitely many possible values, those can't be all distinct... Say the  n+k+1^st  is equal to the n+1^st   (for some k>0). Let's put   u = y^k

x,    ( x u x ) yⁿ⁺¹  =   x y^n+k+1 x yⁿ⁺¹   =   0

As there are no zero-divisor, the bracket must vanish, so x u = x.  Likewise, u x = x. Thus,  u  is neutral for multiplication; the ring is unital  ( 1 = u ).

Now, for any nonzero a, the map which sends  x  to a x  is injective: If two distinct elements  x  and  y  had thesame image, the product a (x-y)  would vanishwithout any factor vanishing, which is ruled out here.

Any injection of a finite  set into itself issurjective  (by the pigeonhole principle ). So,  there's an element a' whose image is 1  (unity) this element is thus the right-inverse of a. The existence of a right-inverse for every nonzero elementsufficesto show that all  of them are invertible.

Every nonzero element of a subring S of a finite division ring has its inverse in S.

Indeed,  we only have to remark that such an S  is a finite integral ring.

(2006-03-18)  
Multiplication in afinite division ringis necessarilycommutative.

In other words, every finite  division ring is a field.

In English at least, "fields" are now officiallyrequired to be commutative,but there's no law against memorizing this surprising result the French way:

Every finite "field" is commutative.
Tout corps fini est commutatif.

The theorem was first published in 1905 by the Scottish mathematician JosephWedderburn (1882-1948). After seeing a proof of the theorem byL.E.Dickson (1874-1954),Wedderburn gave two other proofs in the same year... However, Karen H. Parshall points out (in her 1983 study of the issue)  thatWedderburn's first "proof" had a gap which wentunnoticed at the time.  Although Dickson did acknowledge Wedderburn's priority,he should have been given credit for the first valid proof ofwhat's now universally known as Wedderburn's theorem.

 Let K  be a finite  division ring.

Let C(x)  be thecentralizer  (orcommutant) in K  of a nonzero element  x [ this consists of all the elements  y  of K,including 0, for which   x y = y x ]. It's easy to establish that C(x)  is a subring  of K,whichmeansthat it contains the reciprocals of all its nonzero elements. So is thecenter C of K  (which consists of those elements of K  which commute with every  element of K). Since  C  is commutative,it's a field  (of order q ).

K  and C(x)  arevector spaces over C, whose respective dimensions are  n  and  n(x). K  can also be viewed as a module over C(x).    n(x)  divides  n.

Notice that  n cannot  be equal to 2: Otherwise, all the elements of  K  would be of the form x + y, with  x  and  y  in thecenter C, which would make all of them commute  (thus implying that  n  is  1,  not 2).

Let's apply the conjugacyclass formula to the multiplicative group formed by the  qⁿ-1 nonzero elements of K, whosecenter (C-{0})  is of order  q-1. The order of the conjugacy class  of  x  is theindex of C(x)-{0}  in the wholemultiplicative group, namely (qⁿ-1) / (q^n(x)-1). We may enumerate all the conjugacy classes of noncentral elements  (assuming that there are any)  by letting n_i  be n(x_i )  n  for some member  x_i  of the i^th such class:

qn-1   =   q-1  +			qn  - 1
			qni - 1

To establish that multiplication is commutative (K =C)  we must prove that this relationimplies that  n = 1 (i.e.,  the   on the right must  be empty).

Let's first show that the special cases of Zsigmondy's theorem, stated above,  don't apply: We've already observed that  n  cannot be equal to 2. It's not possible either to have  q = 2  and n = 6,  because the sum   would then be equal to  62 while consisting of multiples of  3  (i.e.,  9, 21 or 63).

Therefore, Zsigmondy's theorem tells us that there's a prime  p  which divides qⁿ-1  but not  q^m-1  for any positivevalue of  m  less than  n  (if there are any). Since such a  p  necessarily divides  q-1  because it divides allother terms in the above equation, we must conclude that  n = 1.  

(2015-12-24)  
A finitealternative algebrawithout zero-divisors is necessarily a field.

This is a generalization ofWedderburn's theorem (1905)which states that a field is neccessarily obtained even whith a multiplicationwhich need not be postulated to be associative, alternativity  is strong enough...

This theorem first appeared in the doctoral dissertation (1930)  of Max Zorn (1906-1993) on alternative algebras. Zorn himself credits the above theorem to his doctoral advisor, Emil Artin (1898-1962).

(2006-03-18)  
The order of a finite field is necessarily a power of a prime number.

Evariste Galois (1811-1832)established the existence of a field of order  q (a finite field with q  elements)  whenever  q  is a power of a prime number.

In 1893,E.H.Moore(1862-1932) proved that all  finite fields are necessarily such Galois Fields. All finite fields of the same order are isomorphic !

The essentially unique finite field of order  q = pⁿ  isdenoted  GF(q)  or F_q 

The prime number  p  is the characteristic  of GF(q).  Any sum of  p  identical terms vanishes in  GF(q).

The additive group of  GF(q) =F_q  is isomorphic tothe direct sum C_pⁿ of  n  cyclic groupsof order  p (the  n  components addindependentlymodulo p).

Multiplicatively,  the  q-1  nonzero elements of F_q  form acyclic group.

In particular, if  q  is prime  (q = p) then F_q  is simply isomorphic tothe field of integers modulo  p.  In other words,   GF(p)  = (/p, + , )

If  n > 1,  the Galois field   GF(q) of order  q = pⁿ  may be constructed explicitely fromthe prime field  GF(p),  by adding formally to it a root  of any polynomial of degree  n which happens to be irreducible in  GF(p).

For example, a construction of  GF(8)  is based on either one of the twoirreducible cubic polynomialof  GF(2) = ({0,1},+,)  namely:

x³ + x² + 1      and      x³ + x + 1

Let's use the latter. A root of that polynomial verifies  x³ = x+1 (an element is its own opposite in a field of"characteristic 2"like this one). We may call such a root  "2"  and call its square  "4", so the rules ofbitwise additioncan be used to name  the other elements of GF(8)  after ordinary integers.

When the order  (q-1)  of the multiplicative group of  GF(q) isn't prime, there's a complication, best illustrated with the constructionof  GF(9) : Three of the  9 monic  quadratic polynomials over  GF(3) are irreducible:

x² + 1 ,    x² + x + 2 ,    x² + 2x + 2

However, a root  g  of the first polynomial onlygenerates a cycleof order 4  (namely, g, -1, -g, 1). What we need is a primitive  element of order 8 whichwould generate the entire multiplicative group of  GF(9). A root of either of the last two polynomialshas this property.  (Such polynomials are thuscalled primitive  polynomials.)

Using the last of the above polynomials (whose roots verify  x² = x+1) we may simply proceed as we did for  GF(8) : We just call the new root "3",  and use ternary digit-wise addition to name other elements after integers:

Abstractly,  GF(q) = GF(pⁿ) may also be defined as the set of solutions,of  x^q = x in the algebraic closure of the prime field  GF(p). For example, the following identity holds in F₉ [x] (theringof the polynomials over F₉  ).

x⁹ - x   =  x (x-1) (x-2) (x-3) (x-4) (x-5) (x-6) (x-7) (x-8)

Therefore, allsymmetric functionsof the nonzero elements of  GF(q) vanish, except their product,  which is  -1. (When  q  is even, -1 = +1.)

The automorphism group of  GF(pⁿ) is the cyclic group of order n generated by the  (standard) Frobenius map. Its other elements are also loosely called Frobenius maps. Each of these sends  x  to the Galois k-th power of  x for some  k  which is a power of  p. (There's only one such automorphism when  n = 1.)

k   =  1 ,   p ,   p² ,   p³  ...   p^n-1

Finite fields are essential in theclassificationof finite simple groups.

(2015-12-22)  
Irreducible and primitive polynomials of a Galois field

The ring  of the polynomial functions  over the field /p  whose degrees are less than  n  form avector space  of dimension  n  isomorphic to the Galois field F_q of  (prime)  characteristic  p  and order  q = pⁿ.

The multiplicative group formed by the nonzero elements of  GF(q)  is acyclicgroup of order  q-1.  As such, it can be generated by a single element and by any power of  g  whose exponent is relatively prime to  q-1. Such generating elements are called primitive. There are  (q-1) primitive  elementsin GF(q)  (where    is Euler'stotient function).

The number of primitive polynomials of degree d in GF(q) is equal to:

(q^d-1) / d

(2022-05-08)  
The structure of a field is the structure of itssymmetries.

(2006-04-05)    F₁ = GF(1)
The field with only one  element:   0 = 1.

Thezeroth power of any prime is 1. Arguably, the simplestGalois field is thus of order 1. Its single element is neutral for both addition and multiplication  (1 = 0)  which cannot  happen in a nontrivial field  (with 2 elements or more). Some textbooks rule out fields with only one element.

A few authors observed that some concepts traditionally studiedon their own can be viewed as the specialization to  q = 1 of structures normally defined over a nontrivial finite field of order  q > 1.  On this subject,Christophe Soulé(2003) quotesJacques Tits  (1957),A. Smirnov  (1992) and Y. Manin  (1995).

Such enlightening specializations aren't obvious. For example, theclassical group SL(n,F) is identified with thesymmetric group S_n  when F = F₁

Incidentally,  if we didn't count the trivial field as a legitimate ring,  thenthe function    which counts the number of finite rings of given order would no longer be a multiplicative function. That would be silly. 

Faire des mathématiques,  c'est 
donnerle même nom à des choses différentes.
Henri Poincaré  (1854-1912)

(2006-03-25)   F[x]
The smallestsubfield orextension ofF where P factors completely.

An extension  of a field F  is a field K  of which F  is a subfield. There's no proper subfield of thesplittingfield of P where P can be completely factored  (into polynomials of degree 1).

(2017-12-06)  
Characteristic is either  0  or  p,  if every element has a p-th root.

A perfect field is either a field of characteristic 0  (like the rationals) or a field of characteristic  p  where all elements are  p-th  powers. (This includes all finite fields and all algebraically-closed fields.)

If the ground field is perfect, Galois theory  is simpler, because every finiteextension of the field is separable (that's called Galois' hypothesis).

Some equivalent characterizations of what makes a field K perfect :

• Everyirreducible polynomial over K  has distinct roots.
• Everyirreducible polynomial over K  is a separable polynomial.
• The characteristic of K  is 0, or it's p>0 and every element is a p-th power.
• Everyalgebraic extension of the field K  is separable.
• Everyfinite extensionof the field K  is separable  (Galois' hypothesis).

(2017-08-01)  
The Laurent series over any  given field of coefficients.

A formal Laurent series  (irrespective of its convergence) is uniquely formed by a formal power series f  and a polynomial  P:

f(x)   +   P(1/x) / x.

The field of Laurent series  is to the ring of formal power series  what the field of p-adic numbers  is to the ring of p-adic integers. In either case,  allowing finitely many negative powers makes every element invertibleand turns a ring into a field.

(2006-03-17)  
A multiplication compatible with bitwise  addition of integers.  (1975)

In the seventh chapter (Chapter 6) of his  1976  masterpiece On Numbers and Games (Academic Press, London, ISBN 0-12-186350-6) John Horton Conway (1937-2020)  shows in what sense bitwise  addition is the simplest "addition" we can endowthe natural integers with.  This operation can be described as binary addition without carry.  It's also known as Nim-sum,  orbitwise  "exclusive or".  Under the latter name, this operation iswidely available at the fundamental level of the assembly languages  ofmodern binary computers  (abbreviated "xor" or "eor").

• The Nim-sum  of distinct powers of  2  is their ordinary sum.
• The Nim-sum  of two equal integers is 0.

Conway then introduces the "simplest" multiplication compatible with this addition. This multiplication can be effectively computed for integers usingfield propertiesand two additional statements which parallel those given above for Nim-addition.

• The Nim-product  of distinct Fermat 2-powers is their ordinary product.
• The Nim-product of two equal Fermat 2-powers is their sesquimultiple.

A Fermat 2-power  is 2 raised to the power of a 2-power ( 2²ⁿ )  namely:
2, 4, 16, 256, 65536, 4294967296, 18446744073709551616, ...A001146
The aforementionedsesquimultiples of thoseare half as much again:
3, 6, 24, 384, 98304, 6442450944, 27670116110564327424, ...

Those two operations give natural integers the structure of afieldofcharacteristic 2,which can be generalized to the entire class of ordinal numbers  (the numbers below a Fermat 2-power form a subfield). Conway calls the whole field On₂  (pronounced "onto").

On₂ 

We may call nimbers  the elements of On₂  especially the finite ones...

Note that the Nim-product of a Fermat 2-power and any  lesser number is the same as their ordinary product (:  The lesser number is a sum of 2-powers,each of which is a product of Fermat 2-powers). Thus, using the fact that the Nim-square of 16 is 16+8,Nim-products of factors up to 255 can be "put together" after 5look-ups of the above table and three 4-bit Nim-additions.  Example:

100.200 = (6.16 + 4)(12.16 + 8)        = (6.12)(16+8) + (6.8 + 4.12) 16 + 4.8   [4 look-ups]        = 9 (16+8) + (7+13) 16 + 11        = (9+7+13) 16 + 9.8 + 11                 [1 look-up ]        = (9+7+13) 16 + (5+11)                   [3 Nim-sums]         = 3.16 + 14 = 62

It takes little more than  5^m steps to Nim-multiply two 2^m-bit integers from scratch using the recursive proceduresuggested by the above example.  Asymptotically, this means that theNim-product of two  n-bit  integers can be computed in time O(n) where  k = lg(5) < 2.322.

Denoting '  an arbitrary ordinalsmaller than  , Conway gives tworemarkable one-line definitions of the Nim-operations  which arevery similar to his other one-liners in the realm ofsurreal numbers (the "" sign is used here as a synonymof the "+" sign for aesthetic reasons, in part to reinforce that similarity).

•  is the least ordinal distinct from all numbers  '  and '.
•  is the least ordinal distinct from all numbers  ''''.

Note that, in an additivegroup,   cannot be equal to either '  or ' unless ' or '. Therefore, the above definition is the "simplest" possible definition of addition in some sense.

Likewise, in a field,   can't be equal to  ''''. Otherwise,   (')(')  would be a zero product of nonzero factors.

These "genetic" definitions are also valid for infinite ordinals (they're equivalent to the above practical rulesfor finite  integers)  and domake On₂   afield.

1/  is recursively defined asthe least nonzero ordinal distinct from all numbers

(1/' )[ 1 + (')(1/)' ]

The Nim-reciprocals of nonzero ordinals are 1, 3, 2, 15, 12, 9, 11, 10, 6, 8, 7, 5, 14, 13, 4,170, 160, 109, 107, 131, 139, 116, 115, 228, 234, 92... A051917. One way to compute the reciprocal of a finite ordinal a  is by iterating the function which sends x  to ax (compare this to the computation of ap-adic reciprocal). Starting from 1, we obtain 1 again after anumber of iterations equal to the bit-length of a,roundedup to a 2-power.  The last step revealsthe reciprocal of a.

In a field of characterisic  2,  like this one, the square  functionis a field homomorphism  (the square of a sum is the sum of the squares) which isinjective (:  If x and y have equal squares,then x+y vanishes). Therefore, it's a bijection within any finite additive subgroup (which is a fancy way to say that a nimber and its Nim-square have the same bit length). This shows that Nim-squaring is a field automorphism among finite nimbers (in fact, it's an automorphism of the whole fieldOn₂ .)

Conversely, anynimber  x  has a uniquesquare-root rim (x)  and the rim function is an automorphism  as well (the square-root of a sum is the sum of the square roots). For finitenimbers,  the rim  function can bedefined recursively :

rim (0) = 0       rim (x) = x +rim ( x + x ²)

That's effectively a recursive definition, because x + x ² has fewer  bits than  x.  (A160679)

An homomorphism f  from the multiplicative subgroup of finite ordinals  onto the unit circle ofthecomplex plane (a representation of U(1), the phase group  ofphysicists) can be defined by any sequence of integers  (u)  satisfyingthe following conditions.  (Using Conway's own convention, weput "ordinary" arithmetic between square brackets, wherever needed.)

u_n  <  [ 2²ⁿ ]         f (u_n)  = exp [ 2i / (2²ⁿ-1) ]         u_n+1^{[ 22n+ 1 ]}  =  u_n

Because x <  [ 2²ⁿ ]  just when x ^{[ 22n 1 ]} = 1, the above inequality is true for  n if  it's true for  n+1 (:  Raise both sides of the last equality to a power). As there's clearly a finite  satisfactory sequence of any length,there must be an infinite  one (among finitely many possibilitiesfor the n-th term, at least one must belong to a satisfactory sequence whose lengthexceeds any given bound).
Actually, for any  satisfactory  u_n there are  2²ⁿ  satisfactory values of  u_n+1

---

With thesurrealtransfinite ordinals it contains, On₂  is algebraically complete. Here's just one of many mind-boggling results about the least infinite ordinal  :

³ = 2

(2013-07-22)  
The On₃ field   (Michon, 2013)

Think how far the reasonable person would go, and then go a step further.John H. Conway (introduction to the Atlas of Finite Groups, 1985) 

On₂  was the name given by Conway tothe "curious field" discussed in theprevious article, with infinitely many nested finite subfields (Galois fields)  of characteristic 2. Here, we introduce a counterpart of characteristic 3. (Other prime characteristics are discussed inthenext section.)

Preliminaries :

Consider the Cayley-Dickson construct presentedelsewhere on this site,in the special case of algebras over the field of real numbers.

So to speak, that construct yields the square  ofanalgebraby doubling its number of dimensions  (e.g., complex numbers areobtained from real numbers, quaternions from complex numbers, and so forth).

The original algebra must be endowed a priori with a conjugation  unary operator,traditionally denoted by a postfixed star  (the conjugate of x is x*) having the following axiomatic properties:

• Conjugation is an additive isomorphism:   (x+y)*   =   x* + y*
• It's an involution  (i.e., a bijection equal to its inverse):  (x*)*  =  x
• It endows multiplication with Hermitian symmetry:  (x y)*   =   y* x*

Because a finite  division algebra iscommutative, it's a field andconjugation is a field automorphism. Any such automorphisms must beFroebenius maps,which is to sau that there is an integer k such that:

x ,    x*   =   x^pk

The last two axioms imply that   1*   =   1   because:

1   =   (1*)*   =   (1 1*)*   =   1** 1*  =   1 1*   =   1*

The squared algebra consists of ordered pairs of elements from the original algebra,endowed with addition and multiplication defined as follows:

(a ,b )  +  (c ,d )	=	(a +c  , b +d )
(a ,b )   (c ,d )	=	(ac   db*  , a*d  + cb )

From the fact that   1* = 1   it follows that  (1,0)  is neutral for multiplication. By equating  (x,0)  with  x,  the original algebra is considered tobe included in its Cayley-Dicskon square. Another key remark is that:

(a ,b ) (a* , -b )   =  (a a* +b b* , 0 )

Now, by definition, a division algebra is analgebrawhere every nonzero element has a multiplicative inverse. (If such an algebra is associative and commutative it's afield. If it's only associative, it's askew-field.) Our last relation shows that the square of a division algebra is a divisionalgebra provided  the following postulate  holds:

Postulate :  The quantity a a* +b b*  is nonzero,unless a = b = 0

Over the field of real numbers, the Cayley-Dickson construction can be appliediteratively by defining conjugation over the squared algebra in terms of conjugationover the original algebra:

(a ,b )*   =   (a* , -b )   [for real algebras only]

It's then easy to show, byinduction, thatthe quantity  x x*  is always a positive real number. The above postulate follows from the fact that a sum of nonnegative realscan only vanish if they're all zero.

With other fields  (in particular, finite fields)  the same postulatecould possibly be derived from other ad hoc  properties, like:

u ,    u + u     0 ,     x ,    x x*   =   u

Clearly, this implies that  u = 1  (:  consider  x = u) and that the field is of characteristic 3  (i.e.,  1+1+1=0)  because:

u   =   (u+u) (u*+u*)   =  u u* + u u* + u u* + u u*   =   u + u + u + u

The above holds for the field F₃ = (/3,+,.) with trivial conjugation:

0* = 0 ,    1* = 1 ,    2* = 2 ,    1 . 1*   =  2 . 2*   =  1   =   u

Now, for any division algebras over that field  (starting with the fielditself)  we may define conjugation via:

x 0 ,    x*   =   - x^-1

For successive algebra obtained from the Cayley-Dickson construct, the samedefinition can be given a recursive expression, if needed:

(a ,b )*   =   ( -a* ,b )   [for characteristic-3 fields only]

All such successive Cayley-Dickson algebrasare thus division algebras.  By induction, they can all be proved tobe fields. (:  The Cayley-Dickson square ofa commutative algebra is associative, an associative algebra is a ringanda finite division ring is commutative.)

Therefore, the Cayley-Dickson construction  (with theabove conjugation adapted to characteristic 3) defines a nested sequence of finite fieldswhose orders form a sequence where every term is followed by its square:

3,  9,  81,  6561,  43046721, 1853020188851841,  ... 3²ⁿ ...(A011764)

The sum or product of finite ordinalsis well-defined by working things out in any field fromthat sequence large enough to contain all operands.

Ternary Addition :

Ternary  addition is of  "characteristic 3".  This means:  x,  x+x+x = 0

We understand the ternary sum of two integers as the integer whose n-th ternary digit is the sum of thetwo n-th ternary digits of the operandsmodulo 3. You may call that ternary addition  "without carry" if you must.

Ternary addition  must naturally always be understood "withoutcarry", as addition performed with  carry is the same operationon integers regardless of the numeration radix used (just call that "ordinary addition"). Likewise, binary addition  can only be interpreted asthe operation we've called "Nim-addition"aboveandelsewhere.

H.W. Lenstra (1978,Exercise 15) attributes to Simon Norton a definition of ternary addition à la Conway  (where ' denotes any ordinal below) :

The ternary sum   +  is defined  as the least ordinal not expressible as:

• ' +   or
• +'   or
• ' +'   with  ' +  =  +'

This recursive definition need not be limited to finite ordinals.

Ternary Multiplication :

Let's use the theoretical definitions to work out practical rules,similar to those devised by Conway in thebinary case. In what follows, we use Conway'sconvention of putting ordinary arithmetic between square brackets  [ ].

• 2.2 = 1   and   x, 0.x = 0,  1.x = x.
• If  x < [3²ⁿ]  then  x [3²ⁿ]  =  [x 3²ⁿ] 
• The ternary square of   [ 3²ⁿ]   is   [ 2ⁿ]

(2006-03-22 & 2013-07-24)  
Defining On_p  as a field  of prime  characteristic p.

The approach we used in theprevious section,based on the Cayley-Dickson construct, will not work for characteristic 5 because  1 1* is 1  and 2 2*  is  4.  This makes  1 1* + 2 2* vanish modulo 5.

Likewise, it won't work for any other prime characteristic  p  congruent to 1 modulo 4, because theequation  n²+1 =  has asolution modulo p. For such a solution,  1 1* + n n*  vanishes modulo p.

Ternary Multiplication  (a second look)

Let's take ternary addition  (without carry)  for granted. Considering only finite integers for now, we want to define a compatiblecommutative multiplicationwhich does not break the rules of arithmetic in aring.

This goal can be achieved with the following practical rules,similar to those devised by Conway in thebinary case. In what follows, we use Conway'sconvention of putting ordinary arithmetic between square brackets  [ ].

• 2.2 = 1   and   x, 0.x = 0,  1.x = x.
• If  x < [3²ⁿ]  then  x [3²ⁿ]  =  [x 3²ⁿ] 
• The ternary square of   u_n = [3²ⁿ]   is  u_n+ v_n   where  v_n< u_n.

Any sequence  v  stricly dominated by  u  would do if  all we wanted was a unital ring,  but only special ones will yield a field...

v_n  can't be zero, or else  u_n (u_n-1) would be a zero product of two nonzero factors, which is ruled out in a field. Here are 4 satisfactory initial terms:

v₀ = [¹/₃ u₀]    v₁ = [²/₃ u₁]    v₂ = [²/₃ u₂]    v₃ = [¹/₃ u₃]

The first relation means   3² = 4  and corresponds to a subfield  already presented explicitelyabove as a particular  representation of  GF(9). The next relations yield subfields of orders  81, 6561 and 43046721.

With  v₄ = 10000000   we obtain a subfield of order 1853020188851841  whose nonzero elements are all powers of 43046731 (not 43046721).

The multiplicative group of afinite field beingcyclic, a finite ring is a field if and only if there's a primitive  element in it (for example, the nonzero elements of the aforementioned field of order 43046721are the distinct  powers of 6561).
An element  x  is of order k = [3²ⁿ-1] if and only if:

• x^k = 1
• For anyprimedivisor  p  of  k,   x^[k/p] 1.

(When this holds for some element  x  of amonoid of order  k, then this monoid is a cyclic group  consisting of the k distinctpowers of  x.)

Conversely, if x^k  1 for some nonzero  x, then the ring can't possibly be a field (byLagrange's theorem, the order of an elementwould divide the order of the group, so the  k  nonzero elementscan't form a group).

Usingfast exponentiation,a guess-based search for a primitive root may thus result in anefficient proof that the ring is a field or that it's not (albeitmuch less efficiently in the latter case,so the repeated lack of a firm conclusion is astrong indication that thering is not  a field). There are  (k)  "lucky guesses" for  x which will prove that a subfield of order  k  is just that (where    is Euler'stotient function). This is always a substantial percentage of random guesses.

Judea Pearl(2013-09-25)  
How Galois proved that quintic  equations can't be solved by radicals.