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 0   1   2   3 + 0 1 2 3 1 0 3 2 2 3 0 1 3 2 1 0
 0   1   2   3 ´ 0 0 0 0 0 1 2 3 0 2 3 1 0 3 1 2

# Fields

In France, about 1830, a new star of unimaginable brightness
appeared in the heavens of pure mathematics:   Evariste Galois
.
Felix Klein  (1849-1925)

I need all my courage to die at 20.
Evariste Galois  (1811-1832)

### Related Links (Outside this Site)

Wikipedia:   Field (mathematics)
The Evariste Galois Archives

DMOZ: Field Theory

## Fields  (and Skew Fields)

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

At the very least, the analogy  (attributed to bourbakist  Roger Godement)  is amusing.  It deserves much better consideration than what it received when somebody  (Robinh)  kindly posted it within the Wikipedia article defining a  division ring,  only to see the remark hastily dismissed as "patent nonsense".

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 helpful and 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 the inclusive concept of  division ring  (commutative or not).

We argue that the term  skew field  should only designate a  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 the locution  "skew field"  only in special contexts where noncommutativity is otherwise clearly stated...

(2006-03-16)   Fields
Commutative rings in 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 when the Norwegian  Niels Henrik Abel (1802-1829) and the Frenchman  Evariste Galois (1811-1832) where led to consider  finite fields  in their independent investigations concerning the impossibility of solving "by radicals" a general polynomial of degree greater than 4.  This had been  the  central problem of Algebra ever since Renaissance Italian algebraists gave solutions  by radicals  for equations of the third and fourth degree.

The groundwork for the successful investigations of Abel and Galois was laid by the forefathers of group theory, starting with the first paper ever published by Alexandre-Theophile Vandermonde (1735-1796) :  Mémoire sur la résolution des équations (1770).  This introduced the clever idea of investigating functions which are invariant under the permutations of a polynomial's roots, which Joseph-Louis Lagrange (1736-1813) would soon build on.  Paolo Ruffini (1765-1822) proposed an incomplete proof of what's now called the "Abel-Ruffini theorem".

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

field  is a  commutative ring  in which every element but "zero"  (the neutral element for addition)  has a multiplicative inverse  (a reciprocal).  This means that the properties listed below hold for "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 the field is also an element of the field).

Addition Multiplication "x "y "z x + (y + z)  =  (x + y) + z x (y z)  =  (x y) z x + y  =  y + x x y  =  y x x + 0  =  x x 1  =  x \$(-x),   x + (-x)  =  0 " x¹0, \$ x-1,   x x-1  =  1 x (y + z)   =   x y  +  x z     and(*)     (x + y) z   =   x z  +  y z

(*)   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 multiplicative commutativity.  Two-sided versions of the other multiplicative properties can be derived from their one-sided counterparts  without  assumming commutativity  (see elsewhere on this site for a proof).

The terms  commutative  and  distributive  (French:  commutatif & distributif)  were both introduced in a memoir of Joseph 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 of octonions  does not  have this property...

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

"Definition of a group and a field by independent postulates"  Leonard Eugene Dickson (1874-1954).
Transactions of the American Mathematical Society, 6:198-204, 1905.

(2006-02-06)   Quotient Field
Smallest field containing a given ring  (without any divisors of zero).

In a ring,  a divisor of zero  x  is a nonzero element whose product with some nonzero element  y  is zero.  In a subring of a field,  that never happens because 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 divisors of zero,  then we may 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 of elements from A where the second one is nonzero)  by stating that  (a,b)  and  (c,d)  are equivalent  when:

a d   =   b c

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

All such quotients form the ring's  quotient field  K  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)   =   (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 divisors of zero in A).

Next we can show that any element  x  of  A  is uniquely associated with the class consisting of all pairs  (bx,b)  where  b  is a nonzero element of  A.  Indeed,  all such pairs are  equivalent  and the class associated with  x  can't be associated with another element  y  of  A  (HINT:  otherwise  x-y  would be a divisor of zero).  It's also easy to verify that this one-to-one mapping is an homorphism  (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 a 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 all field properties is just routine.

Some common examples of quotient fields :
RingQuotient Field
Integers   Rationals
(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 of whatever applications it may have.  For example,  a formal polynomial 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.

(2006-03-18)   Wedderburn's Little Theorem  (1905)
Multiplication in a finite division ring is necessarily commutative.

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

In English at least, "fields" are now officially required 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.

In French, a "corps" is a division ring  (it may or may not be commutative).  When applicable, the French may specify "corps commutatif" which is what we simply call a "field" in English.

The theorem was first published in 1905 by the Scottish mathematician  Joseph Wedderburn (1882-1948).  After seeing a proof of the theorem by L.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)  that Wedderburn's first "proof" had a gap which went unnoticed at the time.  Although Dickson did acknowledge Wedderburn's priority, he should have been given credit for the first valid proof of what's now universally known as  Wedderburn's theorem.

"In pursuit of the finite division algebra theorem and beyond:  Joseph H.M. Wedderburn, Leonard E. Dickson, and Oswald Veblen"   by Karen Hunger Parshall.   Archives of International History of Science  33:111, 274-299 (1983).

Proof :   Let  K  be a  finite  division ring.

Let  C(x)  be the centralizer  (or commutant)  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, which means that it contains the reciprocals of all its nonzero elements.  So is the center  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)  are vector 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 + ya,  with  x  and  y  in the center  C,  which would make all of them commute  (thus implying that  n  is  1,  not 2).

Let's apply the  conjugacy class formula  to the multiplicative group formed by the  qn-1  nonzero elements of  K,  whose center  (C-{0})  is of order  q-1.  The order of the  conjugacy class  of  x  is the index of  C(x)-{0}  in the whole multiplicative group, namely  (qn-1) / (qn(x)-1).   We may enumerate all the conjugacy classes of  noncentral  elements  (assuming that there are any)  by letting  ni  be  n(x¹ n   for some member  xi  of the ith such class:

qn-1   =   q-1  +
 å i
qn  - 1
qni - 1

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

There are several ways to do so.  Wedderburn used the special case b=1  (A.S. Bang, 1886)  of Zsigmondy's Theorem (1892)  itself often credited to Birkhoff and Vandiver (1904) and rediscovered by many authors:  Carmichael in 1913, Kanold in 1950, Emil Artin in 1955, etc.  It says that, if a and b are coprime, then   an-bn  has a primitive factor (i.e., a prime factor not dividing that expression for a lower positive value of  n) except with  26-16  or for n=2 when a+b is a power of 2.

In 1931, Ernest Witt proposed instead a celebrated self-contained argument based on the cyclotomic polynomials in the complex plane.  Several authors have modified Witt's proof to shun complex numbers.

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  S  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  qm-1  for any positive value of  m  less than  n  (if there are any).  Since such a  p  necessarily divides  q-1  because it divides all other terms in the above equation, we must conclude that  n = 1.

Proof of Wedderburn's theorem   |   Witt's Proof

(2006-03-26)   Finite Integral Domains
Every  finite  integral domain is a  field.

Proof :   (Ruling out as trivial a single-element ring.)

In this, we understand an  integral domain  to be a ring  (commutative or not)  where the product of two nonzero elements is never zero.  Commutativity will be implied by Wedderburn's theorem  if we just prove that every element is necessarily invertible in such a  finite  structure...

First, we must 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, these can't be all distinct...  Say the  n+k+1st  is equal to the  n+1st   (for some k>0).  Let's put   u = yk

"x,     ( x u - x ) yn+1   =   x yn+k+1 - x yn+1   =   0

As there are no divisors of zero, 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 element  a,  the map which sends  x  to  a x  is  injective:  If two distinct elements  x  and  y  had the same image, the product  a (x-y)  would vanish without any factor vanishing, which is ruled out here.

Any injection of a  finite  set into itself is surjective  (the  pigeonhole principle )  which implies that 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 element is sufficient to establish that  all  of them are invertible.

It's remarkable that the mere absence of divisors of zero in a finite ring makes it necessarily commutative and isomorphic to a Galois field.
(2015-12-24)   Artin-Zorn Theorem   (1930)
A finite alternative algebra without divisors of zero is necessarily a field.

This is a generalization of Wedderburn's theorem (1905) which states that a field is neccessarily obtained even whith a multiplication which need not be postulated to be  associativealternativity  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)   Galois Fields  (Finite Fields)
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  is denoted  GF(q)  or  Fq

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) = Fq  is isomorphic to the  direct sum  Cpn  of  n  cyclic groups of order  p  (the  n  components add independently modulo p).

Multiplicatively,  the  q-1  nonzero elements of  Fq  form a cyclic group.

In particular, if  q  is prime  (q = p)  then  Fq  is simply isomorphic to the 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 from the  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 two irreducible cubic polynomial of  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 of bitwise addition can be used to  name  the other elements of  GF(8)  after ordinary integers.

 x0 = x7 x1 x2 x3 x4 x5 x6 1 x x2 x + 1 x2 + x x2 + x + 1 x2 + 1 1 2 4 3 6 7 5

0 1 2 3 4 5 6 7 + 0 1 2 3 4 5 6 7 1 0 3 2 5 4 7 6 2 3 0 1 6 7 4 5 3 2 1 0 7 6 5 4 4 5 6 7 0 1 2 3 5 4 7 6 1 0 3 2 6 7 4 5 2 3 0 1 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 0 2 4 6 3 1 7 5 0 3 6 5 7 4 1 2 0 4 3 7 6 2 5 1 0 5 1 4 2 7 3 6 0 6 7 1 5 3 2 4 0 7 5 2 1 6 4 3 0 1 3 2 6 4 5

When the order  (q-1)  of the multiplicative group of  GF(q)  isn't prime, there's a complication, best illustrated with the construction of  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 only generates a cycle of order 4  (namely, g, -1, -g, 1).  What we need is a  primitive  element of order 8 which would generate the entire multiplicative group of  GF(9).  A root of either of the last two polynomials has this property.  (Such polynomials are thus called  primitive  polynomials.)

We have no shortcut to predict which irreducible polynomials of degree n over  GF(p)  yield primitive roots of  GF(p)  but many do.

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:

 x0 = x8 x1 x2 x3 x4 x5 x6 x7 1 x x + 1 2x + 1 2 2x 2x + 2 x + 2 1 3 4 7 2 6 8 5

0 1 2 3 4 5 6 7 8 + 0 1 2 3 4 5 6 7 8 1 2 0 4 5 3 7 8 6 2 0 1 5 3 4 8 6 7 3 4 5 6 7 8 0 1 2 4 5 3 7 8 6 1 2 0 5 3 4 8 6 7 2 0 1 6 7 8 0 1 2 3 4 5 7 8 6 1 2 0 4 5 3 8 6 7 2 0 1 5 3 4
0 1 2 3 4 5 6 7 8 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 0 2 1 6 8 7 3 5 4 0 3 6 4 7 1 8 2 5 0 4 8 7 2 3 5 6 1 0 5 7 1 3 8 2 4 6 0 6 3 8 5 2 4 1 7 0 7 5 2 6 4 1 8 3 0 8 4 5 1 6 7 3 2 log3 0 4 1 2 7 5 3 6

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

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

Therefore, all symmetric functions of 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 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   ...   pn-1

Finite fields are essential in the classification of  finite simple groups.

Galois Fields on the HP Prime Calculator

(2015-12-22)   Fq[x] :   The Polynomials in a Galois Field.
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 a vector space  of  dimension  n  isomorphic to  the Galois field  Fq  of  (prime)  characteristic  p  and order  q = pn.

The  multiplicative group  formed by the nonzero elements of  GF(q)  is a cyclic group 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  f(q-1)  primitive  elements in GF(q)  (where  f  is Euler's totient function).

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

f (qd-1) / d

 d 1 2 3 4 5 6 7 8 9 OEIS F2 1 1 2 2 6 6 18 16 48 A011260 F3 1 2 4 8 22 48 156 320 1008 A027385 F4 2 4 12 32 120 288 1512 4096 15552 A027695 F5 2 4 20 48 280 720 5580 14976 99360 A027741 F7 2 8 36 160 1120 6048 37856 192000 1376352 A027743 F8 6 18 144 432 5400 23328 254016 859440 12607488 A027744 F9 4 16 96 640 5280 27648 340704 1966080 15676416 A027745

Introduction to finite fields   |   MathWorld :   Primitive polynomials
Wikipedia :   Primitive element in a Galois field   |   Primitive polynomial

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

The zeroth power of any prime is 1.  Arguably, the simplest Galois 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).  Many textbooks rule out fields with only one element.

A few authors observed that some concepts traditionally studied on 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) quotes Jacques Tits  (1957), A. Smirnov  (1992)  and  Y. Manin  (1995).

Such enligntening specializations aren't obvious.  For example, the classical group  SL(n,F)  is identified with the symmetric group  Sn  when  F = F1

(2006-03-25)   Splitting Field of a Polynomial  P Î F[x]
The smallest subfield or extension of F 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 the splitting field of P where P can be completely factored  (into polynomials of degree 1).

(2017-12-06)   Perfect Fields
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 finite extension of the field is  separable  (that's called  Galois' hypothesis.

There are many equivalent characterizations of perfect fields  K.  Here are a few:

Perfect field

(2017-08-01)   The Field of Laurent Series
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 invertible and turns a ring into a field.

Formal Laurent series   |   The field of Laurent series over finite fields  (Mathematics Stack Exchange,

(2006-03-17)   The Algebraically Complete Nim-Field:  Conway's  On2
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 shows in what sense  bitwise  addition is the simplest "addition" we can endow the natural integers with.  This operation can be described as binary addition  without carry.  It's also known as  Nim-sum,  or bitwise  "exclusive or".  Under the latter name, this operation is widely available at the fundamental level of the  assembly languages  of modern 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 using field properties and two additional statements which parallel those given above for Nim-addition.

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

Fermat power  means 2 to the power of a 2-power  ( 22)  namely:
2, 4, 16, 256, 65536, 4294967296, 18446744073709551616, ... A001146
The aforementioned sesquimultiples of those are half as much again:
3, 6, 24, 384, 98304, 6442450944, 27670116110564327424, ...

Conway coined these learned terms to shun ordinary arithmetic when describing Nim operations  (originally, he wrote "Fermat 2-power").

Pierre de Fermat (1601-1665) once conjectured that a prime number always came after 2 raised to a 2-power.  The conjecture is false; there are probably no such prime numbers beyond the five known to Fermat:  3, 5, 17, 257 and 65537.  In 1796, teenager Carl Friedrich Gauss showed that a regular n-gon is constructible just when  n  is a 2-power multiplied by a square-free product of those "Fermat primes".

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

On is meant to stand for "Ordinal numbers with characteristic 2".

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

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 2 3 1 8 10 11 9 12 14 15 13 4 6 7 5 0 3 1 2 12 15 13 14 4 7 5 6 8 11 9 10 0 4 8 12 6 2 14 10 11 15 3 7 13 9 5 1 0 5 10 15 2 7 8 13 3 6 9 12 1 4 11 14 0 6 11 13 14 8 5 3 7 1 12 10 9 15 2 4 0 7 9 14 10 13 3 4 15 8 6 1 5 2 12 11 0 8 12 4 11 3 7 15 13 5 1 9 6 14 10 2 0 9 14 7 15 6 1 8 5 12 11 2 10 3 4 13 0 10 15 5 3 9 12 6 1 11 14 4 2 8 13 7 0 11 13 6 7 12 10 1 9 2 4 15 14 5 3 8 0 12 4 8 13 1 9 5 6 10 2 14 11 7 15 3 0 13 6 11 9 4 15 2 14 3 8 5 7 10 1 12 0 14 7 9 5 11 2 12 10 4 13 3 15 1 8 6 0 15 5 10 1 14 4 11 2 13 7 8 3 12 6 9

Note that the Nim-product of a Fermat power and  any  lesser number is the same as their ordinary product  (HINT:  The lesser number is a sum of 2-powers, each of which is a product of Fermat 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 5 look-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  5m steps to Nim-multiply two  2m-bit integers from scratch using the recursive procedure suggested by the above example.  Asymptotically, this means that the Nim-product of two  n-bit  integers can be computed in time  O(n k )  where  k = lg(5) < 2.322.

Denoting  a'  an arbitrary ordinal smaller than  a, Conway gives two remarkable one-line definitions of the Nim-operations  which are very similar to his other one-liners in the realm of surreal numbers  (the "-" sign is used here as a synonym of the "+" sign for aesthetic reasons, in part to reinforce that similarity).

• a + b   is the least ordinal distinct from all numbers   a' + b  and  a + b'.
• a b   is the least ordinal distinct from all numbers   a' b + a b' - a' b'.

Note that, in an additive groupa + b   cannot be equal to either  a' + b  or  a + b'  unless  a' = a  or  b' = b.  Therefore, the above definition is the "simplest" possible definition of addition in some sense.

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

These "genetic" definitions are also valid for infinite ordinals  (they're equivalent to the above practical rules for  finite  integers)  and do make  On  a field.

1/a  is recursively defined as the least nonzero ordinal distinct from all numbers

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

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  ax2  (compare this to the computation of a p-adic reciprocal).  Starting from 1, we obtain 1 again after a number of iterations equal to the bit-length of  a, rounded up to a 2-power.  The last step reveals the reciprocal of  a.

In a field of characterisic  2,  like this one,  the  square  function is a field homomorphism  (the square of a sum is the sum of the squares)  which is injective  (HINT:  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  field On.)

Conversely, any nimber  x  has a unique square-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 finite nimbers,  the rim  function can be defined  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)

 n rim(n) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 3 2 7 6 4 5 14 15 13 12 9 8 10 11 30 31 29 28 25

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

u <  [ 22n ]           f (un )  =  exp [ 2pi / (22n-1) ]           un+1[ 22n+ 1 ]  =  un

Because  x <  [ 22n ]  just when  x [ 22n- 1 ] = 1,  the above inequality is true for  n  if  it's true for  n+1  (HINT:  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 possibilities for the n-th term, at least one must belong to a satisfactory sequence whose length exceeds any given bound).
Actually, for  any  satisfactory  un  there are  22n  satisfactory values of   un+1

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

w3 = 2

On Numbers and Games (1975)  J.H. Conway   (1976, Academic Press, London, ISBN 0-12-186350-6)
On the Algebraic Closure of Two   by  Hendrik W. Lenstra, Jr.   (Leiden University, 1977).
Nim Multiplication   by  Hendrik W. Lenstra, Jr.   (IHES, Bures-sur-Yvette, February 1978).
On2Transfinite Number Hacking   &   Conway's Nim-Arithmetics   by  Lieven Le Bruyn   (January 2009).
Conway's Nimbers   by  Alasdair McAndrew   (2010-10-09).
Conway's Nim Field   by  Peter Cameron   (2011-10-29).
Video :   Lexicographic codes of minimal distance 3 are vectors over nimbers  by  J.H. Conway  (2014).

(2013-07-22)   One step beyond  Conway's  On2 :
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 to the "curious field" discussed in the previous 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 in the next section.)

### Preliminaries :

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

So to speak, that construct yields the  square  of an algebra by doubling its number of dimensions  (e.g., complex numbers are obtained 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 is commutative, it's a field and conjugation is a field automorphism.  Any such automorphisms must be Froebenius maps, which is to sau that there is an integer k such that:

" x ,     x*   =   xpk

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 ) = ( a c  -  d b*  ,  a*d  +  c b )

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 to be 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 an algebra where every nonzero element has a multiplicative inverse.  (If such an algebra is associative and commutative it's a field.  If it's only associative, it's a skew-field.)  Our last relation shows that the square of a division algebra is a division algebra  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 applied iteratively by defining conjugation over the squared algebra in terms of conjugation over the original algebra:

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

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

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

\$ u ,     u + u   ¹   0 ,     " x ,     x x*   =   u

Clearly, this implies that  u = 1  (HINT:  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 field itself)  we may define conjugation via:

" x ¹ 0 ,     x*   =   - x-1

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

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

All such successive Cayley-Dickson algebras are thus division algebras.  By induction, they can all be proved to be  fields.  (HINT:  The Cayley-Dickson square of a commutative algebra is associative, an associative algebra is a ring and a finite division ring is commutative.)

Therefore, the  Cayley-Dickson construction  (with the above conjugation adapted to characteristic 3)  defines a nested sequence of finite fields whose 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 ordinals is well-defined by working things out in any field from that sequence large enough to contain all operands.

The ternary field structure so given to nonnegative integers mirrors Conway's binary structure, although he went much further with inductive definitions encompassing surreal ordinals  to form the great  algebraically complete  field described in his exciting masterpiece  "On Numbers and Games"  (1976).

The above preliminaries provide theoretical specifications for the "ternary" operations, which are described below in practical terms.

0 1 2 3 4 5 6 7 8 + 0 1 2 3 4 5 6 7 8 1 2 0 4 5 3 7 8 6 2 0 1 5 3 4 8 6 7 3 4 5 6 7 8 0 1 2 4 5 3 7 8 6 1 2 0 5 3 4 8 6 7 2 0 1 6 7 8 0 1 2 3 4 5 7 8 6 1 2 0 4 5 3 8 6 7 2 0 1 5 3 4
0 1 2 3 4 5 6 7 8 0 x* 0 0 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 7 8 2 0 2 1 6 8 7 3 5 4 6 0 3 6 2 5 8 1 4 7 7 0 4 8 5 6 1 7 2 3 8 0 5 7 8 1 3 4 6 2 3 0 6 3 1 7 4 2 8 5 4 0 7 5 4 2 6 8 3 1 5 0 8 4 7 3 2 5 1 6

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 the two n-th ternary digits of the operands modulo 3.  You may call that  ternary addition  "without carry" if you must.

Ternary addition  must naturally always be understood "without carry", as addition performed  with  carry is the same operation on integers regardless of the numeration radix used  (just call that "ordinary addition").  Likewise,  binary addition  can only be interpreted as the operation we've called "Nim-addition" above and elsewhere.

Ternary  addition (without carry) pertains to a version of Moore's Nim where a player may take from  one or two  heaps at each turn.

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

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

• a' + b   or
• a + b'   or
• a' + b'   with   a' + b  =  a + b'

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 the binary case.  In what follows, we use Conway's convention 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 ]

Simon P. Norton (1952-)   |   Simon Norton   by Frances Hubbard  (2011-09-10).

(2006-03-22 & 2013-07-24)   Beyond  On2  and  On3...
Defining  Onp  as a  field  of  prime  characteristic p.

The approach we used in the previous 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 the equation  n2+1 =  has a solution 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 compatible commutative multiplication which does not break the rules of arithmetic in a ring.

This goal can be achieved with the following practical rules, similar to those devised by Conway in the binary case.  In what follows, we use Conway's convention 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   un = [32n ]   is   un+ vn   where   vn< un.
A slight generalization would be to let the square of  un  be  wnun+vn  where both sequences v and w are dominated by u.

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...

vn  can't be zero, or else  un (un-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 explicitely above 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 a finite field being cyclic, 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 43046721 are the  distinct  powers of 6561).
An element  x  is of order  k = [32n-1]  if and only if:

• xk = 1
• For any prime divisor  p  of  k,   x[k/p] ¹ 1.

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

Conversely, if  xk ¹ 1  for some nonzero  x,  then the ring can't possibly be a field  (by Lagrange's theorem, the order of an element would divide the order of the group, so the  k  nonzero elements can't form a group).

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

In the case where we're actually faced with a field, the above proof has a good chance to work with a random  x  (we just need a factorization of  k  into primes, which can be a significant problem for very large values of k).  If the above proof fails with a specific x, then all we know is that  x  is not a primitive root...  However, repeated failures within a field are highly unlikely because there are so many primitive roots in it.  Such repeated failures thus indicate that we're not dealing with a field...

 n Prime Factorization  of   32n-1+ 1   ( if  n > 0 ) 2 2 . 2 2 . 5 2 . 41 2 . 17 . 193 2 . 21523361 2 . 926510094425921 2 . 1716841910146256242328924544641 2 . 257 . 275201 . 138424618868737 . 3913786281514524929 . 153849834853910661121 2 . 12289 . 8972801 . 891206124520373602817 . 707275264749309881405141965802671548079179711820351316861777644606207216944972589404100097 2 . 59393 . 448524289 . 847036417 ... (466-digit  composite  factor)

A Recursive Definition of p-ary Addition without Carry (1999)   by  François Laubie.
ON Onp   by  Dr. Joseph M. DiMuro   (2011-08-08).

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