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...
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:
A 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).
(*) 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 (1874-1954) 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.
Actually, in a unital ring, distributivity by itself implies that 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
QED
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 the ring axioms which are included in the above field axioms).
"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.
How distributivity makes addition commutative (0:57)
by Michael Penn (2023年04月11日).
In a ring, a zero-divisor 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 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 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 (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 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 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 :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.
We call integral ring (anneau intègre in French) a ring (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+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 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 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 (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 element suffices to show that all of them are invertible. QED
Corollary : 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.
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 call "corps commutatif" what's just dubbed a "field" in English. Some French authors write in English as if the English word was defined the French way (Cédric Milliet is one of them).
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(xi ) ¹ n for some member xi of the ith such class:
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. QED
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 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).
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) = ( Z/pZ/pZ/pZ, + , ´ )
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.
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(pn ) 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:
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 F9 [x] (the ring of the polynomials over F9 ).
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 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 ... pn-1
Finite fields are essential in the classification of finite simple groups.
The ring of the polynomial functions over the field Z/pZ 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
[画像: Come back later, we're still working on this one... ]
Introduction
to finite fields
|
MathWorld :
Primitive polynomials
Wikipedia :
Primitive element in a Galois field
|
Primitive polynomial
[画像: Come back later, we're still working on this one... ]
The Fundamental Theorem of Galois Theory (25:09)
by (Aleph 0, 2021年11月25日).
The Insolvability of the Quintic (10:18)
by (Aleph 0, 2021年02月20日).
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). Some 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 enlightening specializations aren't obvious. For example, the classical group SL(n,F) is identified with the symmetric group Sn when F = F1
Incidentally, if we didn't count the trivial field as a legitimate ring, then the function r which counts the number of finite rings of given order would no longer be a multiplicative function. That would be silly. Wink
Faire des mathématiques, c'est
donner
le même nom à des choses différentes.
Henri Poincaré (1854-1912)
Les variétés sur le corps à
un élément (2003, 2018) by
Christophe Soulé
https://en.wikipedia.org/wiki/Field_with_one_element
https://en.wikipedia.org/wiki/Field_with_one_element
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).
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).
Some equivalent characterizations of what makes a field K perfect :
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.
Example
|
Formal Laurent series
|
Pierre Laurent (1813-1854; X1830)
The
field of Laurent series over finite fields (Mathematics Stack Exchange, 2011年05月14日).
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 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").
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.
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 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.
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 2-power form a subfield). Conway calls the whole field On2 (pronounced "onto").
On2 is meant to stand for "Ordinal numbers with characteristic 2".
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 (HINT: 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 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).
Note that, in an additive group, a + 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 On2 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 On2 .)
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)
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.)
un < [ 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).
On2 :
Transfinite
Number Hacking &
Conway's
Nim-Arithmetics by Lieven Le Bruyn (January 2009).
Conway's Nimbers
by Alasdair McAndrew (2010年10月09日).
Conway's
Nim Field by Peter Cameron (2011年10月29日).
Video :
Lexicographic codes of minimal distance 3
are vectors over nimbers by J.H. Conway (2014).
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.)
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:
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:
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 = (Z/3Z,+,.) 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.
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:
This recursive definition need not be limited to finite ordinals.
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 [ ].
[画像: Come back later, we're still working on this one... ]
Simon P. Norton (1952-) | Simon Norton by Frances Hubbard (2011年09月10日).
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.
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 [ ].
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:
(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.) Joseph-Louis Lagrange 1736-1813
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...
[画像: Come back later, we're still working on this one... ]
A Recursive Definition of p-ary
Addition without Carry (1999) by
François Laubie.
ON Onp by
Dr.
Joseph M. DiMuro (2011年08月08日).
