supplement xi synopsis

7/31/2019 Supplement Xi Synopsis

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ALGEBRAIC NUMBER THEORY IN DEDEKINDS

SUPPLEMENT XI

ANSTEN KLEV

In the following we aim to give a summary of some main sections in DedekindsSupplement XI to the 4th edition of Dirichlets Lectures on Number Theory (Dirich-let, 1894). The summary leads the reader to the theorem of unique factorizationof ideals in 179. Along the way Dedekind has introduced some fundamental al-gebraic notions such that of a field, field isomorphism, vector space, module, ring(ordering), and algebraic integer, and he has developed their theory. As part of

his investigations into fields and field isomorphisms Dedekind gives a very moderndevelopment of Galois theory. This has been summarized, and the relevant sectionstranslated, by Dean (2009), and so is omitted here. On the other hand, we have in-cluded a summary of some results from 180 where Dedekind lifts classical resultsregarding Eulers totient function to ideal theory. Throughout we have insertednotes with commentary on results and definitions.

Ideally, the following should give a sense of the mathematics developed in theSupplement. It has been written partly with the Germanless reader in mind, but itshould hopefully also serve as a rough guide for those who wish sit down and workthrough the Supplement.

To achieve at least some historical authenticity, we follow as much as possibleDedekinds terminology (translated into English), and thus avoid as much as possi-ble terms such as set and homomorphism. Our choice of symbolism will, however,sometimes diverge from Dedekinds.

159. Motivation for the theory to be developed is found in the theory of theGaussian integers, or whole complex numbers as Dedekind calls them, introducedand studied by Gauss in 1832:

Z[i] := {a + bi : a, b N}.An element Z[i] is said to be divisible by Z[i] if there is a Z[i] suchthat = . An element Z[i] is a unit if 1 is divisible by . The units are 1,1, i, i. Numbers , Z[i] are said to be associated if there is a unit suchthat = . A prime number Z[i] is a number whose divisors are either unitsor associated with . Analogously with Z[i] one defines Q[i], which is a field (thisconcept is introduced formally in the next section).

The operation of complex conjugation is defined by

a + bi := a bi;for , Q[i] we have + = + , = . The key notion of the normN() N of an element Z[i] is defined by

N() := .It satisfies N() = N()N().

1


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2 ANSTEN KLEV

Lemma. For any , Z[i] with = 0 there are , Z[i] such that = + and N() < N().

To prove this we note first that any Q[i] can be written as a sum + ,where Z[i] and Q[i] satisfies N() < 1. Hence, given , Z[i], there are Z[i] and Q[i] with N() < 1 such that

= + ; since = , we

have Z[i]; moreover, N() = N()N() < N(). Given the notions of divisibility and prime, and by means of the properties of

the norm N() one may carry through the Fundamental Theorem of Arithmeticto the Gaussian integers. From the definition of prime number it follows that anyGaussian integer = 0 is either a unit, a prime, or is divisible by a number whichis neither a unit nor associated with . In the latter case is called composite,and we have N() < N(). Repeated application of this fact yields that everynumber decomposes into a product of primes. Two numbers , Z[i] are said tobe relatively prime if they have no common divisors apart from units. The Lemma

above yields an Euclidean algorithm for Z[i], and from that algorithm it followsthat , Z[i] are relatively prime if and only if there are , Z[i] with + = 1.

Given this it is easily established that if a prime divides a product , thenit divides one of the factors , , . . .. It follows then that the decomposition intoprimes is unique in the following sense.

Proposition. If 1, . . . , n and 1, . . . , m are primes and1in

i =

1im

i,

then n = m and for every i there is a j and a unit such that i = j .

It is now possible to classify all the primes in Z[i]: for any prime number

Z[i]

there is a natural prime number p N such that either = p or = p for someunit .

160.

Definition. A field (Korper) is a system A of real or complex numbers such thatthe sum, difference, product, and quotient of any two numbers of A is again amember of A.

Note. This definition is found already in Dedekind (1871). In a footnote to thecurrent definition, Dedekind says that he employed the concept of a field already inhis lectures at Gottingen (1857-58), and he gives an explanation for the choice ofthe name Korper (English near equivalent: body; Latin: corpus). This explanationis worth quoting (Dedekind, 1894, p. 20):

This name shall here, as it does in the natural sciences, in ge-ometry, and in the life of human society, designate a system whichpossesses a certain completeness, perfection, closure, through whichit appears as an organic whole, as a natural unity.

It may be noted that in an unpublished manuscript discussed by Dedekind in aletter to Cantor 1879 and reprinted in (Dedekind, 1932a, ch. XXXV), outliningbasic concepts of topology (open set, exterior, limit point), Dedekind uses the termKorper for what we today would call an open set (in a metrical space).


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ALGEBRAIC NUMBER THEORY IN DEDEKINDS SUPPLEMENT XI 3

Dedekind remarks that Kroneckers notion of domain of rationality (Rationalitats-bereich) agrees essentially with his own conception of a field. Kroneckers notionis:

A domain of rationality (R, R, R, . . .) contains every one of thosequantities which are rational functions of the quantities R, R,R,. . . with integer coefficients.1

Hence a domain of rationality may be viewed as a domain of rational expressionsin the quantities R, R, R,...; these Rs may be indeterminates or roots ofequations. Kiernan (1971, p. 129) sums up the difference between Kronecker andDedekinds notion in two points: firstly, fields, unlike domains of rationality, areassumed to be completed totalities (so that one may, as did Dedekind, performoperations on the fields themselves); secondly, when treated as Dedekindean fields,domains of rationality (at least as initially defined) are restricted to finite fields(in Dedekinds sense, i.e., finite extensions of the rational numbers). Kronecker(1882, p. 3) says that his notion originated already in 1853 (the year he learned

of Galoiss work). Dedekind reports he employed the notion of field (under thename of rational domain (rationale Gebiet)) in his lectures at Gottingen in 1857-58.2 van der Waerden (1930, p. 41) defines Korper oder Rationalitatsbereich,witnessing debt to Dedekind as well as to Kronecker.

The system of rational numbers Q forms the smallest field: it is contained inany other field. If is the solution of a quadratic equation x2 + ax + b = 0 overQ, then {a + b : a, b Q} is a field. The real numbers and the complex numbersare fields. Let A and B be fields. If A B, then A is said to be a divisor ofB, and B a multiplum of A. The intersection A B is a field, the least commondivisor of A and B. More generally, the intersection of an arbitrary number (eveninfinite) of fields is again a field. If G is any system (finite or infinite) of numbers,

the intersection of all fields A such that G A is hence itself a field [G], indeed itis the smallest field containing the numbers in G. The field [G] is the closure of Gunder the operations of multiplication, addition, subtraction, and division.

Note. One should note the close relationship between the notion of closure of afield and the notion of chain of a system. Recall that the latter is defined in Wassind und was sollen die Zahlen? (Dedekind, 1888) art. 44 as follows. Given asystem (that is, a set) A S and a mapping of S, then the chain of the systemA, denoted by A0, is the intersection over all sets B S such that A B and[B] B, where [B] := {(b) : b B}. Thus, the chain of a system is simply itsclosure under a given mapping. The field [G] is the closure of the set G under theoperations of addition, subtraction, multiplication, and division. These two notionsof closure can be treated in a uniform way if we think in terms of a set operator

on S, that is a mapping sending subsets of S to subsets of S, and not in termsof a mapping of S. In general, given such an operator , we may define furtheroperators n by recursion as follows:

0(B) := Bn+1(B) := (n(B))

1Cf. Kiernan (1971, p. 127) and Kronecker (1882, pp. 35).2Dedekinds notes from these lectures have been published in Scharlau (1981).


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4 ANSTEN KLEV

Then A0 =n


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Dedekind writes the mapping to the right of the argument, presumably becausethat is more natural when considering the composition of mappings.

If A is a field, and a permutation of A, then A is a field; the fields A andA are said to be conjugated. The identical mapping : a a on a field A isa permutation of A; complex conjugation is a permutation of the field Q[i]. Theidentical mapping is the only permutation of Q. Any permutation of a field issimilar (ahnlich), or distinct (deutlich): if = then = . 162. If is a permutation of A and a permutation of A, then , definedby := () is a permutation of A; the permutation is called the resultantof the components , and denoted by . It follows that ifA and A each areconjugated with a field A, then A and A are conjugated with each other. Theresultant 1 is the identical mapping ofA, and assuming that is a permutationof A and a permutation of A, we have () = ().

163. If is a permutation ofA and B

A, then B, the restriction of to B,

is said to be divisor of, and a multiple ofB. If is a system of permutationsof A then A is n-valued w.r.t. if

card({() : }) = n.Any rational number is 1-valued w.r.t. any class of permutations of a field A, andthus the identity mapping ofQ is a divisor all field permutations. More generally,the following holds.

Proposition. If is a system of permutations of a field A, then

A := { A : () is 1-valued w.r.t. }is a subfield of A; moreover,

:= A for some (any) is the greatest

common divisor of the permutations in , that is to say, if divides all ,then divides . The field A is called the field of the system .

It is clear that the sum, difference, product, and division of two 1-valued elementsis again 1-valued, hence A is a field. If is a common divisor of the permutationsin , then must be a permutation of some field B A and

B = .

Two permutations are said to be in agreement or in harmony if they have acommon multiple. If , a permutation of A, and , a permutation of B, are inagreement, and is a common multiple of and , then AB is completelydetermined: for we must have a = a and b = b for any a A, b B. Hence,AB is the least common multiple of and .

Note. The Proposition above tells us how to find the greatest common divisor ofpermutations of the same field, but not of permutations in general. Dedekind mighthave left it as an exercise to the reader to work this out. In fact, Ferreiros (1999,

ch. VII.5) argues that, less trivially, in Was sind Dedekind left it to the reader towork out what is now known as the Cantor-Schroder-Bernstein Theorem from thetheorem of art. 63 there (whose proof itself was explicitly left it to the reader). Tofind the greatest common divisor, then, of two arbitrary permutations and , oneconsiders their common domain A = dom() dom(), which is an intersection

ist eine Abbildung (der abbildende Maler), die aus dem System (Original) Sdas Bild (S) erzeugt.


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6 ANSTEN KLEV

of fields and hence itself a field; then one applies the Proposition to the system{A, A}.

164. A system {1, . . . , n} of numbers is said to be reducible over A, if thereare 1, . . . , n A, not all of which are = 0, such that

1in

ii = 0,

otherwise the system is said to be irreducible over A. The elements of a reduciblesystem are said to be dependent over A, and those of an irreducible system inde-pendent over A. A number is algebraic over A if there are 1, . . . n A suchthat

n + 1n1 + + n1 + n = 0.

If is algebraic over A, then there is an equation of least degree which witnessesthis; this number is said to be the degree of over A.

If 1, . . . , n are independent over A, then the system of numbers of the form1in

ii

is called a flock (Schaar) over A, and the system {1, . . . , n} a basis for the flock.Given a basis for a flock , any element is uniquely determined by coordinates(1, 2, . . . , n), namely such that

=

1in

ii;

this follows from the fact that the basis is independent over A.

Proposition. The following are characteristic properties of the notion of a flock

with a basis of n elements:

(1) The sum and the difference of two elements of is again in .(2) The product of a number in and a number in A is again a number in .(3) There are n numbers in independent over A, but any n +1 such numbers

are dependent over A.

If is a flock over A, then it is clear that it has properties (1) and (2). Theproof that also has property (3) is by induction on n. It is easily seen to hold forn = 1. So suppose n > 1 and it established for n 1. Then, given n + 1 elements, 1, . . . , n of with a basis of n elements, we may assume that one of thosenumbers, say , is = 0 (otherwise they are clearly dependent). Hence, for somek n, the k-th coordinate of is = 0. Thus there are numbers c1, . . . , cn A suchthat the k-th coordinate of each of the numbers

1 + c1 , . . . , n + cn

vanishes. These n numbers therefore belong to a flock with a basis ofn1 elements,and thus are dependent by the induction hypothesis. Hence there are b1, . . . , bn Asuch that

b1(1 + c1) + . . . + bn(n + cn) = 0,

and this proves that the numbers , 1, . . . , n are indeed dependent over A.


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Conversely, suppose that and A are systems of numbers possessing the prop-erties (1)(3). Then by (3) there are n numbers 1, . . . n in independent over Aand every number of has the form

1in

ii

for i in A. Furthermore, from (1) and (2) it follows that all numbers of this formare in .

Note. Since A is assumed to be a field, properties (1) and (2) in the Propositionexpress that is a vector space over A. Property (3) restricts the space to beingof finite dimension. Thus, Dedekind points out: with every passage from onebasis to another is obviously related a transformation of the coordinates of everynumber, as in analytical geometry, which indicates that he saw the relation ofvector spaces of the kind he considers, namely over subfields of C, and those ofthe type

R

n overR

. Dedekind also proves the now well-known result that given abasis B = {1, . . . n} for , then any system of n numbers in is a basis if andonly if the determinant formed from the coordinates of these numbers relative toB does not vanish. The use of linear algebra in Galois theory is a hallmark of thepresentation of Artin (1942), but it is recognized in the literature (e.g. Kiernan,1971) that this is in effect found already in Dedekind. For Dedekinds place in thehistory of vector space theory, see (Dorier, 1995).

The Proposition in effect shows that even if the notion of a flock is intro-duced through a form of representation, namely as all the numbers of the form

1in ii, a representation-free definition is possible, namely in terms of the

properties (1)(3). In light of Dedekinds decision for the inner against the outer,it is clear that he would prefer a definition of the latter kind, and this preferencemight be one reason why he found and proved this Proposition. Finally, the simi-

larity of the above Proposition with art. 48 in Was sind may be noted. From thedefinition of A0, the chain of the system A, Dedekind infers in articles 4547 ofthat work three properties, and these are then said in art. 48 to be characteristicof the notion of the chain of a system: if B is a system possessing these properties,then it is the chain of the system A (relative to the mapping in question).

Proposition. A flock over A is a field if and only if any product of its basiselements is again a member of the flock. If is a field, then every is algebraicover A and of degree at most n.

For the proof of the first assertion, if the product of two basis elements is againin , then is clearly closed under multiplication. Let 1, . . . n be a basis for .For any = 0 in , the numbers 1, . . . , n are over A, hence from the previousProposition, any

can be written as (11 +

+ nn) for i in A; thus

. The only-if direction is immediate.For the second assertion, let . By the previous Proposition the numbers

n, n1, . . . , , 1 are dependent over A, which is to say that is algebraic over Aand of degree at most n.

If there is a system ofn numbers in B which is irreducible over A, but no systemof n + 1 numbers in B which is irreducible over A, then we write:

[B : A] = n


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8 ANSTEN KLEV

and say that B is of degree n over A. If [B : A] = n for some n, then B is also saidto be finite over A.

Note. In the foregoing, Dedekind does not require that A be a divisor (i.e., asubfield) of B. Hence, his notion of a field Bs being finite over A is more generalthan that of being a finite extension, namely that of Bs being of finite dimensionas a vector space over A.

Suppose [B : A] = n, and let 1, . . . n in B be independent over A. Then it isreadily seen that the field AB is equal to the flock over A with {1, . . . n} as abasis. It follows from this that [AB : A] = [B : A].

Proposition. If B is finite over A, and C finite over AB, then BC is finite overA, and

[BC : A] = [C : AB][B : A].

This is clear from the readily verified fact that if 1, . . . , n is a basis for C over

AB and 1, . . . , m a basis for B over A, then the ij s form a basis for ABC overA. A consequence of this is what is now called the tower rule: if A B C, then

[C : A] = [C : B][B : A].

Another important consequence is that the numbers algebraic over A form a field:given and algebraic over A, the fields A() and A() are finite over A; thus bythe Proposition, their product A(, ), which contains the sum, difference, product,and quotient of and , is also finite over A, hence any A(, ) is algebraicover A.

165.

Proposition. Suppose B is finite over A and a permutation of A. Let

:= { : AB C | is a permutation of AB and A = }.Then

[B : A] = ||and, moreover, A = (AB) and =

.

In prose the first part of this Proposition says that the degree of B over A isequal to the number of permutations of AB which are multiples of . The proofis by induction on the degree. For a degree n > 1 there are two cases to consider,according as to whether or not there exists a C such that A C AB. In casesuch a C exists, then the Proposition follows from the tower rule and the inductionhypothesis. In case no such C exists, then some (any) B\A has degree n overA and we must have AB = A(). Since there are c1, . . . , cn A such that

(1)

n

= c1

n1

+ c2

n2

+ + cn1 + cn,it is clear that if = for some , then we must have(2) n = c1

n1 + c2n2 + + cn1 + cn,

where ci := ci. Conversely, suppose satisfies this equation (2). Now A() is aflock over A with basis {1, , . . . , n1}, that is for any A() there are uniquea1, . . . , an A such that

= a1n1 + a2

n2 + + an1 + an.


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Hence a permutation ofAB which at the same time is a multiple of is determinedby setting

(3) := a1n1 + a2n2 + + an1 + anwhere ai = ai. Indeed, it is then straightforward to see that we have ( + ) = + for , AB; moreover, from (1) and (2) and consideration of thedefinition (3), it is clear that n = n, whence one sees that () = ()(),and thence more generally that () = ()(). It remains to show that thereare indeed n distinct numbers satisfying the equation (2). Recall that is a multipleroot of a polynomial p(x) if and only if p() = 0, where p(x) is the derivative of

p(x). Since 1, , . . . , n1 form an irreducible system over A this condition is notfulfilled, hence we have || = n.

For the second part of the Proposition, set [AB : A] = n, [AB : (AB)] = p and[(AB), A] = q. Then, by the tower rule, n = pq, thus n p. Next, let :=

and

:= { : AB C | is a permutation of AB and (AB) = }.Then ||=p by the part of the Proposition already proved. Moreover, by thedefinition of , any is a multiple of , hence . Therefore p n,whence q = 1, so (AB) = A and moreover = as A = .

Note. Dedekind calls this theorem a Fundamentalsatz in light of its role in Galoistheory.

Proposition. Let be a system ofn permutations ofA. Then there are infinitelymany elements of A that are n-valued w.r.t. .

The proof is by induction on n. For 1, . . . , n and a A, define ai := ai.For n = 2, notice that if a1

= a2, then (aq)1

= (aq)2 for all q

Q. For n > 2,

we may assume, on the basis of the induction hypothesis, that there is some a Asuch that a2, . . . , an are all distinct, but a1 = a2. Let b A be such that b1 = b2and define y = ax + b for x Q. Hence we have

yr ys = (ar as)x + (br bs),whence one sees that y1 = y2 for all x and, indeed, that for each pair r, s there isat most one x for which y is such that yr = ys. Disregarding these xs one obtainsinfinitely many n-valued ys.

Note. In this proposition, which we have included here although it was proved in 161, Dedekind does not make use of his technical notion of an infinite set, definedin (Dedekind, 1888, art. 63). The following corollary was also given in 161.

Corollary. Given n permutations 1, . . . , n of a field A, then there are n numbersa, a, . . . , a(n) A such that the determinant formed by the numbers aij is non-zero.

For the proof, let y be n-valued w.r.t. the permutations 1, . . . , n. In the nota-tion of the previous proposition, let

D :=r


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One easily convinces oneself that D is equal to the determinant

(y1)n1 (y1)

n2 . . . y1 1

(y2)n1

(y2)n2

. . . y2 1......

. . ....

...(yn)

n1 (yn)n2 . . . yn 1

Hence one may put a(k) := y

nk.

Note. The determinant considered here is a so-called Vandermonde determinant,a determinant whose rows are the elements of a geometric progression. The proofthat it equals D is by induction on n, and is omitted here.

Suppose [AB : A] = n and let 1, . . . , n AB, and 1, . . . , n be the permuta-tions ofAB that leave A fixed. Consider the matrix

T :=

11 21 . . . n112 22 . . . n2

......

. . ....

1n 2n . . . nn

Denote its determinant by (T).

Corollary. The system := {1, . . . , n} is irreducible over A if and only if(T) = 0.

For if is irreducible over A, and given 1, . . . , n AB, any such i has theform

jn

ai,ji.

Let U be the matrix formed from these is in the same way that T is formed from

the is. Then, by matrix multiplication we have U = [ai,j ]T, where [ai,j ] is thematrix with ai,j in the i-th column and j-th row; hence, recalling that the determi-nant of a matrix product equals the product of the corresponding determinants, wehave for the corresponding determinants (U) = (ai,j)(T). If we now take the isas given by the previous Corollary, we have (U) = 0 and therefore also (T) = 0.For the converse, assume that is reducible over A. Hence there are non-zeroa1, . . . , an A such that

in

aii = 0.

Hence the matrix T is such that, by scalar multiplication of its i-th column by ai andsubsequent addition of the resulting columns with each other, one will obtain the 0matrix. By the properties of determinants it therefore follows that

in ai (T) = 0,

hence (T) = 0.

168.

Definition. A system a of real or complex numbers is called a module when a isclosed under subtraction.

Note. As the starting point of a very general consideration, the notion of modulewas introduced in Dedekind (1871). There it was defined as a system of real orcomplex numbers closed under subtraction and addition. Dedekind would then


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later have realized that the condition of closure under addition follows from closureunder subtraction, and hence left it out of the definition. As Dedekind says in 171below, the term module was modeled upon Gausss term modulus, introducedin article 1 of the latters Disquisitiones Arithmeticae (Gauss, 1801). Recall thattwo integers, and , are said to be congruent modulo an integer, , called themodulus, and one writes (mod ), when divides the difference . Insaid 171 below, Dedekind will extend this to a relation on C where modules takethe role of the modulus.

The notion of module current today, as a commutative group closed under scalarmultiplication with an underlying ring seems to occur for the first time in Noether(1921). A brief history of the notion of module can be found in section 15 of Moore(1995).

For any module a, we have 0 a; if a then a; if 1, 2 a then1 + 2 a; if a then n a for n Z. If T C, then

[T] := { : T},the system obtained by taking sums and differences of the elements of T, i s amodule. The system T is called a basis for the module, and the elements of T theelements of the basis. If T = {1, . . . n}, then we write [1, . . . n] for [T]. If the module a is equal to [1, . . . n] for some 1, . . . n a, then a is said to be afinite module, otherwise infinite. The system {0} is a module, the only module offinite cardinality. Other examples of modules are [1] = Z, [2] = [2, 6, 10] = 2Z, and[1, i] = Z[i].

Note. Dedekind writes 0 for the module {0}. This is in accordance with hispractice in Was sind, namely of writing s for the system consisting solely of theelement s, a practice which was criticized by Frege (1893, p. 2). Dedekind himselfwas aware of the danger of this practice, as is clear from Dedekind (1899). Judgingfrom that note, he seems to have sensed possible problems already around the timeof publication of Was sind (which, interestingly, Dedekind here dates to ChristmasEve 1887), for he mentions his plan to write a Selbstanzeige of Was sind for theGottinger gelehrten Anzeigen in which he would comment on this practice. Howeverthat may be, in the note (Dedekind, 1899), which was completed no earlier than1899 (and perhaps as late as 1911, in preparation for the 3rd edition of Was sind?),Dedekind shows how contradiction can ensue if one is not careful to distinguish asingleton system from its sole element; he also discusses there the empty system(not included in Was sind), or the null system, as he calls it, proves that it is uniqueand part of any system, and denotes it by 0.

169. Ifa b, then a is said to be divisible by b, and b a divisor ofa.

Note. Dedekind writs a > b or b < a for this relation, but we will avoid that hereand keep the -notation. Dedekind remarks in a footnote that the word divisibleas applied to modules has a sense directly opposite to what it has when appliedto fields, but this should be of no concern as confusion is avoided with a bit ofawareness. The motivation for this terminology will become clear below, but onemay notice right away that, for integers, m, n, we have [m] [n] if and only if ndivides m in the ordinary sense. Of course this does not explain why Dedekind saidof fields F G that F divides G.


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12 ANSTEN KLEV

For illustration, we have [4] [2] and {0} a C for any module a. We definea + b := { + : a, b}.

This module is the greatest common divisor of a and b, that is, the -least modulecontaining both a and b. The greatest common divisor of an infinite system ofmodules is the module with the union of these modules as basis:

iN

ai := [iN

ai].

Ifai aj whenever i < j, then [iNai] = iNai, which we may therefore denoteby a. For instance, ifan := [2

n], then a := {k2n : k Z, n N}. This is anexample of an infinite module.

The least common multiple of modules a and b is their intersection a b, whichindeed is a module. The least common multiple ofa and b is denoted by a b. Wehave a a = a, a b = b a, and (a b) c = a (b c). Moreover, ifa a andb

b then a

a

b + b.

Proposition. Ifm d and a is an arbitrary module thenm + (a d) = (m + a) d.

Since, by assumption, m d, it is clear that that the module on the left-handside is a subsystem of the one on the right-hand side. Suppose (m+a)d; then m + a, so = + for m, a; by the hypothesis of the Proposition,we have d, hence also = d, whence a d, and therefore = + m + (a d).

Let a, b, c be arbitrary modules. By setting m = b and d = b + c, the conditionof the previous Proposition is satisfied, and we obtain that, in general,

(a + b) (b + c) = b + (a (b + c)).

Likewise, setting m = b c and d = b yields(a b) + (b c) = b (a + (b c)).

Note. The foregoing equations, exhibiting a peculiar duality, whose underlyingreason may be difficult to comprehend,4 will feature prominently in Dedekindstheory of dual groups, or lattices, developed in (Dedekind, 1897, 1900). A dualgroup is defined in the former as A system A of whatever things , , , . . . onwhich is defined two operations , that from two things , produce two things likewise contained in A, and that at the same time satisfy the conditionsof commutativity, associativity, and absorption (cf. Dedekind, 1932a, p. 113). Itseems that Dedekind intended to ground the theory of modules in the more generaltheory of dual groups, presumably by viewing module theory as obtained from dualgroup theory by addition of the foregoing so-called modular laws

( + ) ( + ) = + ( ( + ))( ) + ( ) = ( + ( )).

For a more detailed account of the relation of module theory to dual group theory,see Mehrtens (1979, ch. 2.1) and Corry (1996, ch. 2.3).

4Dedekind (1932b, pp. 6566): Zwischen den Begriffen des groten gemeinsamen Teilers und

des kleinsten gemeinsamen Vielfachen beliebiger Moduln besteht ein eigentumlicher Dualismus,dessen l e t z e r Grund schwer zu erkennen sein mag.


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170. For two modules a, b, define their product ab to be the system of all sumsthe summands of which are products for a, b; that is

ab := {in

ii : n N, a, b}.

The products of two modules is a module. Ifb = [] for some C, then we setab := a. Multiplication of modules is associative and commutative. Moreover,there is an identity element: for any a, we have aZ = a, and Z is unique with thisproperty (in contemporary language, modules form a monoid under multiplication).In interaction with addition of modules relations such as the following obtain:

a(b + c) = ab + ac(a + b + c)(bc + ca + ab) = (b + c)(c + a)(a + b)

Division of modules is defined as follows:

b

a:= { C : a b}.

It is easy to verify that ba

is a module. Moreover, from the definition it is clear thatwe have

ac b c ba

.

The division ofa by itself is called the ordering (Ordnung) ofa and denoted by a0;that is,

a0 :=a

a= { C : a a}.

This is a module which is closed under multiplication: given , a0 and a,then a, whence a, therefore a0. Moreover, as one easily sees,Z a0. In fact, a converse holds as well: ifa is closed under multiplication and wehave Z a, then a0 = a, that is a is already an ordering; for, since aa a, we havea a

0

, and for any a0

, a a, in particular 1 a, whence a0

a.Note. What Dedekind calls an ordering is therefore a ring; in particular a0 is thering over which a is a module in the modern sense. It was Hilbert who introducedthe term ring for this notion in his (Hilbert, 1897, 31).

Exponentiation of modules is defined recursively from multiplication. One canalso define negative exponentiation:

an :=a0

an.

171. Given a module m, we say that the numbers and are congruent over m,written

(mod m),if m. Gausss notion is obtained by restricting and to integers andletting m = [] for an integer (as Dedekind will show in the next section, allsubmodules ofZ have the form []). Let (mod m); then, if (mod m)it follows that + + (mod m); and if m0, it follows that = (mod m).

Congruence over m is an equivalence relation on C, and hence partitions C intoresidue classes m + . Ifm a and a, then m + a, hence a is partitionedby residue classes m + . Ifa and b are arbitrary modules, then a b is a subset


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14 ANSTEN KLEV

ofa, so by the foregoing a is partitioned by residue classes (a b) + . In fact, for, a we have

(mod a b)if and only if

(mod b),as one easily verifies. Let {i}iJ be a system of numbers such that {i + (a b) :i J} is a partition of a, and such that i and j are incongruent over a bwhenever i = j. The system {i}iJ is then called a residue system, or a systemof representatives ofa over b, and we define

(a, b) :=

card(J) if J is finite0 otherwise.

It is easy to see that (a, b) = 1 if and only if a

b, and that, if c

b

a, then

(a, c) = (a, b)(b, c). Moreover, from the above we have (a, b) = (a, a b). Finally,it holds that (a, b) = (a + b, b); for the representatives {i}iJ are all elements ofa + b and incongruent over b, while every number in a + b has the form + for a, b, and is thus congruent to (mod b), whence it is also congruent toone of the representatives i.

Proposition. Let a, b be modules and a; then (a, b) 0 (mod a b), hence(a, b)a b.

Namely, let {i}iJ be a residue system of a over a b; then { + i}iJ islikewise such a residue system, hence we have

i + i

i i (mod a b), and

the Proposition follows by subtraction of

i on both sides. The following generalizes a theorem of ordinary number theory.

Proposition. The following simultaneous congruences

(mod a) (mod b)

have a common root if and only if

(mod a + b).In this case, the solutions form a residue class a b + .

If is a solution of the given congruences, then both and are membersof a + b, hence a + b as well. On the other hand, if a + b, thenthere are

a,

b such that

= + ; then :=

= + solves

the congruences simultaneously. Finally, if is a solution, then clearly + for a b is a solution as well; and if is another solution, then (mod a) and likewise (mod b), hence (mod a b).

If c a, and (a, c) > 0, then

card({b : c b a}) 2(a,c)1;for any such b is a union of residue classes c + a one of which is c + 0.


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172. Ifa = [1, . . . n], then a is said to be n-membered.

Proposition. Ifa is one-memberd, then b a is also one-membered, andb = (a, b)a.

Towards proving this, let a = [], that is a = Z, and set m := {k Z : k b}.Then m is a module, indeed m = [k0] where k0 is the least positive integer in m; forif m m, then m = pk0 + q with 0 q < k0, so as m pk0 m, we have q = 0,whence m = pk0. It follows that b = k0a, hence it remains to show that k0 = (a, b).But this follows from the fact that k0 = (Z,m) and the fact that m1, m2 a arecongruent over b if and only if m1, m2 Z are congruent over m.

It follows from this Proposition that any finite module a Q is one-membered;for, since a = [q1, . . . , qn], there is a positive integer k such that ka Z, whenceka = [m] for some m Z, and it follows that a = [ m

k].

We proceed to generalize the previous Proposition to n-membered modules.

Lemma. Let a and b be arbitrary, and m a one-membered module. Then there is

a one-membered module m such that

a (b + m) = (a b) + m.For the proof, let m = [] and set k := (m, a + b) = (m,m (a + b)). We then

have m (a + b) = km = [k] by the previous Proposition. Hence, k = ,where a, b. We show that [] is the module we seek, that is we show

a (b + m) = (a b) + [].: If x a (b + m), then x = = + l with a, b, l Z. Wetherefore have = l (a + b) m; the latter module was shown to equal[k], whence we get = nk = n( ) for some n Z; by which we have n = n a b. Hence x = = ( n) + n (a b) + [].: Clearly, a (b + m) (a b); and = k + for b, whence a (b + m). Proposition. Ifb is n-membered and a b, then a is n-membered as well.

The proof is by induction on n. The case n = 1 is covered by the previousProposition. Suppose b is n-membered, n > 1, and a b. Then we have adecomposition b = c+m, where m is one-membered and c is (n 1)-membered. Bythe Lemma there is a one-membered module m such that (c+m)a = (ca) +m.By the induction hypothesis, ca is (n1)-membered, hence a = ba = (c+m)a =(c a) + m is n-membered. Proposition. Every finite module has a basis which is irreducible over Q.

For suppose {1, . . . , m} is a basis for a which is reducible over Q. We mayassume {1, . . . , n} to be irreducible, but {1, . . . , n, s} for s > n to be reducible.Hence, for every j m j =

in

ejii,

with eji Q. We can find a positive integer k such keji Z for all i, j. Thus,

if we set i :=ik

, then a [1, . . . , n], and {1, . . . , n} is irreducible over Q.By the previous Proposition a = [1, . . . , n] for some 1, . . . , n, and these mustlikewise form an irreducible system over Q, for otherwise every n-membered systemof numbers in a would be reducible.


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16 ANSTEN KLEV

Note. This Proposition is not a trivial consequence of the results on flocks in 164 above, for the notion of basis is not the same in the two cases. In the case ofa flock the scalars are taken from some field A, whereas in the case of modules thescalars are integers.

Suppose that we have two modules with irreducible bases, a = [1, . . . , n] ando = [1, . . . , n], such that the following equations hold

(4) j =in

cjii,

where cji Z. Dedekind gives an algorithm for transforming the given basis for ainto a basis {1, . . . , n} satisfying equations of the following type

1 = a111

2 = a211 + a

222

......

...

n1 = an11 1 + an12 2 + . . . + an1n1n1n = a

n1 1 + a

n2 2 + . . . + a

nn1n1 + a

nnn

Each stage of the procedure takes one (irreducible) basis for a and yields another.The idea is simply to replace, e.g., 1 by 1 + k2 where k is such that c

1n + kc

2n 0, for some n, by the last Proposition of the previous section 172. Hence is integral.

We say that a finite module a is a folder of the number if a0.Corollary. A number is integral if and only if it possesses a folder.

For if is integral then ()n is a folder of for suitable n, and to say that possesses a folder is to say that a0 for a finite a.


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18 ANSTEN KLEV

Note. In an earlier proposition, Dedekind proves the last part of this Corollarythat if possesses a folder, then it is integralby means of the theory of deter-minants. But he notes immediately that the latter theory lies quite far from theproper content of the proposition in question, whereupon he goes on to state andprove the above Proposition. With its application of the ascending chain condi-tion, that Proposition may indeed be said to lie quite far apart from the theory ofdeterminants.

Note that ifa is a folder of and b is a folder of , then ab is finite and indeeda folder of both and , that is a common folder of these two numbers.

We are now in a position to see that the algebraic integers satisfy some importantclosure properties.

Proposition.

(i) The algebraic integers are reproduced through multiplication, addition, andsubtraction.

(ii) If satisfies the equation:

n + p1n1 + + pn1 + pn = 0

with p1, . . . pn integral, then is integral.(iii) If A is integral and : A C is a permutation, then is integral as

well.

To prove (i), let , be integral. Hence they possess a common folder a; that isto say , a0. Hence , + , a0, thus these are all integral by theprevious Proposition.

For (ii) it is sufficient to prove that possesses a folder. The integers p1, . . . pnpossess a common folder a. We claim that a()n is a folder of . Note first that as()n

()n+1, we have

(6) a()n a()n+1 = a()n + an.But if a, then as n = 1inpini and pi a, we have n a()n,whence, an a()n, and so in light of (6), a()n is indeed a folder of .

Given the assumption in (iii), satisfies some equation

n + k1n1 + + kn1 + kn = 0,

for ki Z. Since is the identity mapping on Q, it follows that()n + k1()

n1 + + kn1() + kn = 0,witnessing that is integral.

Note. Part (i) of this important Proposition was something Dedekind earlierproved by means of the theory of determinants, for instance, in Dedekind (1877, 13). Now it is rather a consequence of the more general theory of modules. Thus,cf., Dedekind (1877, 12, p. 102):

It is preferable, as in the modern theory of functions, to seek proofsbased immediately on fundamental characteristics, rather than oncalculation. . .


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Note that, from (ii) above it follows that the n-th root n

of an algebraic integer is again an algebraic integer.

The Lemma below will become important in 177 below.Lemma. Ifm = 0 is a finite module consisting of algebraic numbers, then there isa module n such that

(1) the members of n are obtained by rational operations on the members ofm;

(2) the module mn consists solely of integers;(3) we have Z mn.

Its proof is by induction on the number of elements in the basis of m. Form = [] it suffices to take n := [1]. For the case of two basis elements, m = [, ],Dedekind defines numbers i through rational operations on and and showsthat the module n = [1, . . . , n] satisfies the required properties. The details are

omitted here. For the case of m > 2 basis elements one makes use of the modularidentity

(a + b + c)(bc + ca + ab) = (b + c)(c + a)(a + b).

Partition the basis ofm into three non-empty sets obtaining modules a, b, c suchthat m = a+ b+ c. By the induction hypothesis, there are modules a, b, c whoseelements are obtained through rational operations on numbers ofm and such thateach of the three modules (b+ c)a, (c+a)b, (a+b)c, and hence also their product(cf. the above identity)

m(bc + ca + ab)abc

contain integers only and moreover is a superset ofZ. Hence

n := (bc + ca + ab)abc

is the required module.

174. If and are (algebraic) integers, then is said to be divisible by ,written | , if there is an integer such that = . Using the fact that theintegers are closed under multiplication, addition, and subtraction, one sees that if and are divisible by , then is divisible by , and if is divisible by and by , then is divisible by . A number is said to be a unit if it divides 1 (thatis, if it has an inverse in the integers). The units are closed under multiplication,division, and root extraction.

If there is a unit such that = , then and are said to be associated.Associated numbers behave the same with respect to divisibility: if is divisibleby , then all numbers associated with are divisible by , and is divisible byall number associated with .

Two integers , are said to be relatively prime if there are integers and such that

+ = 1.

From this definition it follows that if is relatively prime to as well as to thenit is also relatively prime to the product . Moreover, if and are relativelyprime and divides , then divides . Finally, if is a common divisor of and , then is a unit.


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20 ANSTEN KLEV

Note. Dedekind notes of his definition of relative primality that it is somewhatformal and therefore perhaps not to be marked as the best. It is not clear to mewhat he here means by formal. In any case, he says that a simpler definition isforthcoming. Namely, in 181, he will prove that if the only divisors of and areunits, then and are relatively prime; together with the last fact noted abovethis makes it possible to define and are relatively prime if and only if theironly common divisors are units, in line with the ordinary definition.

Note. Dedekind calls a field A finite if there is an n N such that for all Athere is an equation of degree n with coefficients in Q of which is a root. Thecurrent section ends with Dedekind motivating why the focus in what follows willbe on finite fields. Namely, in the system of algebraic numbers every element canbe decomposed into non-units in an infinite amount of ways (recall that the root ofan algebraic integer is again such an integer). Dedekinds main interest, however,is in questions of decomposition, and such questions are non-trivial only for finitefields.

175. Let A be a finite field, that is a field such that [A : Q] = n for somen Z. Let = {1, . . . , n} be the system of permutations : A A extendingthe identity mapping on Q (which is the only permutation of Q). We define thefollowing numbers

S() :=

(), the trace of

N() :=

(), the norm of

(1, . . . , n) :=

11 21 . . . n112 22 . . . n2

......

. . ....

1n 2n . . . nn

2

The latter is called the discriminant of 1, . . . , n, and notice that it is the squareof the determinant. By the elementary properties of determinants one has

(7) (1, . . . , n) = (N())2(1, . . . , n)

If , then , that is is closed under composition. It follows,since these are permutations, that, for any we have (S()) = S(),(N()) = N(), and ((1, . . . , n)) = (1, . . . , n). Therefore, by the firstProposition of 165 above, we have S(), N(), (1, . . . , n) (A) = Q. Inother words, the norm, the trace, and the discriminant are always rational, more-over the discriminant, being a square, is 0.

Suppose now that {1, . . . , n} is a system of elements of A irreducible over Q.From the properties of flocks proved in 164 it follows that they also form a basisfor A over Q. Let a := [1, . . . , n], the module generated by these basis elements.Suppose that {1, . . . , n} is likewise a basis for a. Then equations of the followingkind obtain, where the coefficients cji Z:(8) j = c

j11 + c

j22 + . . . + c

jnn.

Letting (cji ) be the determinant formed from these coefficients, and noting that all

permutations leave the cji s fixed, it follows from the elementary properties


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of determinants that

(1, . . . , n) = (cji )

2(1, . . . , n).

But from the last Propositions of 172, we have (cji ) = (a, a) = 1. Hence, thediscriminant (1, . . . , n) is the same for all irreducible bases for a, and thus wemay set (a) := (1, . . . , n).

For what follows it is well to recall from 172 that any finite module has anirreducible basis. Let b be a module in A with an irreducible basis {1, . . . , n}such that equations of the type (8) obtain (hence, a b). Then, by the lastProposition of 172, we have

(a) = (b, a)2(b).

Next suppose that b is any module in A with an irreducible basis {1, . . . , n}. Asbefore, equations of type (8) will obtain, but with c

ji Q. There is an m Z

such that mcji Z for all j, i, and for this m one has mi a b for all i. Hence

a b contains n independent numbers, namely mi, whence any irreducible basisfor a b must consist of n elements, and so (a b) is defined. Indeed, recallingthat (a, a b) = (a, b), one sees that

(a b) = (a, b)2(a) = (b, a)2(b).Hence we get

(b, a)

(a, b)=

(a)

(b)= (cji )

The special case ofb = a will be of interest later. In light of (7) above, one seesthat

(9)(a, a)

(a, a)=

(a)

(a)= N()

Let o be the class of algebraic integers in A. From the closure properties of suchintegers it follows that o is closed under subtraction and multiplication, and sinceQ A, we have Z o, whence o is an ordering (cf. 170).Proposition. The class of algebraic integers o in a finite field A is a finite modulewhose basis is also a basis for A.

To prove this fundamental Proposition, let {1, . . . , n} be a basis for A. Foreach i there is a ki Z such that i := kii o (cf. 173), and clearly{1, . . . , n} is a basis for A. Consider the module a := [1, . . . , n]. Ifa = o, thenwe are done; otherwise, let o a. As A, there are m, m1, . . . , mn Z,m > 0 relatively prime such that

=m11 + m22 +

+ mnn

m

Let b := a + []. It is then clear that (b, a) = m. Moreover, b being a finitemodule containing 1, . . . , n has an irreducible basis of n elements, whence (b)is defined. Indeed,

(a) = m2(b),

in particular (b) < (a). Therefore, the process of adding elements to a in theabove manner will come to an end after finitely many steps by which time one


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22 ANSTEN KLEV

has reached o, which is therefore a finite module, and any irreducible basis for itcontains n elements, and which is therefore also a basis for A.

176. We fix a finite field A. Let o be the module of integers in A. If o isdivisible by o, that is if = for an integer , then o, since A.Lemma. The norm N() of a number o has the following properties.

(1) | N()(2) N() = N()N()(3) If | then N() | N()(4) N() = 1 if and only if is a unit

Properties (1)(2) are immediate from the definition of the norm, and (3) followsfrom these. For property (4), suppose first that N() = 1. Then | 1, which isto say that is a unit. Conversely, suppose that is a unit. Since N(1) = 1 itthen follows from (2) that N() | 1. Since N() Z, we have N() = 1.

An integer o is said to be decomposable if there are integers ,

o which

are not units, not associated with , and are such that = . In the contrary case is said to be indecomposable. Now if o is decomposable with = , thenthe absolute values of their norms satisfy |N()|, |N()| < |N()|, by the Lemma.Hence the decomposition of must reach an end:

Proposition. Let o be the class of integers in a finite field A. Then any o canbe written as a product of indecomposable numbers from o.

From elementary number theory we know that in the case of Z, this decompo-sition is essentially unique, that is unique up to the order of the factors and thesubstitution of associated numbers. The same result holds for Z[i]. On the otherhand, since

6 = 2 3 = (1 + 5)(1 5)and 2,3, 1 + 5, and 1 5 can all be seen to be indecomposable in Z[5],it follows that unique decomposition does not hold in this domain.

An integer o is a prime number if it is not a unit and is such that( and ) .

If is decomposable then = where and , hence is not prime inthis case; in other words, all prime numbers are indecomposable. The inclusion inthe other direction is equivalent to unique factorization:

Proposition. Every number o has a unique factorization if and only if allindecomposable numbers in o are prime.

For, firstly, if is not prime, then there are , such that , but | .Hence, if in addition is indecomposable, then the number = = must

have two essentially different decompositions, since is indecomposable and notassociated with any factor of either or . Conversely, if any indecomposablenumber is prime, the proof of uniqueness in Z (or in Z[i]) adapts to show that ohas unique factorization.

Hence, if one wants to regain unique factorization it is natural to see whetherone can in some way introduce ideal prime numbers so that all indecomposablenumbers are also prime. Dedekind now outlines, within the framework he hasdeveloped, the ideas that guided Kummer in his introduction of such ideal numbers.


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Assume that we are in a module of integers o in which unique factorization fails.Then we have a number o which is indecomposable in o but not prime, hencethere are ,

o such that

and

, but

|. This gives rise to

two essentially different factorizations of the same number = and = .Consider the contrary case where Z. Here, as one readily sees, we would havenumbers 1, 2, 1, 2 Z, all non-units, with gcd(i, i) = 1 and

= 12, = 12, = 12, = 21.

Notice now that for any Z we have1 | | | .

Hence we can determine whether the factor 1 of divides a number Z sim-ply by consideration of the initially given numbers ,,,. Returning to o, byassumption there are no 1, 2 o satisfying the above, since is assumed tobe indecomposable. But one can introduce ideal numbers of this kind and definewhat it means for a number in o to be divisible by such an ideal number simplyin terms of numbers originally in o. Then one can say that an ideal number isprime if whenever it divides a product, it divides one of the factors. We do notneed to define an arithmetic of these ideal numbers, in particular we do not need todefine multiplication of ideal numbers, but we can introduce powers of prime idealnumbers (1)

r with divisibility conditions of the following type

(1)r | r | r r | r

In this way unique factorization can be regained; it can be shown that every number o, in virtue of the divisibility relations that obtain, can be viewed as a productof powers of ideal or actual prime numbers (1)

r1 , . . . , (m)rm such that for any

o, if | then (i)ri | for all i.Note. For the working out of these ideas in a concrete example, the reader is

referred to Dedekind (1996, 712). 177. For , o we have

| o o.Thus divisibility of numbers in o is mirrored by divisibility (inclusion) of modulesof the form o for o. At the end of the previous section we saw that divisibilityof numbers in o by ideal numbers is equivalent to a condition of the type

| .Numbers satisfying this conditions form a module m o with the additionalproperty

o m m.It is easily seen that modules of the form o also have this property. These consid-erations motivate the following definition.

Definition. A module m o is said to be an ideal ifm = 0, and om m.Note. The similarity between Dedekinds treatment of ideal numbers and his treat-ment of irrational numbers is of course rather striking. The starting point in bothcases is the same: a certain number domain is incomplete in some sense. Thus inStetigkeit und irrationale Zahlen (Dedekind, 1872, p. 322), it is remarked:


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24 ANSTEN KLEV

The above comparison of the domain R of rational numbers with aline has lead to the recognition of the gappiness (Luckenhaftigkeit),incompleteness, or discontintuity of the former, while we ascribe tothe line completeness, gaplessness (Luckenlosigkeit) or continuity.

The current section 177 of the XIth Supplement opens with the following remark:The example treated of right now suggests that the characteristicgaps that emerge in the investigation of the divisibility of the num-ber within the domain o, and that let one recognize a certainincompleteness in said domain, could be filled by the introduction,not of individual numbers , but rather of whole systems of suchnumbers.

These gaps (in the former, order-theoretic, in the latter, arithmetical) are filled bythe introduction of certain sets of numbers from the original gappy domain. Inboth cases these sets are of two kinds, one corresponding to elements in the originaldomain, and the other to the essentially new elements. Thus, in the case of cuts,there will be those cuts ({p Q : p < q}, {p Q : p q}) generated by rationalnumbers q, and others not thus generated, corresponding to irrational numbers; inthe case of ideals there will be those ideals o generated by elements of o andother not thus generated, corresponding to ideal numbers.

From the last Proposition of 175 we know that o is a finite module with anirreducible basis which is also a basis for A. Hence, any ideal m o is a finitemodule; in fact, since om m, there are n independent numbers in o hence mhas an irreducible basis of n integers. Since Z o, one sees that om = m. Theproduct of two ideals a and b is an ideal since firstly ab a b o, and secondlyo(ab) = (oa)b = ab.

Recall from 170 that for a module a we defined

a

0

:= { C : a a}a1 := { C : a a0}A module a is said to be proper if aa1 = a0. From these definitions one sees bycomputation that a is proper if there is a module b such that a0 = ab; for in thatcase we have

a0 = a0a0 = (ab)a0 = a(ba0),

hence ba0 a1; on the other hand,a1 = a0a1 = (ab)a1 = b(aa1) ba0.

Thus a1 = ba0, and so

aa1 = aba0 = a0a0 = a0

Proposition. An ideal m is a proper module and m0 = o.

For, as m is a finite module there is, according the Lemma proved in 173,a module n such that Z mn o. Multiplying with o we get a squeezingo mno o, hence mn = o. Now, as mm0 = m, we get

om0 = mnm0 = mn = o,

hence m0 o. And since m is an ideal we have o m0. Hence m0 = mn, and so itis proper by the above.


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Corollary. Given ideals m a there is an ideal b such that m = ab, namelyb := ma1.

For as a and m are ideals, we have by the foregoing ab = aa1m = om = m.So it remains to see that b is indeed an ideal; but, firstly, since m a, we haveb = ma1 aa1 = o; and, secondly, ob = oma1 = ma1 = b, hence b is indeedan ideal.

Note. With this Corollary, which is used again and again in the following, Dedekindhas established equivalence between the notions of being a factor and that of beinga divisor in the domain of ideals. Let us say that a module a is a factor of anothermodule m if there is a module b such that m = ab. Let us say that a module a isa divisor of another module m ifm a. We have already seen that for ideals a, b,their product ab is a again a module, and a is one of its divisors, since ab a. TheCorollary shows that if an ideal a is a divisor of an ideal m, then it is also a factorofm, and indeed, there is another ideal b such that ab = m.

Establishing this connection, Dedekind points out in a footnote to the Corollary,was the greatest difficulty yet to be overcome in the ideal theory presented inDedekind (1877). Indeed, in 23 in the latter work one reads (Dedekind, 1996, p.126)

It is very easy to see that any product of a by an ideal b is divisbleby a, but it is by no means easy to show the converse, that each idealdivisible by a is the product ofa by an ideal b. This difficulty, whichis the greatest and really the only one presented by theory, cannotbe surmounted by the methods we have employed thus far, and it isnecessary to examine more closely the reason for this phenomenon,because it is connected with a very important generalization of thetheory.

In said footnote to the Corollary, Dedekind also points out the connection with a

generalization of ideal theory (I believe the generalization in question is the theoryof ideal classes).

Ideals of the type o are called principal ideals. It follows from the Corollarythat for any ideal a there is an ideal b such that ab is principal; for take any a;then o a, so there is an ideal b such that ab = o.

In light of the Corollary above, let us say that a factor of an ideal a is any idealm such that a m o.Proposition. The number of factors of an ideal is finite.

For given an ideal a, let us take some non-zero a. Then, by equation (9) in 175 above, we have (o, o) = N() > 0. As was noted at the end of 171, thereare therefore finitely many modules b such that a

b

o. But a

a

o, and

the Propositions follows. 178. If and a and b are ideals, then so are a b and a+ b, as one readily verifies(for the non-emptiness of a b recall that ab a b). For a class of modules ailet us write iai instead ofa1 a2 a3 . . ., +iai instead ofa1 + a2 + a3 + . . ., and

i ai instead ofa1a2a3 .Proposition. Given ideals ai, there are ideals bi such that +ib1 = o, and suchthat for any index j, iai = ajbj .


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26 ANSTEN KLEV

For, setting m := iai, and noting that m aj it follows from the Corollary ofthe previous section 177 that there is an ideal bj such that m = ajbj . Settingn := +

ib, and noting that b

j n, it follows from the same Corollary that b

jn1 is

an ideal. Hence mn1 = ajbjn1 aj , and therefore mn1 m, whence m mn,

so by multiplication with m1 we get n = o. In particular, given two ideals a and b, there are ideals a and b such that

a b = ab = ab

a + b = o

With a little bit of computation one can see from this that

(a + b)(a b) = ab,which is the ideal theory correspondent to the number-theoretic proposition thatgcd(m, n) lcm(m, n) = mn.

Two ideals a, b are said to be relatively prime ifa + b = o. It is then straight-forward to check that the following obtains.

(1) Ifa and b are relatively prime, and is c any other ideal, then a+bc = a+ c.(2) If a is relatively prime to both b and c, it is also relatively prime to the

product bc. This generalizes to more factors.(3) If a and b are relatively, then so are any factors a and b of a and b

respectively.(4) Ifai is a family of relatively prime ideals, then

i

ai =i

ai.

In particular, (4) is proved by induction: Suppose that {a} {bi}in is a family ofrelatively prime ideals. Then

ibi

= ibi

by the induction hypothesis, moreoverone sees that a+ibi = o. Hence (aibi) = (a

i bi)o = (a

i bi)(a+

i bi) =

a

i bi.

Lemma. If the ideal a is not a subset of any of the ideals c1, . . . , cn, then there isa number a, such that / 1in ci.

The proof of this Lemma is by induction on n, and is immediate in the case ofn = 1. So let c1, . . . , cn be given such that a ci for all i. Since a ci a, thereis, by the Corollary of the previous section 177, an ideal bi such that a ci = abi,and as a ci, we have abi = o. We show that there is a number a (iabi).There are two cases to consider. In the first case, assume that b1 + b2 = o. Thena a(b1 + b2), hence by the induction hypothesis there is an a such that /

a(b1 + b2)

i>2abi whence it follows that /

i1abi. In the second case,

assume that all the bis are relatively prime. Then (bi +

j=i bj) = o. Setting,bi :=

j=i bj , we have b

i bi since bi = o; hence abi abi. Let i abi abi

and define :=

i i. Then a, so it remains to show that / abi. Nowbj bi for j = i, so abj abi, whence j abi for j = i; on the other hand, bythe choice of i, we have i / abi, and therefore / abi. Proposition. Let a and b be any ideals. Then there is a number o such thatab + o = a.


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The claim of the Proposition is clear when b = o, so suppose otherwise. Thenab a. Let J be the system of modules c such that ab c a. By the lastProposition of

177, J is a finite system. Hence, by the Lemma there is an

a

such that / J. Hence, ab + o = c for any c J. On the other hand we haveab ab + o a. Hence, by the definition of J, ab + o = a.

As a consequence, we have that any ideal a is the sum of two primary ideals: Fortake some a; then there is an ideal b such that o = ab, so by the Propositionthere is an such that o + o = a.

179. An ideal p is said to be a prime ideal if, for any ideal a, p a implies a = por a = o. In the terminology of divisibility: a prime ideal is an ideal whose onlydivisors are o (=1) and p itself. An ideal = o is said to be composite if it is notprime. Ifp is prime and a any ideal, then p a + p, whence either a p or elsea + p = o, that is to say, a and p are relatively prime. Hence, given a family ofideals ai and a prime ideal p such that, for all i we have ai p, then ai + p = o,

and therefore

i ai + p = o. Contrapositively, if

i ai p, then for at least one iwe have ai p. In the language of divisibility: if a prime divides a product, then itdivides at least one of the factors. Now suppose that m is a composite ideal. Thenthere are a, b = o such that m = ab; we therefore have a, b m, and so there is an a and a b such that , / m but, of course, m.Note. The definition Dedekind gives of a prime ideal is the one that today is said todefine maximal ideals. A prime ideal, on the other hand, is in contemporary algebraan ideal a for which a implies a or a. Even though every maximalideal is prime, it is not in general the case that every prime ideal is maximal. Forthe rings Dedekind considers, however, the two notions coincide. Indeed, Dedekindproved in the foregoing that every prime ideal in the modern sense is a maximalideal, that is a prime ideal in Dedekinds sense; for he shows that every compositeideal m contains a product such that , /

m. The standard proof that every

maximal ideal is prime in the modern sense looks at the quotient structure R/J forthe ideal J; it is proved that R/J is a field if and only if J is maximal, and thatR/J is an integral domain if and only ifJ is prime. Since every field is an integraldomain, it follows that every maximal ideal is prime. Now Dedekind does notconsider quotient structures in these investigations, so it is natural to ask whetherthere is a more elementary proof in the current setting that every maximal ideal isprime. The following purports to be one. Suppose that m is not prime in the modernsense, that is, that there are , / m for which m. We want to prove thatm is not maximal. And indeed we have m m + o o. The first strict inclusionis clear in light of the fact that / m. For the second strict inclusion, supposeotherwise that m+o = o. Then (mo) = (mo)(m+o) = m(o) = (mo)m.By assumption we have m o, and thus also m. Hence there is a m such that = , and therefore m contrary to the assumption. Lemma. Any ideal a = o is divisible by at least one prime ideal.

This is clear ifa is itself a prime ideal, so suppose that a is composite. It thenpossesses a factor b = o. Now if this factor b is prime, we are done; if, on the otherhand, b is not prime, then it possesses a factor c, which is either prime or itselfcomposite. Continuing in this way we obtain a chain (Kette) a b c ofdistinct ideals = o each of which is a factor ofa. Since there are only finitely many


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28 ANSTEN KLEV

factors ofa, this chain must be finite, and by construction it must end in a primeideal dividing a.

Proposition. Any ideal a = o has a unique factorization into primes.By the Lemma there is a prime p1 dividing a, hence by the Corollary of 177, we

have a = p1a1. Ifa1 = o, then a is itself prime and serves as its own factorization,which obviously is unique. If a is not prime, then a1 = o, and so a1 is dividedby a prime ideal p2, that is a1 = a2p1, where a2 is either o or divisible by prime.Continuing in this way we obtain a chain a a1 a2 of distinct ideals eachof which is a factor of a. Accordingly, this chain is finite, and with the chain weobtain a factorization

(10) a =

1in

pi.

To see that this factorization is unique, note first that if p divides a, then we musthave p = pi for some pi occurring in the factorization (10). Note secondly thatif e is the number of times the prime p occurs in the factorization (10), then wehave a = bpe where p does not divide b; hence if we had bpe = (cp)pe, then bymultiplication of pe we would have had b = cp, and therefore b p contrary tothe assumption. Hence pe does, but pe+1 does not divide a. This means that thefactorization (10) determines both which primes divide a, and how many times sucha prime divides a, and therefore this factorization must be unique.

180. We define the norm of an ideal a to be

N(a) := (o, a).

This definition is motivated by the fact that (o, o) = N(), and the fact that forany m we have o m o, and thus (o, o) = (o,m)(m, o). In consequence,N(a) is always > 0 and hence counts the number of residue classes a +

o.

With N(m) = n, let R := {1, . . . n} be a residue system over m. Note that if(11) (mod m),then

(12) m + o = m + o;

for if m, then we have m + o, and likewise m + o. Conversely,if (12) holds, then = + for some m, whence (11) holds. The system{1, . . . n} therefore corresponds to n ideals m + oi, and these are independentof the choice of residue system. Given this, we can define the number

(m) := card({i R : m + oi = o}).A number is said to be relatively prime to a module m ifm and o are relatively

prime, that is ifm+o = o. Hence, (m) is the number of representatives i whichare relatively prime to m. When o = Z the function coincides with Eulers totientfunction: any ideal m Z is of the form [], hence for integers , we have (mod m) if and only if (mod ); and for any integer such that 0 < ,we have m + o = o if and only if + = 1 for some , Z, and this lattercondition is equivalent to and s being relatively prime.

When m is an ideal, congruences may be multiplied; that is, if

, (mod m),


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then

(mod m).

Now let m = ab. By the last Propositions of 178 there is then a number suchthat a = m + o. Since o a, there is an a such that o = aa, and this ais relatively prime to b; namely, a(a + b) = aa + ab = m + o = a, whence bymultiplication with a1 we have a + b = o. This gives the possibility of division ofcongruences as follows:

(mod m) (mod b).For the left to right direction, note that if ( ) m, then o( ) m,that is aa( ) ab, and therefore

o( ) = (a + b)( ) = a( ) + b( + ) b.For the right to left direction, if (mod b), then (mod b), thusalso (mod m)

Again let m = ab, and let {1, . . . n} be a system of residues over m. Wewant to find the number of representatives j such that m + oj = a. By the lastProposition of 178 we know that there is at least one i such that m + oi = a.We have m + oi = m + oj if and only if there are ,

such that

i j (mod m), j i (mod m).In case this obtains, then i

i (mod m), hence by division, 1 (mod b),

and this is equivalent to s begin relatively prime to b. On the other hand, supposethat is a number relatively prime to b and let j be the representative such that

(13) j i (mod m).Given that is relatively prime to b, there is an such that

1 (mod b),

and this yields i i (mod m), thus(14) i j (mod m).The congruences (13) and (14) together yield m + oj = m + oi = a. Hence wemay find all representatives j with m + oj = a by setting j i (mod m) forevery o which is relatively prime to b. By division,

i i (mod m) (mod b),hence there are just as many residue classes m + i as there are residue classesb + such that is relatively prime to b, that is to say, this number is equal to(b). Now we may partition the residue system {1, . . . n} by relating i andj if and only if m + oi = m + oj . For each i there is an ideal b such thatm = (m + oi)b and the foregoing shows that the corresponding equivalence class

[i] contains (b) elements. On the other hand, for each b m there is at leastone i such that b = m + oi. Hence if we sum over all such (b) we will get n,that is N(m). In symbols, we have

ma

(a) = N(m).

This lifts the well-known fact about Eulers totient function that

k|n (k) = n.

The Chinese Remainder Theorem also knows an ideal theoretic version:


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30 ANSTEN KLEV

Proposition. Let ai be relatively prime ideals. Then there is a number satisfyingthe following simultaneous congruences

1 (mod a1), 2 (mod a2), 3 (mod a3) . . .Moreover, these solutions form a residue class modulo the product

i ai.

For two ideals a1, a2 the Proposition is immediate from the last Proposition of171, in view of the fact that a1 a2 = a1a2 and a1 + a2 = o. The general case nowfollows by induction, noting that if a, b, and c are relatively prime then so are aband c.

It follows from this Proposition that there is a one-one correspondence betweenthe residue classes

i ai+ and the n-tuples of residue classes (a1 +1, a2 +2, . . .).

Now if is relatively prime to

i ai, then is relatively prime to each of the factorsai; and if i is relatively prime to ai for each i, then the Proposition gives an relatively prime to each of the ais, and hence relatively prime to their product

i ai. It follows that(i

ai) =i

(ai)

This corresponds to the fact of ordinary number theory that the totient functionis multiplicative. Combined with unique decomposition into prime powers, thisequality reduces the problem of computing (a) to computing (pn) for primeideals p. Now any number is relatively prime to pn if and only if / p. Thus,to compute (pn) one has only to subtract (p, pn), that is the number of residueclasses pn + p, from N(pn), the total number of residues classes pn + . Since(o, pn) = (o, p)(p, pn), we get the formula

(pn) = N(pn)(1 1N(p)

).

Again, this corresponds to the formula computing the totient function for primepowers in ordinary number theory.

Finally, Eulers generalization of Fermats Little Theorem may be lifted to idealtheory; this theorem says that () 1 (mod ) for relatively prime integers, Z. Namely, let be relatively prime to m, with (m) = n and let {1, . . . n}be those elements of a given residue system that are relatively prime to m. Thenthe products i are likewise all relatively prime to m and incongruent with eachother. Hence

i i

i i (mod m). Since the product

i i itself is relatively

prime to m, it has an inverse modulo m, and multiplying by this inverse we get

(m) 1 (mod m).

References

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y permute de toutes les manieres possibles les quantites quelle renferme. Journal de l Ecole

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supplement xi synopsis

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