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AN EXPLORATION OF MINKOWSKI THEORY AND ITS
APPLICATIONS
DANIEL COMEAUX
Abstract. This is a paper that examines the area of number theory laid outby Herman Minkowski in his explorations of the ”geometry of numbers,” herereferred to as Minkowski Theory. It assumes the reader will have a basicfamiliarity with algebraic concepts, such as groups, rings, fields, extensions,etc. If the reader is unfamiliar with these, Dummit and Foote is a useful ex-planatory resource. After a discussion of the necessary preliminaries (algebraicconcepts, gaussian integers, integrality, ideals, and lattices), we establish theframework of Minkowski’s Lattice Point Theorem, as well as its implicationsin Minkowski Theory. We then use these results to prove the finiteness ofthe ideal class group and Dirichlet’s Unit Theorem, and as a conclusion makea logical jump to consider the extension of Dedekind domains. This paperis based on Algebraic Number Theory by Jurgen Neukirch, but provides ad-ditional examples and expands upon the often dense proofs presented in hisbook. Explanations of concepts were drawn from Dummit and Foote, Lang,and Ash.
Contents
1. Preliminaries 21.1. Basic Algebra 21.2. Relevant Algebra 22. Gaussian Integers 33. Integrality 64. Ideals 74.1. Preliminaries 74.2. Results 95. Lattices 106. Minkowski Theory 137. Class Numbers 177.1. Preliminaries 177.2. Results 188. Dirichlet’s Unit Theorem 189. Extending Dedekind Domains 2010. Acknowledgments 21References 21
Date: August 24, 2012.
1
2 DANIEL COMEAUX
1. Preliminaries
1.1. Basic Algebra. Readers comfortable with algebraic concepts such as mod-ules, embeddings, and extensions may skip to subsection 1.2.
Definition 1.1. Consider a ring R. From now on, assume than any such R containsthe nontrivial multiplicative identity, i.e. 1 6= 0, 1 2 R. An R-module (technicallya left R-module, but we will only consider left and not right modules) is a set Mpaired with some binary operation + on M under which M is an abelian group,as well as some map R ⇥M ! M denoted (r,m) 7! rm for all r 2 R and for allm 2 M with the following conditions:
(i) (r + s)m = rm+ sm for all r, s 2 R,m 2 M(ii) (rs)m = r(sm) for all r, s 2 R,m 2 M(iii) r(m+ n) = rm+ rn for all r 2 R,m, n 2 M(iv) 1m = m for all m 2 M
Note that this object is actually a more general form of the familiar notion ofthe vector space, itself a special case of modules where the foundational ring isactually a field. Just as in our considerations of vector spaces, we can considersubsets of modules that inherit the properties of their parent modules, denoted assubmodules.
Definition 1.2. With R a ring and M an R-module, a subgroup N of M whichis closed under the action of ring elements (i.e. rn 2 N , for r 2 R and n 2 N) isknown as a R-submodule of M .
Other terms that are likely familiar to most but may be unfamiliar to someinclude:
Definition 1.3. The field F is known as an algebraic closure of F if both F isalgebraic over F and F contains all elements algebraic over F .
Definition 1.4. We say that a field K is algebraically closed if each polynomialwith coe�cients in this K has a root in K.
Definition 1.5. Consider two fields K and L. An embedding of K in L is ahomomorphism � : K ! L. A K-embedding is an embedding of L in the closure ofK which maps any element of K to itself.
Definition 1.6. A separable extension of a field is a field extension K|F in whichthe minimal polynomial for each element of K is separable over F , i.e. this poly-nomial has no multiple roots (as they are all distinct) in an algebraic closure of K.See [3] pp. 178, 243.
1.2. Relevant Algebra. In this paper, we will often consider the trace, norm,and discriminant of various elements. These operations are likely familiar, butthe interpretations we will work with may be novel to the reader. For clarity, thedefinitions are listed below.
Definition 1.7. Considering the field extension L|K, the trace of an element x 2 Lis defined to be the trace of the endomorphism
Tx
: L ! L with Tx
(↵) = x↵
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 3
of the K-vector space L:
(1.8) TrL|K(x) = Tr(T
x
).
Fact 1.9. If L|K is a separable extension and � : L ! K varies over di↵erent K-embeddings of L into an algebraic closure K of K then
(1.10) TrL|K(x) = Tr(T
x
) =X
�
�x
Definition 1.11. Considering the same field extension and endomorphism, thenorm is characterized as
NL|K(x) = det(T
x
)
Fact 1.12. If L|K is a separable extension and � : L ! K varies over di↵erentK-embeddings of L into an algebraic closure K of K then
(1.13) NL|K(x) = det(T
x
) =Y
�
�x
Definition 1.14. The discriminant of a basis ↵1, . . . ,↵n
, where the basis (a K-basis, i.e. some linearly independent points ↵
i
2 L such that when multipliedby coe�cients in K, they span L) of some separable extension L|K is defined tobe d(↵1, . . . ,↵n
) =det((�i
(↵j
)))2 where �i
with i = 1, . . . , n is variant over K-embeddings L ! K.
Remark 1.15. Alternatively, we may write d(↵1, . . . ,↵n
) =det(TrL|K(↵
i
↵j
)).
2. Gaussian Integers
Before we begin our discussion of general algebraic number fields, we will considerone of the most standard such fields Q[i] and its ring of integers, Z[i]. This ring isnormally known as the Gaussian Integers:
Definition 2.1. A Gaussian Integer is an element of Z[i] = {a + bi | a, b 2 Z}where i is
p�1.
While a familiarity with the concept of prime numbers is expected, it will be use-ful to consider an alternate characterization than that normally given in elementaryinstruction.
Definition 2.2. If for all a, b 2 Z such that p | ab we have that either p | a or p | b,then p is prime.
Definition 2.3. A is said to be a Euclidean Domain if there is a norm N on A,i.e. a function N : A ! N [ 0 with N(0) = 0 such that for any a, b 2 A with b 6= 0there are q, r 2 A with a = qb+ r and r = 0 or N(r) < N(b).
We can now show:
Proposition 2.4. The ring Z[i] is a Euclidean Domain.
4 DANIEL COMEAUX
Proof. In order to show this result, we will show that Z[i] is Euclidean with respectto the Euclidean norm function, N : ↵ 7! |↵|2. We must prove that there existgaussian integers �, ✏ such that ↵ = �� + ✏ and |✏|2 < |�|2. Through algebraicmanipulation, it will be enough to prove the existence of � 2 Z[i] with |↵
�
� �| < 1.
Now note that the gaussian integers form a lattice in the complex plane (definedmore fully later), with each point having integer coordinates from the basis (1, i)serving as a lattice point. We know that ↵
�
is complex and thus lies in some squareof the lattice. At most, it is half the distance of the diagonal of the lattice meshfrom at least one lattice point, i.e. a distance of 1
2
p2. Thus, choose the point that
minimizes the distance and we have � 2 Z[i] such that |↵�
� �| 12
p2 < 1 as we
wanted. ⇤Therefore, as the ring is Euclidean, it belongs to a larger class of Principal Ideal
Domains (discussed later at 4.5), which we know by [1] p. 273. Since an ED isa PID, it is also a Unique Factorization Domain, i.e. it possesses the property ofunique factorization (see [1] p. 287). This property, while very desirable, is notuniversally possessed, as shown in section 4.
Definition 2.5. Consider a ring A. An element u of A such that there is somev 2 A with uv = vu = 1 is known as a unit. Notationally, the set of units in A isdenoted A⇤.
Definition 2.6. If we have a 2 A with a 6= 0 and not a unit, then a is calledirreducible in A if whenever a = bc with b, c,2 A, at least one of b or c must be aunit in A.
Definition 2.7. An nth root of unity in C is a complex number z satisfying theequation zn = 1.
Now, we return to our considerations of Z[i].Lemma 2.8. For any ↵ in Z[i], ↵ is a unit if and only if N(↵) = 1
Proof. Let ↵ be a unit. By definition, some � 2 Z[i] exists such that ↵� = 1. Thus,we have that � = 1
↵
. Since � 2 Z[i], we know that � must be of the form x+ iy forx, y 2 Z, giving us 1,�1, i,�i as the only solutions. For each of these ↵, N(↵) = 1and we are done.
If ↵ = a+ bi we have that a2 + b2 = 1. Since a and b are integers, we know thateither a = ±1 and b = 0 or a = 0 and b = ±1, yielding the above units. ⇤
Hence, the group of units of the ring Z[i] consists of the fourth roots of unity,Z[i]⇤ = [1,�1, i,�i].
Also using this lemma, we can prove the following result:
Theorem 2.9. For all prime numbers p 6= 2 we have, for a, b in Z,p = a2 + b2 if and only if p ⌘ 1 mod 4.
Proof. Assuming that p = a2 + b2, it is clear that p ⌘ 1 mod 4 since any perfectsquare a is either congruent to 0 mod 4 or 1 mod 4, and any prime p could not becongruent to 0 mod 4 or 2 mod 4.
Now assume that p ⌘ 1 mod 4. We must show that it is not a prime element inZ[i]. This su�ces because, if it is not prime, then we have some p = ↵� where ↵and � are non-units in the gaussian integers. Now, use the norm function
N(x+ iy) = (x+ iy)(x� iy) = x2 + y2
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 5
and thus p2 = N(↵)N(�) with N(↵), N(�) 6= 1, so we have, with ↵ = a+ bi
p = N(↵) = a2 + b2
Note that the congruence �1 ⌘ x2 mod p has a solution, x = (2n)!. Now, byWilson’s Theorem we know that
�1 ⌘ (p� 1)! mod p
and thus
�1 ⌘ (p� 1)! = [(1)(2) . . . (2n)][(p� 1)(p� 2) . . . (p� 2n)]
⌘ [(2n)!][(�1)2n(2n)!] = [(2n)!]2 mod p.
Hence, p|x2 + 1 = (x + i)(x � i). However, since x
p
± i
p
is not in Z[i], p does not
divide x+ i or x� i and is thus not prime in Z[i]. Thus, we are done. ⇤Definition 2.10. Two elements b, c 2 A are associated if they di↵er by only a unitfactor; this is denoted b ⇠ c
Every element associated to an irreducible element is also irreducible. Giventhis, it is possible to prove the following (the result is interesting, but the proof isomitted for brevity. It relies on the above proofs).
Fact 2.11. The prime elements ⇡ of Z[i], up to associates, are:
(1) ⇡ = 1 + i(2) ⇡ = a+ bi with a2 + b2 = p, p ⌘ 1 mod 4, a > |b| > 0(3) ⇡ = p, p ⌘ 3 mod 4.
Remark 2.12. In general, irreducibility does not necessarily imply primality orvice versa. However, in the case of Principal Ideal Domains (defined at 4.5), anonzero element is prime if and only if it is irreducible (see [1] p.284). The failureof this statement in some non-PID is clear. For example, consider Z[
p�5] = R,and consider one of its elements, 3. 3 is irreducible in this ring (shown in a methodsimilar to that employed in section 4), but 3 is not prime since (2+
p�5)(2�p�5) =32 is divisible by 3 but neither of the factors is divisible by 3 in the ring.
As we mentioned earlier, Z[i] serves as the ring of algebraic integers of Q[i](defined below at 4.2). We now prove this:
Proposition 2.13. Z[i] consists only of those elements of the extension field Q(i)of Q which satisfy, with a, b 2 Z, a monic polynomial equation x2 + ax+ b.
Proof. Any element ↵ 2 Q(i) is of the form ↵ = c + id with c, d 2 Z and is a zeroof the polynomial
x2 + ax+ b 2 Q[x] with a = �2c, b = c2 + d2
If c, d 2 Q then a, b 2 Q. Also, if a, b 2 Z, then 2c, 2d 2 Z.Note that (2c)2 + (2d)2 = 4b ⌘ 0 mod 4. Since b is an integer and squares are
always ⌘ 0 or ⌘ 1, (2c)2 ⌘ (2d)2 ⌘ 0 mod 4, c and d are integers and we aredone. ⇤
Thus, we have examined the units, prime elements, and unique factorization ofthis special case. If all algebraic number fields behaved in this manner, there wouldbe no need to proceed. However, they do not - unique factorization fails, unitsbehave unpredictably, and prime elements take a di↵erent form than in this case.
6 DANIEL COMEAUX
In the following pages, we will establish the foundations necessary to consider thegeneral subject of algebraic number fields.
3. Integrality
We now proceed to define the more specifically applicable terms to our explo-ration of Minkowski Theory.
Definition 3.1. An algebraic number field is a finite field extension K of Q, i.e. thedegree (or index ) of K|Q, normally denoted [K : Q] and representing the dimensionof K as a vector space over Q, is finite. The elements of such a K are known asalgebraic numbers, and if these are zeroes of monic polynomials f(x) 2 Z[x] theyare called integral.
Definition 3.2. A subring S of the ring R is a subgroup of R that is closed undermultiplication, and contains the multiplicative identity 1.
Definition 3.3. Let A ✓ B be an extension of rings (where A is a subring of B).An element b 2 B is called integral over A if it satisfies a monic equation
xn + a1xn�1 + · · ·+ a
n
= 0, n � 1
with coe�cients ai
2 A. This definition can be generalized to the ring B, which iscalled integral over A if all elements b 2 B are integral over A.
Definition 3.4. If we have some �1, . . . ,�n
2 B with the property that each b 2 Bcan be expressed uniquely as a linear combination b = ↵1�1 + · · · + ↵
n
�n
withcoe�cients ↵
i
2 A, then we call this set an integral basis of B over A. Note thatthis integral basis is automatically a basis of L|K as above, n is equal to the degree[L : K]. If such an integral basis exists, then B is a free A-module of rank n
In order to ensure that the concept of an integral basis is clear, we have includedthe following example, discussing the integral basis of Q( 3
p2).
Example 3.5. (1, 3p2, 3
p22) is an integral basis for Q( 3
p2). Since Q( 3
p2) is a degree
3 extension of Q, it must have an integral basis of size 3. We will now show that
1, 3p2, 3
p22are linearly independent over Q, as suppose not. Then we have some
a, b, c 2 Q such that a + b 3p2 + c 3
p22= 0, but then a = � 3
p2(b + c 3
p2) 2 Q i.e.
� 3p2(b + c 3
p2) = p
q
with p, q 2 Z. Thus we have p = �q 3p2(b + c 3
p2) = �bq 3
p2 �
cq 3p22= �d 3
p2� e 3
p22 2 Z for some d, e 2 Q. Thus we have � 3
p2(d+ e 3
p2) 2 Z,
but � 3p2f 2 Z only when f is some integer multiple of 3
p22or such a multiple
plus 13p2
. Since neither case holds, we can conclude that p /2 Z, a contradiction. So
(1, 3p2, 3
p22) is linearly independent. Moreover, it spans the set since it is a linearly
independent set of size 3 in an extension of degree 3, due to basic properties ofbases. Hence, it serves as an integral basis for the set it spans, namely Q( 3
p2).
While the following result is not proved here, the transitivity of integrality isuseful for our considerations and will be employed several times in the comingpages.
Proposition 3.6. If A ✓ B ✓ C are two ring extensions, C is integral over B, Bis integral over A, then C is integral over A.
Proof. See [2] ⇤
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 7
3.7-3.9 may seem tangential; however, they will serve a crucial role in later proofs.
Definition 3.7. The set of elements A = {b 2 B | b integral over A} in somering extension A ✓ B is known as the integral closure of A in B, and A is calledintegrally closed in B if we have that A = A.
Lemma 3.8. Consider the integrally closed integral domain A, its field of frac-tions K, and its integral closure B in the finite separable extension L|K. If we let↵1, . . . ,↵n
be some basis of L|K contained in B, i.e. with ↵i
2 B and spanningL|K with coe�cients in K, with discriminant d = d(↵1, . . . ,↵n
), then we will havethat dB ✓ A↵1 + · · ·+A↵
n
Proof. Note that the concept of a field of fractions will be clarified later (section7.1). To prove this, we will show that if ↵ 2 B then we have that
d↵ 2 A↵1 + · · ·+A↵n
So if ↵ = k1↵1 + · · · + kn
↵n
2 B, kj
2 K, then these kj
are a solution of thefollowing system of linear equations: Tr
L|K(↵i
↵) =Pj
TrL|K(↵
i
↵j
)kj
. Moreover,
since we know that TrL|K 2 A, each such k
j
is the quotient of an element of Agiven by the value d = det(Tr
L|K(↵i
↵j
)). Multiplying, we obtain that dkj
2 A,and since A is integrally closed, d↵ 2 A↵1 + · · ·+A↵
n
. ⇤
Proposition 3.9. If L|K is separable and A is a principal ideal domain, then everyfinitely generated B-submodule M 6= 0 of L is a free A-module of rank [L : K]. Morepractically, B admits an integral basis over A.
Proof. See [2] ⇤
4. Ideals
4.1. Preliminaries. In section 2, we considered the ring of integers Z[i] of Q[i].But what do we call the more generalized ring of integers of some algebraic numberfield, an extension K|Q? How do we define this ring of integers?
Definition 4.1. Consider an extension field K|Q. Recall that if ↵ 2 K is analgebraic integer, it is the solution of some monic polynomial with coe�cients inZ. The ring of all such ↵, i.e. the integral closure of Z in K is called the ring ofintegers of K and is denoted O
K
.
Our discussion will now require a motivating example. Unlike the rings Z andZ[i], the ring O
K
of integers of an algebraic number field K does not inherently pos-sess a general uniqueness of prime factorization. For example, consider Q(
p�47).Its ring of integers is O
K
= Z+Z( 12 +12
p�47), which we know by [1], p. 229. Nowconsider
54 = (2)(33) = (13 +
p�47
2)(13�p�47
2)
Unlike in Z or Z[i], each of these factors is irreducible in OK
. We will show this bycontradiction. So suppose that 2 = ab where a and b are non-units. Therefore, wehave 2 = ab which gives us that 4 = N
K|Q(a)NK|Q(b). This implies that, withoutloss of generality, N
K|Q(a) = ±2, but remember that we have
NK|Q(x+ y(
1
2)(1 +
p�47)) = ((x+1
2y) + y
p�47)((x+1
2y)� y
p�47)
8 DANIEL COMEAUX
which simplifies to x2 + xy + 12y2 = ±2, and this has no integral solutions, soat least one of a or b is a unit. The other factors can be similarly proven to beirreducible. It is because of this failure of unique factorization that we have insteadbegun our discussion of ideals, with which we attempt to solve the problem of uniquefactorization in integer rings of algebraic number fields. The underlying intuitionbehind this progression is that each of these irreducible factors is a product of idealprime numbers, such that
2 = P1P2, 3 = P3P4, (13 +
p�47
2) = P1P3, (
13�p�47
2) = P2P4
and thus we have P1P2P3P4 = 54. Before we establish prime ideals, however, wemust define the concept of ideals in general and outline their basic properties.
Definition 4.2. Let R be a commutative ring, and I a subgroup (an abelian groupunder addition) of this R. Such an I is an ideal of R if, for all r in R and a in I,ar is contained in I.
Definition 4.3. For any r in a ring R, we will denote (r) as the smallest ideal of Rcontaining r, and this is the ideal generated by r. An ideal generated by only oneelement is a principal ideal, and an ideal generated by a finite number of elementsis a finitely generated ideal
Hence, in the ring of integers of Q, which is Z, the ideal generated by 1 is (1) = Z,the ideal generated by 2 is all even numbers, by 3 is all multiples of 3, etc. Also,note that if an ideal is generated by, for example, 2 and 3, this is actually the sameas (1) as we can subtract 2 from 3 to obtain the principal ideal (1) = Z.
Definition 4.4. An ideal B in R is a prime ideal if B 6= R and when, with a, b 2 R,we have ab 2 B then either a, b or both are elements of B.
Definition 4.5. A Principal Ideal Domain is an integral domain in which everyideal is principal.
Also, note that if a given domain is Euclidean, it is automatically a PrincipalIdeal Domain. Hence, Z[i] is a Principal Ideal Domain, in addition to being Eu-clidean.
Definition 4.6. An ideal C in R is a maximal ideal if C 6= R and the only idealsof R containing C are itself and the whole ring, i.e. C and R.
Note that if C is an ideal of R such that R/C is a field, then C is maximal, andin any Principal Ideal Domain, a nonzero prime ideal is maximal.
Definition 4.7. A commutative ring R with an identity is noetherian if each ofits ideals is finitely generated.
Definition 4.8. A noetherian, integrally closed integral domain in which eachnonzero prime ideal is also maximal is known as a Dedekind Domain.
We now establish basic operation on ideals:
Remark 4.9. Take A,B 2 O where O is an arbitrary Dedekind Domain.
(1) The divisibility relation A|B is defined by B ✓ A(2) Summation (or union) of ideals is defined by A+B = {a+ b | a 2 A, b 2 B},
i.e. the smallest ideal containing both A and B, or the greatest common divisor.
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 9
(3) Intersection of ideals is defined by the least common multiple of A and B.(4) The product of ideals is defined by AB = {P
i
ai
bi
| ai
2 A, bi
2 B}
4.2. Results. We can now prove the following facts about OK
.
Theorem 4.10. The ring OK
is noetherian, integrally closed, and every primeideal P 6= 0 in O
K
is a maximal ideal.
Proof. We know that OK
is noetherian because each ideal A is a finitely generatedZ-module by 3.6. Thus, it is a finitely generated O
K
-module, or noetherian.As it is the integral closure of Z in K, it is integrally closed by results proved in
section 3.Finally, we must show that each prime ideal P 6= 0 is a maximal ideal. We
know that P \ Z is a nonzero prime ideal (p0) in Z, since it is clearly prime and ify 2 P, y 6= 0, and yn+a1y
n�1+ · · ·+an
= 0 (with ai
2 Z, an
6= 0) then an
2 P \Z.We also know that the integral domain O = O
K
/P is generated from A = Z/p0Zby adjoining algebraic elements, and this is thus a field, since A[↵] = A(↵) if ↵ isalgebraic. Thus P is maximal. ⇤
We will now consider ideals on an arbitrary Dedekind domain O.
Lemma 4.11. For every ideal A 6= 0 of O there exist nonzero prime ideals P1, . . . , Pn
such that A ◆ P1 . . . Pn
.
Proof. We prove this by contradiction. Suppose that the set � of ideals withoutsuch nonzero prime ideals is nonempty, i.e. there are ideals B 6= 0 of O such thatthey do not contain any products of nonzero prime ideals. Since O is noetherian,every ascending chain of ideals becomes stationary. Thus, � has a maximal elementA0, as if it did not, there would be an infinite chain of ideals in contradiction toour hypothesis. Since A0 is not a prime ideal, we can find some b1 and b2 in O suchthat neither b1 nor b2 are in A, but their product b1b2 is in A.
Now set A1 = (b1) + A0 and A2 = (b2) + A0. A0 ( A1 and A0 ( A2 andA1A2 ✓ A0. Since A0 is the maximal element of � and A1 and A2 are both propersupersets of A0, both A1 and A2 contain a product of prime ideals. However, theproduct of these products is contained in A0, so A0 contains a product of primeideals, a contradiction. ⇤
We now define the inverse operation on prime ideals. We can also prove thatany ideal multiplied by an inverse prime ideal is not equal to itself:
Lemma 4.12. Let P be a prime ideal of O and define
P�1 = {x 2 K | xP ⇢ O}
Then we have AP�1 := {Pi
ai
xi
| ai
2 A, xi
2 P�1} 6= A for all ideals A 6= 0.
Proof. Let s 2 P, s 6= 0 and P1P2 . . . Pr
⇢ (s) ⇢ P , minimizing r. P1 is contained inP which implies that P1 = P since P1 is a maximal ideal. Now, since P2 . . . Pr
* (s),there exists t 2 P2 . . . Pr
with t /2 sO, which implies that s�1t /2 O. We also have,however, that tP ✓ (s) giving that s�1tp ✓ O, hence giving us that s�1t 2 P�1,and therefore P�1 6= O.
10 DANIEL COMEAUX
Following [2], assume that AP�1 = A, and thus P�1 = O which contradicts ourprevious statement. ⇤
We can now resolve the problem of unique factorization encountered earlier.We say that Euclidean domains are contained within the larger domain of UniqueFactorization Domains [UFDs], which have unique factorization at the element level.However, our next consideration will be of a special type of non-UFD domains thathave unique factorization at the ideal level.
Theorem 4.13. Every ideal A of O which di↵ers from (0) and (1) admits a fac-torization A = P1 . . . Pr
into nonzero prime ideals Pi
of O which is unique up tothe order of the factors.
Proof. First we show the existence of such a prime ideal factorization. Similar toour proof above, let � be the set of all ideals di↵erent from (0) and (1) which donot admit a prime ideal decomposition. Suppose it is nonempty, then there is somemaximal element A in �, itself contained in a maximal ideal P . We know thatO ✓ P�1, and thus A ✓ AP�1 ✓ PP�1 ✓ O.
By the previous lemma, A ( AP�1 and P ( PP�1 ✓ O. Since P is a maximalideal, PP�1 = O. Also, we know that A is maximal in � so since AP�1 6= O, AP�1
admits some prime ideal decomposition AP�1 = P1 . . . Pr
. Since A = AP�1P =P1 . . . Pr
P , A admits a prime ideal decomposition, a contradiction.Now, we aim to show such a factorization is unique. Let A = P1P2 . . . Pr
=Q1Q2 . . . Qs
be two prime ideal factorizations of A. Without loss of generality, P1
divides Q1, and since P1 is maximal, P1 = Q1. Now multiply this by P�1 and thenP2 . . . Pr
= Q2 . . . Qs
. Continuing, r = s and each P and Q can match, i.e. withrenumbering P
i
= Qi
. ⇤
The following theorem will be employed later and is thus included for complete-ness. However, its proof is not instructive relative to our discussions.
Theorem 4.14. Chinese Remainder Theorem. Let A1, . . . , An
be ideals in a ring
O such that Ai
+Aj
= O for i 6= j. Then, if A =nT
i=1A
i
, one has O/A ⌘nL
i=1O/A
i
Proof. See [2]. ⇤
5. Lattices
While we considered the motivating example of Q[i], we mentioned the conceptof a lattice in the complex plane. In that case, the lattice was merely the integralpoints spanned by the basis (1, i). However, we will now extend this definition tothat of all algebraic number fields.
Definition 5.1. Consider V , an n-dimensional R-vector space. A subgroup in Vof the form � = Zv1 + · · ·+Zv
m
with linearly independent vectors v1, . . . , vm of Vis known as a lattice. We call (v1, . . . , vm) a basis.
Note that although the lattices which we will consider will be over real vectorspaces, this definition can be generalized to lattices over C, for example, or forany field. In the above definition, replace R with some arbitrary field K, and
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 11
thus V is an n-dimensional vector space over K. Instead of Z, we will use some� = (b1, . . . , bn), a K-basis for V . Now for some ring R in K, the lattice is
� = {nX
i=1
ri
bi
| ri
2 R, bi
2 B}
Definition 5.2. The set � = {x1v1+ · · ·+xm
vm
| xi
2 R, 0 xi
< 1} is known asthe fundamental mesh of the lattice, and the lattice is said to be a complete latticeif m = n.
For example, in the case of the first lattice we considered, the basis was (1, i)and therefore the fundamental mesh was {x1 + ix2 | x
i
2 R, 0 xi
< 1}. This isequivalent to the square of edge 1 in the complex plane with its lower left corneron the origin; though it is only a fraction of the entirety, we can learn a great dealabout the lattice by examining the fundamental mesh.
Several useful properties of lattices are as follows:
Proposition 5.3. A subgroup � ✓ V is a lattice if and only if it is discrete
Proof. See [2]. The discreteness of an arbitrary lattice is clear from the definition.Briefly, in the other direction, the proof proceeds as follows: assuming that � isdiscrete, the proof continues by showing that � is closed. Then, considering thesubspace V0 of V spanned by �, a basis of V0 of size m and contained in � is chosento form a complete lattice �0. After proving that (� : �0) is finite, the proof usesthis fact and the main theory on finitely generated abelian groups to conclude that� is indeed a lattice. ⇤Lemma 5.4. A lattice � in V is complete if and only if there exists some boundedsubset M ✓ V such that the collection of all translates M + �, � 2 �, covers theentire space V .
Proof. If � = Zv1 + · · · + Zvn
is complete then taking M to be the fundamentalmesh � is su�cient.
To prove in the other direction, let M be such a bounded subset. Let V 0 be thesubspace which is spanned by �. We must show that V = V 0. Take v 2 V andwe will show that it belongs to V 0. V =
S�2�
(M + �), so with v 2 V , av
2 M and
�v
2 � ✓ V 0, we havevv = av
+ �v
. Now divide this by v and take its limit goingto 1, then we have
v = limv!1
1
vav
+ limv!1
1
v�v
= limv!1
1
v�v
2 V 0
since M is bounded and V 0 is closed. ⇤Remark 5.5. In general, it will be useful to understand the volume of a lattice. Thismay seem nonintuitive, as we have shown a lattice is necessarily a discrete set. Wewill consider another measure of its volume, best thought of as the volume of theparallelepiped spanned by the fundamental mesh. This volume is calculated as
vol(�) = | detA|,with A equivalent to (a
ik
), i.e. the matrix of the base change from the identity tov1, . . . , vn so that v
i
=Pk
aik
ek
. Equivalently, we can write that
vol(�) = [det(hvi
, vj
i)]1/2 = vol(�).
12 DANIEL COMEAUX
With the following definitions, we will finally be ready to discuss the underlyingconcepts of Minkowski Theory.
Definition 5.6. A subset X of V is called centrally symmetric, if, given any pointx 2 X, the point �x also belongs to X.
Definition 5.7. A subset X of V is called convex if, given any two points x, y 2 X,the whole line segment {ty + (1� t)x | 0 t 1} is contained in X.
Theorem 5.8. Minkowski’s Lattice Point Theorem. Let � be a complete lattice inthe euclidean vector space V and X a centrally symmetric and convex subset of V .Now suppose that vol(X) > 2nvol(�). Then X contains at least one nonzero latticepoint � 2 �.
Figure 1. Picture Proof
First, we will demonstrate the intuition behind this theorem. Consider the casewhere V = R2, and then the lattice points are as pictured above (with distance 1between each point on the axes). Any centrally symmetric and convex subset of Vwill be centered on the origin, for example the squares pictured above. In this case,vol(�) = 1. Note that the blue square has a volume of approximately 3, whereasthe green square has a volume of approximately 9.
Now, vol(Blue Square) = 3 < 4vol(�), but vol(Green Square) = 9 > 4vol(�).According to the theorem, this would imply that the green square contains at leastone nonzero lattice point, and in fact it does.
Proof. It will su�ce for us to show that we can find two distinct lattice points�1, �2 2 � such that
(1
2X + �1) \ (
1
2X + �2) 6= ;
So we choose some element in this intersection, and we obtain, with x1, x2 2 Xthat 1
2x1 + �1 = 12x2 + �2 is such an element. Define
� = �1 � �2 =1
2(x2 � x1).
Now note that this � is the center of the line segment joining x2 and �x1 (part ofX through central symmetry) and thus contained in X \ � (through convexity).
Now, suppose that we cannot find such lattice points, i.e. that the sets 12X + �
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 13
are pairwise disjoint for all � 2 �. Then the same property would hold for theintersections �\ ( 12X + �) since an intersection can only remove elements of a set.Then we have
vol(�) �X
�2�
(� \ (1
2X + �)).
However, remember that we can translate � \ ( 12X + �) by �, or in this case, ��,which yields the set (�� �) \ 1
2X. Note that this set has the same volume as theinitial set. Also, � � �, � 2 � cover the entire space V since with the addition ofone �, we have the fundamental mesh and its lattice translate. In particular theycover the set 1
2X. This yields the inequality chain:
vol(�) �X
�2�
vol((�� �) \ 1
2X) = vol(
1
2X) =
1
2nvol(X)
but this violates our initial hypothesis. Thus, two such lattice points exist, and Xcontains at least one nonzero lattice point. ⇤
6. Minkowski Theory
Essentially, the intuition behind Minkowski Theory is the interpretation of degreen algebraic number fields K|Q as points in some n-dimensional space. The resultsof the theory are based upon the above Lattice Point Theorem, but before we canderive the desired results, we must discuss the definitions of Minkowski Space andits associated operations. First among these is the space KC, based upon a familyof complex embeddings.
Definition 6.1. KC is a C-vector space defined by a function j which maps fromK to this KC:
j : K ! KC =Y
⌧
C, a 7! ja = (⌧a)
where ⌧ are the n complex embeddings ⌧ : K ! C. Thus, j(a) = (⌧1a, ⌧2a, . . . , ⌧na)is some element of KC.
Remark 6.2. KC is equipped with the hermitian scalar product hx, yi = P⌧
x⌧
y⌧
Having defined this KC, we now restrict it as follows:
Definition 6.3. The Minkowski Space is the Euclidean vector space
KR = K+C = [
Y
⌧
C]+
where the points in KR are the points (z⌧
) 2 KC such that z⌧
= z⌧
.
Remark 6.4. The scalar product in the Minkowski Space is known as the canonicalmetric, and it is denoted h , i : KR ⇥KR ! KR.
Remark 6.5. We can more explicitly describe Minkowski spaces. Note that of then embeddings ⌧ , the embeddings into C of K, we have some which already mapto R. We will denote the real embeddings as ⇢1, . . . , ⇢r : K ! R. The remainingembeddings are complex, and come in pairs, denoted �1,�1, . . . ,�s
,�s
: K ! C.Thus n = r + 2s and
KR = {(z⌧
) 2Y
⌧
C | z⇢
2 R, z�
= z�
}.
14 DANIEL COMEAUX
We now want to be able to consider the relation of the Minkowski Space to Rn.
Proposition 6.6. There exists an isomorphism
f : KR !Y
⌧
R = Rr+2s = Rn
f : (z⌧
) 7! (x⌧
)
with x⇢
= z⇢
, x�
= Re(z�
), x�
= Im(z�
). This isomorphism transforms thecanonical metric into the scalar product (x, y) =
P⌧
↵⌧
x⌧
y⌧
with ↵⌧
= 1 or 2 if ⌧ is
real or complex, respectively.
Proof. The map is an isomorphism as it is clearly injective and surjective, as wellas structure preserving. Now to prove the claim in regards to scalar products, takez = (z
⌧
) = (x⌧
+ iy⌧
) and z0 = (z0⌧
) = (x0⌧
+ iy0⌧
) 2 KR. First, consider hz, z0i when⌧ is real. We have (z, z0) = z
⇢
z0⇢
= x⇢
x0⇢
which fits our proposition. Now given thaty�
= x�
, x�
= y�
, y0�
= x0�
, and x0�
= y0�
,
z�
z0�
+ z�
z0�
= z�
z0�
+ z�
z0�
= 2Re(z�
z0�
) = 2(x�
x0�
+ x�
x0�
).
⇤
Remark 6.7. Since this scalar product takes the canonical measure from KR toRr+2s, it di↵ers from the standard Lebesgue measure by the relation
volcanonical(X) = 2svolLebesgue(f(X)).
Now we show an alternative method for determining the volume of the funda-mental mesh, and thus a lattice:
Proposition 6.8. If A 6= 0 is an ideal of OK
then � = jA is a complete lattice inKR with fundamental mesh of volume
vol(�) =p|d
K
|(OK
: A).
Proof. Let a1, . . . , an be a Z-basis of A, and thus � = Zja1 + · · · + Zjan
. Nowchoose a numbering of our embeddings ⌧ : K ! C, ⌧1, . . . , ⌧n, and then form amatrix B = (⌧
s
ai
). Using [2] 2.12,
d(A) = d(a1, . . . , an) = (detB)2 = (OK
: A)2d(OK
) = (OK
: A)2dK
.
However,
(hjai
, jak
i) = (nX
s=1
⌧s
ai
⌧s
ak
) = BBt
.
We conclude:
vol(�) = | det(hjai
, jak
i)|1/2 = | detB| =p
|dK
|(OK
: A).
⇤
The following result relies heavily on the above proofs, and will be an importanttool as we proceed.
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 15
Theorem 6.9. Let A 6= 0 be an integral ideal of K, and let c⌧
> 0, with ⌧ 2Hom(K,C), be real numbers such that we have
c⌧
= c⌧
andY
⌧
c⌧
> �(OK
: A),
with � = ( 2⇡
)sp|d
k
|. Then there exists a 2 A, a 6= 0 such that
|⌧a| < c⌧
for all ⌧ 2 Hom(K,C).
Proof. Consider the set X = {(z⌧
) 2 KR | |z⌧
| < c⌧
} which is centrally symmet-ric and convex. Note that its volume, denoted vol(X), can be calculated by theisomorphism we defined in 6.6. By 6.7, we have that
vol(X) = 2svolLebesgue(f(X)) = 2svolLebesgue({(x⌧
) 2Y
⌧
R | |x⇢
| < c⇢
, x2�
+x2�
< c2�
})
sovol(X) = 2s
Y
⇢
(2c⇢
)Y
�
(⇡c2�
) = 2r+s⇡s
Y
⌧
c⌧
.
Now we apply 6.8:
vol(X) = 2r+s⇡s
Y
⌧
c⌧
> 2r+s⇡s(2
⇡)sp
|dK
|(OK
: A) = 2nvol(�)
and we are done. Hence, there is some lattice point ja 2 X, a 6= 0, a 2 A, jA 2 �,and furthermore |⌧a| < c
⌧
. ⇤The following is a necessary lemma for Theorem 6.11, but its proof is not in-
structive for our purposes.
Lemma 6.10. Consider the convex, centrally symmetric set
X = {(z⌧
) 2 KR |X
⌧
|z⌧
| < t}.
X has volume vol(X) = 2r⇡s
t
n
n! .
Proof. See [2], III.2.15. ⇤We now prove the existence of an upper bound for the norm of some element of
any nonzero ideal of a ring of integers OK
.
Theorem 6.11. In every ideal A 6= 0 of OK
, there exists an a 2 A, a 6= 0 suchthat
|NK|Q(a)| M(O
K
: A),
where M is referred to as the Minkowski Bound and M = n!n
n (4⇡
)sp|d
K
|.Proof. We proceed in a manner similar to Theorem 6.9, using 6.10. Consider theset X = {(z
⌧
) 2 KR | P⌧
|z⌧
| < t} which has volume 2r⇡s
t
n
n! . Since X is convex and
centrally symmetric, we know by the Minkowski Lattice Point Theorem that somea 2 A is a lattice point of �, and we also know that vol(X) = 2r⇡s
t
n
n! > 2nvol(�),
where vol(�) = 2�sp|d
K
|(OK
: A). So now, using the known inequality betweenarithmetic and geometric means,
1
n
X
⌧
|z⌧
| � (Y
⌧
|z⌧
|)(1/n)
16 DANIEL COMEAUX
we see that
|NK|Q(a)| =
rY
k=1
|⇢k
(a)|sY
k=1
|�k
(a)2| n�n(rX
k=1
|⇢k
(a)|+ 2rX
k=1
|�k
(a)|)n
which, given the conditions for elements of X, and substituting for tn is
|NK|Q(a)| tn
nn
2n2�s
p|dK
|(OK
: A)n!
2r⇡snn
= (2
⇡)sp|d
K
|(OK
: A)
nn
(4
⇡)sp|d
K
|(OK
: A)
nn
⇤
Example 6.12. The Minkowski bound for Q[p�5] can be calculated using the
above. n = 2 since Q[p�5] is a quadratic extension, therefore of degree 2. Also,
s = 1 because for each real embedding, we have exactly one corresponding pair ofcomplex embeddings. The discriminant is �20, and as the ring of integers for thisfield is O
K
= Z[1,p�5], there exists some a 6= 0, a 2 A (with A 6= 0) of O
K
suchthat |N
K|Q(a)| 4⇡
p5(O
K
: A).
There also exists a multiplicative version of Minkowski theory. Similar to thehomomorphism j from K to KC, we now examine the restriction
j : K⇤ ! K⇤C =
Y
⌧
C⇤.
This multiplicative group K⇤C admits the homomorphism N : K⇤
C ! C⇤ which canbe calculated by taking the product of the coordinates. Note that N � j is the usualnorm of K|Q.
Since we are building a lattice, we must use logarithms to pass from multiplicativeto additive groups. Take l : C⇤ ! R, z 7! log |z|. This map induces the surjectivehomomorphism l : K⇤
C ! Q⌧
R, and this gives us a commutative diagram.
K⇤ j //
NK|Q
✏✏
K⇤C
l //
N
✏✏
Q⌧
R
Tr
✏✏Q⇤ // C⇤ l // R
As we did previously, we restrict these operations to elements fixed under conju-gation. Note that in the below diagram, the plus sign in the exponent of the termon the upper right, [
Q⌧
R]+, restricts the term inside the brackets to elements fixed
under complex conjugation. This notation will be employed similarly from now on.
K⇤ j //
NK|Q
✏✏
K⇤R
l //
N
✏✏
[Q⌧
R]+
Tr
✏✏Q⇤ // R⇤ l // R
Furthermore, we can explicitly describe [Q⌧
R]+ using our existing terminology
as follows.
[Y
⌧
R]+ =Y
⇢
R⇥Y
�
[R⇥ R]+
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 17
Note that the second factor can be identified with R by the map (x, x) 7! 2x, whichgives us an isomorphism [
Q⌧
R]+ ⇠= Rr+s. Thus, the trace now becomes a map from
Rr+s to R as we expect, and l : K⇤R ! Rr+s is explicitly given by the equation
l(x) = (log |x⇢1 |, . . . , log |x⇢r |, log |x�1 |2, . . . , log |x�s |2)
with x 2 K⇤R ✓ Q
⌧
C⇤ since x = (x⌧
)
7. Class Numbers
7.1. Preliminaries. We are now ready to apply the results gained through ourexploration of Minkowski Theory, first by proving the finiteness of the ideal classgroup. Before we proceed, however, we must establish the preliminary definitionsand foundations of this group.
Definition 7.1. Consider the field of fractions K of O, a Dedekind domain. Afractional ideal of K is a finitely generated O-submodule A 6= 0 of K.
Proposition 7.2. The fractional ideals form an abelian group, the ideal groupwhich we will denote J
K
of K. The identity element is (1) = O and we define theinverse of A to be A�1 = {x 2 K | xA ✓ O}Proof. Associativity, commutativity, and the fact that A(1) = A are clear.
For some prime ideal P , we know that P ( PP�1 by 4.12. Thus, since Pis maximal, PP�1 = O. Thus, if A = P1 . . . Pr
is an integral ideal, then B =P�11 . . . P�1
r
is an inverse, and thus BA = O implies that B ✓ A�1.Also, if xA ✓ O then xAB ✓ B, showing that x 2 B since AB = O. Therefore,
B = A�1.Finally, if A is some arbitrary fractional ideal and C 2 O is a nonzero ideal such
that CA ✓ O then (CA)�1 = C�1A�1 is the inverse of CA, which implies thatAA�1 = O. ⇤
We now generalize 4.13 into the broader category of fractional ideals.
Corollary 7.3. Every fractional ideal A admits a unique representation as a prod-uct A =
Qp
pvp , where vp
2 Z and vp
= 0 for almost all p. Namely, this implies that
JK
is the free abelian group on the set of nonzero prime ideals p of O.
Proof. See [2] I.3.9 for the proof that A is a quotient of two integral ideals B andC. Uniqueness comes from 4.13. ⇤
Similarly, we must generalize the concept of principal ideals.
Definition 7.4. The fractional principal ideals (A) = AO with A 2 K⇤ are asubgroup of the ideals J
K
, and we will denote this subgroup PK
.
With fractional ideals established, it is now logical for us to consider the resultof modding out the fractional principal ideals from the ideal group, since J
K
� PK
.
Definition 7.5. The quotient group ClK
= JK
/PK
is known as the ideal classgroup, or class group of K.
Definition 7.6. The absolute norm of an ideal A in the ring OK
is N (A) =(O
K
: A)
18 DANIEL COMEAUX
7.2. Results.
Proposition 7.7. Let A be an ideal of a ring OK
, and let p1, . . . , pr be prime idealsof this O
K
. If A = pv11 . . . pvrr
, is the prime factorization of an ideal A 6= 0, thenN (A) = N (p1)v1 . . .N (p
r
)vr
Proof. See [2]. The proof is based upon the Chinese Remainder Theorem. ⇤With this, we can finally apply Minkowski Theory to the problem of the ideal
class group.
Theorem 7.8. The ideal class group ClK
= JK
/PK
is finite. We will denote itsorder as h
K
= (JK
: PK
) and call it the class number of K.
Proof. If P 6= 0 is a prime ideal of OK
such that P \ Z = pZ, then OK
/P is afinite field extension of Z/pZ of arbitrary degree f � 1. Further, N (P ) = pf , since(pZ : Z) = p. Now, for a given p, only finitely many such prime ideals P existbecause P |(p). Thus we have only finitely many P for each p and there are onlyfinitely many prime ideals P with bounded absolute norm.
Now, since each integral ideal admits some representation A = P v11 . . . P vr
r
withvi
> 0 8 i and since N (A) = N (P1)v1 . . .N (Pr
)vr , we can conclude that there areonly finitely many ideals A of O
K
with a bounded absolute norm N (A) < M .Therefore, to show our claim, it will be enough to show that each class [A] 2 Cl
K
contains some integral ideal A1 satisfying the inequality
N (A1) M = (2
⇡)sp
|dK
|.So we will choose some arbitrary A in the class, along with b 2 O
K
, b 6= 0, such thatwe have B = bA�1 ✓ O
K
. By the above proof of the existence of the Minkowskibound, we can choose some � 2 B,� 6= 0 with |N
K|Q(�)| MN (B) which we canrearrange to the following
|NK|Q(�)|N (B)�1 = N (�)B�1 = N (�B�1) M.
Thus, the ideal A1 : = �B�1 = �b�1A 2 [A] has the required property, and ClK
isfinite. ⇤Example 7.9. For quadratic fields with discriminant 5, 8, 11,�3,�4,�7, and �8,the class number is 1. We know this because of Minkowski’s Bound, which givesus some element of the ideal with norm less than his namesake bound. When weplug in discriminants, along with n = 2 because it is a quadratic field, s = 0 forthe positive cases because there are no complex embeddings and s = 1 for thenegative cases as there are complex embeddings, we obtain, respectively, M ⇡1.12, 1.41, 1.66, 1.10, 1.27, 1.68, and 1.80, all less than 2. Thus, we have that thereexists some a 2 A for each of these ideals such that N (a) M , i.e. that N (a) 1.80 < 2, and since this must be an integer, we can only conclude that N (a) = 1for these discriminants, i.e. that the class number is 1.
8. Dirichlet’s Unit Theorem
Having discussed the ideal class group, we can now discuss other interestingaspects of O
K
, namely the group of units O⇤K
. Although O⇤K
itself is not necessarilyfinite, it contains the finite group of the roots of unity lying in K, denoted µ(K),Recall the commutative diagram from above.
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 19
K⇤ j //
NK|Q
✏✏
K⇤R
l //
N
✏✏
[Q⌧
R]+
Tr
✏✏Q⇤ // R⇤ l // R
Now, looking at the upper half of the diagram, we find three subgroups of note.O⇤
K
= {✏ 2 OK
| NK|Q(✏) = ±1}, the group of units
S = {� 2 K⇤R | N(�) = ±1} the norm-one surface
H = {x 2 [Q⌧
R]+ | Tr(x) = 0}, the trace-zero hyperplane
Note that our existing functions serve as homomorphisms, i.e. j : O⇤K
! S andl : S ! H gives us a composite homomorphism, � := l � j : O⇤
K
! H, and its image� = �(O⇤
K
) ✓ H.
Definition 8.1. A sequence of groups and group homomorphisms, denoted by
· · · ! Xn�1
↵n! Xn
↵n+1! Xn+1 ! . . . , is said to be exact if, for each X
i
, image↵n
=ker ↵n+1.
Proposition 8.2. The sequence 1 ! µ(K) ! O⇤K
�! � ! 0 is exact.
Proof. Exactness at µ(K) and � is clear. Thus, we must show that µ(K) is thekernel of �. Note that for ⇣ 2 µ(K) and ⌧ : K ! C some embedding, we can findlog |⌧⇣| = log 1 = 0, and thus µ(K) ✓ ker(�).
Now let ✏ 2 O⇤K
be in the kernel, i.e. �(✏) = l(j✏) = 0. We can then concludethat |⌧✏| = 1 for each embedding, and thus j✏ = (⌧✏) lies in some bounded domainof KR. However, remember that by 6.8, j✏ is a point of the KR lattice jO
K
. Thus,the kernel of � contains finitely many elements, and, as it is a finite group, containsonly roots of unity in K⇤ (see [2] I.7.1). Thus, µ(K) ◆ ker(�). ⇤Lemma 8.3. Up to multiplication by units, there are only finitely many elements↵ 2 O
K
of a given norm NK|Q(↵) = ↵
Proof. Take some a 2 Z, a > 1. Now consider OK
/aOK
. In each of the finitelymany cosets of this quotient, there exists up to multiplication by units at mostone element ↵ such that |N(↵)| = a. We know this as, if � = ↵ + a�, � 2 O
K
is
another such element, we have ↵
�
= 1 ± N(�)�
� 2 OK
since N(�)/� 2 OK
. Also,�
↵
= 1 ± N(↵)↵
� 2 OK
, which gives us that � is associated to ↵. Hence, up to themultiplication of units, we have at most (O
K
: aOK
) elements of norm ±a, and itis finite as we wanted. ⇤Theorem 8.4. The group � is a complete lattice in the (r + s � 1)-dimensionalvector space H. As a consequence, it is isomorphic to Zr+s�1.
Proof. The proof is rather unwieldy - see [2] for full details. ⇤Theorem 8.5. This is known as Dirichlet’s Unit Theorem. The group of unitsO⇤
K
of OK
is the direct product of the finite cyclic group µ(K) and a free abeliangroup of rank r + s� 1.
Proof. By the above, we know that � is a free abelian group of rank t = r + s� 1.Now, let v1, . . . , vt be a Z-basis of �, and let ✏1, . . . , ✏t 2 O
K
be the preimages of
20 DANIEL COMEAUX
each vi
. Finally, let A ✓ O⇤K
be the subgroup generated by these ✏i
. Thus, we canconclude that A is mapped isomorphically onto � by �. Therefore µ(K)\A = {1},which shows O⇤
K
= µ(K)⇥A. ⇤Remark 8.6. This implies that there exist ✏1, . . . , ✏t called fundamental units suchthat any other unit ✏ can be written uniquely as a product ✏ = ⇣✏v11 . . . ✏vt
t
where ⇣is a root of unity and v
i
are integers.
Example 8.7. We will now apply this result to consider the group of units of bothreal and imaginary quadratic field extensions. First, we will consider real quadraticfield extensions. For an arbitrary Q[
pD], D square-free and positive, note that
r = 2, s = 0, and thus r+ s� 1 = 1. For example, Q[p3] has two real embeddings,
where 1 +p3 can be mapped to either 1 +
p3 or 1 � p
3. This implies that thegroup of units of O
K
= Z+Zp3 is the direct product of group of the roots of unity
of Q[p3], which are only 1 and �1, and a free abelian group of rank 1. Thus, the
group of units is isomorphic to {�1, 1} ⇥ Z, and in this specific case, calculationshows that the group of units is (2 +
p3)Z where the exponent Z refers to the
integer multiples of this fundamental unit, and their respective isomorphisms tothe integers (2 +
p3 maps to 1, �4�p
3 maps to �2, etc.).Now we consider the imaginary quadratic field extensions. For an arbitrary
Q[pD], D square-free and negative, note that r = 0, s = 1, and thus r + s� 1 = 0
as there are no real embeddings and only one pair of complex embeddings. Thus,any imaginary quadratic field has a group of units for its ring of integers that isonly composed of the group of the roots of unity, and in all cases but two, this willbe (1,�1). The two exception are Q[i] as described above, as well as Q[
p�3]. Formore details on the latter, see [4], Chapter 6.
9. Extending Dedekind Domains
We can now take some arbitrary Dedekind domain and extend its characteristicsto some related integrally closed domain.
Proposition 9.1. Let O be a Dedekind domain with field of fractions K, L|K afinite extension of K, and O the integral closure of O in L. Then O is also aDedekind domain.
Proof. As in an earlier proof, we know that since O is the integral closure of O, itis also integrally closed.
The set of nonzero prime ideals P of O is maximal. Choose some p 2 P, p 6=0, P 2 P. Given this p, we can find a polynomial
c0 + c1p+ · · ·+ cn
pn = 0 ci
2 O, c0 6= 0
Therefore, c0 = �p(c1 + · · · + cn
pn�1 and c0 2 P \O. Hence, P \O is a nonzeroprime ideal in O, and since O is a Dedekind domain, P \ O is maximal. Thus,the integral domain O/P is an algebraic extension of the field O/P , and is hence afield. Thus, P is a maximal ideal.
Now we will show that O is noetherian in the case that L|K is a separableextension (for the inseparable case, see [2] 12.8). Take some basis of L|K called↵ = a1+· · ·+a
n
contained in O. The discriminant of this basis is D = d(a1, . . . , an)and by [2] I.2.8 and Lemma 3.4 above, D 6= 0 and O ⇢ Oa1/D + · · · + Oa
n
/D.Thus, each ideal of O is contained in this same O-module, and is thus itself a finiteO-module, i.e. it is also a finitely generated O-module, hence noetherian. ⇤
AN EXPLORATION OF MINKOWSKI THEORY AND ITS APPLICATIONS 21
Though the proof is beyond the scope of this paper, we will conclude with avery interesting identity, known as the fundamental identity. This identity weavestogether many of the concepts discussed in this paper, and while it is not a resultof Minkowski Theory, the conclusions we have drawn through the applications ofMinkowski’s theorems have led us to this point.
With our above notation, note that for P 2 O, we have PO 6= O. More-over, if P 6= 0 then PO decomposes into a unique product of prime ideals, PO =Pe11 . . .Per
r
, where P = P \ O. This ei
is known as the ramification index, of Pi
over P and the degree of the field extension fi
= [O/Pi
: O/P ] is known as theinertia degree of P
i
over P . The fundamental identity is:If L|K is separable, then
rX
i=1
ei
fi
= n = [L : K].
Proceeding based upon the Chinese Remainder Theorem, the proof connects thedegree of the larger field extension with the product of the respective exponents ofprime ideals and the degrees of each sub-field extension.
10. Acknowledgments
It is a pleasure to thank my mentor, Vaidehee Thatte, for her invaluable helpon this project. In addition, I would like to thank the UChicago Department ofMathematics, and especially Dr. Peter May, for providing this wonderful opportu-nity, both to myself and other students - it’s an extraordinary program that hasdeveloped my ability to discuss and write about mathematics. Finally, thank youto the people, far too numerous to name here, who tolerated me as I blabbed on andon about Minkowski Bounds, Class Numbers, and lattice points - I truly appreciateyour patience.
References
[1] David S. Dummit and Richard M. Foote. Abstract Algebra, 3rd ed. John Wiley and Sons,Inc., 2004.
[2] Jurgen Neukirch, translated from the German by Norbert Schappacher. Algebraic NumberTheory. Springer, 1999.
[3] Serge Lang. Algebra. Springer, 2002.[4] Robert B. Ash. A Course in Algebraic Number Theory. Robert Ash, 2003. Available at
http://www.math.uiuc.edu/⇠r-ash/ANT.html.
