Chapter 1
Elementary Number Theory
1.1. The Unique Factorization of Integers
In this section, we will state some properties of Z, the set of integers.
Lemma 1.1.1. If a and b are relatively prime, then we can find integers x and y such that ax+by = 1.
By Lemma 1.1.1, we can easily derive the following.
Proposition 1.1.2. If p is a prime and p | ab, then p | a or p | b.
We have the Fundamental Theorem of Arithmetic.
Theorem 1.1.3. Z has unique factorization.
The existence part of Theorem 1.1.3 can be proved by the fact that every positive integer greater than 1 has a prime divisor and the uniqueness part can be proved by using Proposition 1.1.2 Definition 1.1.4. For any integer n, we define φ(n) to be the number of integers less than n which are co-prime to n. This is known as the Euler φ−function
By interpreting φ(n)/n as the probability that a random number chosen from 1, . . . , n is co- prime to n, we have that
φ(n)
n =Y
p|n
¡1 −1 p
¢.
For a prime number p, we have φ(p) = p − 1. Fermat’s Little Theorem says that if a ∈ Z and p6 | a, then ap−1≡ 1 (mod p). This can be proved by using induction and the fact that the binomial coefficient ¡p
k
¢, for k ≤ p − 1 is divisible by p. However, by using elementary group theory, the following proposition which is due to Euler, generalizes Fermat’s Little Theorem
Proposition 1.1.5. Given a, n ∈ Z, aφ(n)≡ 1 (mod n) when gcd(a, n) = 1.
Consider prime of the form an− 1. If n has a factor, say k, then ak− 1 | an− 1. If a > 2, then an− 1 is divisible by a − 1 and is therefore not prime. Therefore if an− 1 is a prime, then a = 2 and n is a prime. Numbers of this form are called Mersenne numbers.
1
2 1. Elementary Number Theory
On the other hand, consider primes of the form 2n+ 1. If n has an odd factor, say n = rs with r odd, then 2rs+ 1 is divisible by 2s+ 1 and is therefore not prime. Hence, we have the following definition.
Definition 1.1.6. We call integers of the form Fn= 22n+ 1, the Fermat numbers
Fermat made the conjecture that these numbers are all primes. Indeed, F0= 3, F1 = 5, F2 = 17, F3 = 257, and F4 = 65537 are all primes, but unfortunately, F5 is divisible by 641. It is unknown if Fn represents infinitely many primes. It is also unknown if Fn is infinitely often composite.
1.2. The ABC Conjecture
Given a nature number n, let n = pα11· · · pαkk. Define the radical of n, denoted rad(n), to be the product p1· · · pk.
In 1980, Masser and Oesterl´e formulated the following conjecture. Suppose that we have three mutually co-prime integers A, B and C satisfying A + B = C. Given any ε > 0, it is conjectured that there is a constant κ(ε) such that
max(|A|, |B|, |C|) ≤ κ(ε)(rad(ABC))1+ε. This is called the ABC Conjecture.
Proposition 1.2.1. Assuming the ABC Conjecture, if xyz 6= 0 and xn+yn= znfor three mutually co-prime integers x, y and z, then n is bounded.
A nature number n is called squarefull, if n has no squarefree part (i.e. if p | n then p2| n).
Erd¨os conjectured that we cannot have three consecutive squarefull natural numbers. Assuming that the Erd¨os conjecture is true, we can show that there are infinitely many primes p such that 2p−16≡1 (mod p2).
Exercise
(1) Let a1, . . . , an for n ≥ 2 be nonzero integers. Suppose that there is a prime p and positive integer h such that ph| ai for some i and ph does not divide aj for all j 6= i. Then show that
S = 1
a1 + · · · + 1 an is not an integer.
(2) Show that n | φ(an− 1) for any a > n.
(3) Show that n6 | 2n− 1 for any nature number n > 1.
(4) Let π(x) be the number of primes less than or equal to x (a) Show that pk < 22k, where pk denotes the k-th prime.
(b) Prove that π(x) ≥ ln(ln x).
(5) Prove Wilson’s Theorem. Thus (p − 1)! ≡ −1 (mod p) if and only if p is a prime. Use Wilson’s theorem to prove that for a prime p, x2 ≡ −1 (mod p) is solvable if and only if p = 2 or p ≡ 1 (mod 4).
(6) Prove that if f (x) ∈ Z[x], then f (x) ≡ 0 (mod p) is solvable for infinitely many primes p.
(7) Let q be prime. Show that there are infinitely many primes p so that p ≡ 1 (mod q). (Hint:
Consider the polynomial f (x) = (xq− 1)/(x − 1) = 1 + x + · · · + xq−1.)
Exercise 3
(8) Prove that if p is an odd prime, any prime divisor of 2p− 1 is of the form 2kp + 1, with k a positive integer.
(9) Let Fn= 22n+ 1 be the Fermat numbers.
(a) Show that Fndivides Fm− 2 if n < m, and from this deduce that Fnand Fm are relatively prime if m 6= n.
(b) Prove that every divisor of Fn is of the form 2n+1k + 1.
(c) Show that there are infinitely many primes p such that p ≡ 1 (mod 2r) for any given r.
(10) Let p be an odd prime and let d be the order of 2 modulo p (i.e. d is the least positive integer such that 2d≡ 1 (mod p).)
(a) Suppose that 2n≡ 1 (mod p) and 2n6≡1 (mod p2). Show that 2d6≡1 (mod p2).
(b) Suppose that 2n≡ 1 (mod p) and 2n6≡1 (mod p2). Show that 2p−16≡1 (mod p2).
(c) Assuming the ABC conjecture, show that there are infinitely many prime p such that 2p−16≡1 (mod p2).
(11) Assuming the ABC conjecture, prove that there are only finitely many n such that n − 1, n and n + 1 are squarefull.
Chapter 2
Euclidean Rings
2.1. Unique Factorization Domain
In this chapter R is a commutative ring with identity 1. If a, b ∈ R, we will write a | b (a divides b), if there exists some c ∈ R such that ac = b. Any divisor of 1 is called a unit. We will say that a and b are associate and write a ∼ b, if there exists a unit u ∈ R such that a = bu. It is easy to verify that ∼ is an equivalent relation.
Definition 2.1.1. We say that π ∈ R is irreducible if for any factorization π = ab, one of a or b is a unit. We say that τ ∈ R is a prime if τ | ab implies that τ | a or τ | b.
We remark that in general, an irreducible element is not necessary a prime.
Definition 2.1.2. Suppose that there is a map N : R → N such that:
(1) N (ab) = N (a)N (b);
(2) N (a) = 1 if and only if a is a unit.
Then we call such a map a norm map with N (a) the norm of a.
Proposition 2.1.3. Suppose that R is an integral domain with a norm map. Then R can be written as a product of irreducible elements.
Definition 2.1.4. We say that R, an integral domain, is a unique factorization domain if:
(1) every element of R can be written as a product of irreducibles;
(2) if a = π1· · · πr = τ1· · · τs with πi and τj irreducibles, then r = s and after a suitable permutation, πi ∼ τi.
Example 2.1.5. Let R = Z[√
−5]. Then 2, 3, 1 +√
−5 and 1 −√
−5 are irreducible in R and they are not associate. Observe that 6 = 2 · 3 = (1 +√
−5)(1 −√
−5), so that R is not a unique factorization domain.
Definition 2.1.6. An ideal I ⊆ R is called principal if it can be generated by a single element of R. A domain R is then called a principle ideal domain, if every ideal of R is principle.
Theorem 2.1.7. If R is a principle ideal domain, then R is a unique factorization domain.
5
6 2. Euclidean Rings
The proof of Theorem 2.1.7 is similar to the proof Theorem 1.1.3. We first prove that every element of R can be written as a product of irreducibles for the existence and then prove that an irreducible element is a prime for the uniqueness.
Now, we describe an important class of principle ideal domains:
Definition 2.1.8. If R is a domain with a map ψ : R → N and given a, b ∈ R, there exist q, r ∈ R such that a = bq + r with r = 0 or ψ(r) < ψ(b), then we call R an Euclidean domain.
Theorem 2.1.9. If R is an Euclidean domain, then it is a principle ideal domain.
Definition 2.1.10. Let R be a unique factorization domain and f (x) ∈ R[x]. Then we define the content of f to be the greatest common divisor of the coefficients of f , denoted by C(f ).
Let R be a unique factorization domain and let K be its field of fraction. The following result, called Gauss’ Lemma, allows us to relate the factorization of polynomials in R[x] with that in K[x].
Lemma 2.1.11 (Gauss’ Lemma). Let R be a unique factorization domain. Then, for f (x) and g(x) ∈ R[x], we have C(f · g) = C(f )C(g).
Because for a field K, K[x] is an Euclidean domain, we have the following.
Theorem 2.1.12. If R is a unique factorization domain, then R[x] is a unique factorization domain.
2.2. Gaussian Integers and Eisenstein Integers Let Z[i] = {a + bi | a, b ∈ Z, i =√
−1}. This ring is often called the ring of Gaussian integers. It can be shown that Z[i] is an Euclidean domain and the only units in Z[i] are ±1 and ±i .
Let ρ = (−1 +√
−3)/2. Notice that ρ2+ ρ + 1 = 0 and notice that ρ2 = ρ. The set Z[ρ] = {a + bρ | a, b ∈ Z} is a ring which is called the ring of Eisenstein integers. Notice that Z[ρ] is closed under complex conjugation. Z[ρ] is also an Euclidean domain and the only units in Z[ρ] are ±1,
±ρ and ±ρ2.
Example 2.2.1. We have (x−ρ)(x−ρ2) = x2+x+1, so that 3 = (1−ρ)(1−ρ2) = (1+ρ)(1−ρ)2=
−ρ2(1 − ρ)2. Because 1 − ρ is irreducible, we have a factorization of 3.
We can use the unique factorization property of Z[i] and Z[ρ] to solve certain Diophantine equations for integers. For instance, we can apply the arithmetic of Z[i] to show that the equation y2+ 1 = x3 has no integer solutions with xy 6= 0.
Exercise
(1) Let D be squarefree. Consider R = Z[√
D]. Show that every element of R can be written as a product of irreducible elements
(2) Find all the prime elements of the ring Z[i].
(3) Show that a positive integer a is a sum of two squares if and only if a = b2c where c is not divisible by any prime p ≡ 3 (mod 4).
(4) Show that Z[ρ]/(1 − ρ) has order 3.
(5) Consider the ring Z[√
2] = {a + b√
2 | a, b ∈ Z}.
(a) Show that Z[√
2] is Euclidean.
Exercise 7
(b) Let ² = 1 +√
2. Write ²n= un+ vn√
2. Show that u2n− 2v2n= ±1.
(c) Show that there is no unit η in Z[√
2] such that 1 < η < ε. Deduce that every unit of Z[√
2] is of the form ±εn for n ∈ Z.
(6) Prove that Z[√
−10] is not a unique factorization domain.
(7) Show that R = Z[(1 +√
−11)/2] is Euclidean.
(8) Show that for considering the norm map, Z[(1 +√
−19)/2] is not Euclidean.
(9) Prove that Z[√
6] is Euclidean.
(10) Show that there are only finitely many rings Z[√
d] with d ≡ 2 or d ≡ 3 (mod 4) which are norm Euclidean.
Chapter 3
Algebraic Numbers and Integers
3.1. Basic Concepts
a number α ∈ C is called an algebraic number if there exists a polynomial f (x) ∈ Q[x] such that f (α) = 0. If f (x) is a monic polynomial with coefficients in Z, we say that α is an algebraic integer.
Clearly all algebraic integers are algebraic numbers. However, the converse is false.
Example 3.1.1. √
2/3 is a root of f (x) = 9x2− 2, so it is an algebraic number. Because there is no monic polynomial g(x) ∈ Z[x] such that g(√
2/3) = 0, √
2/3 is not an algebraic integer.
Proposition 3.1.2. Let α be an algebraic number. There exists a unique polynomial p(x) ∈ Q[x]
which is monic, irreducible and of smallest degree such that p(α) = 0. Furthermore, if f (x) ∈ Q[x]
and f (α) = 0, then p(x) | f (x) in Q[x].
The degree of p(x) is called the degree of α and is denoted deg(α); p(x) is called the minimal polynomial of α.
The set of algebraic numbers is countable. Since the set of complex numbers, C, is uncountable, it follows that there exist complex numbers which are not algebraic. These numbers are called transcendental and the set of transcendental numbers is uncountable.
3.2. Liouville’s Theorem and Generalizations
In 1853, Liouville showed that algebraic numbers cannot be too well approximated by rational numbers.
Theorem 3.2.1 (Liouville). Given a real algebraic number α of degree n 6= 1, there is a constant c = c(α) such that for all rational numbers p/q, (p, q) = 1, the inequality
¯¯
¯¯α −p q
¯¯
¯¯ > c(α) qn holds.
9
10 3. Algebraic Numbers and Integers
Using Theorem 3.2.1, we can give specific examples of transcendental numbers. For instance, we can show that
X∞ n=0
1 10n!
is transcendental.
In 1909, Thue was able to improve Liouville’s inequality.
Theorem 3.2.2 (Thue). If α is algebraic of degree n, then there exists a constant c(α) so that for
all p/q ∈ Q, ¯
¯¯
¯α − p q
¯¯
¯¯ > c(α) qn/2+1. Theorem 3.2.2 has immediate Diophantine applications.
Proposition 3.2.3. Let f (x, y) be an irreducible polynomial of degree n ≥ 3. For any fixed m ∈ Z∗, f (x, y) has only finitely many solutions.
Over a long series of improvements upon Liouvilles’s Theorem, in 1955, Roth was able to improve the inequality.
Theorem 3.2.4 (Roth). Given an algebraic number α, for any ε > 0, there exists a constant c(α, ε) so that for all p/q ∈ Q, ¯
¯¯
¯α − p q
¯¯
¯¯ > c(α, ε) q2+ε .
This improved inequality gives us a new family of transcendental numbers. For example, P∞
n=12−3n is transcendental.
3.3. Algebraic Number Fields
A field K ⊂ C is called an algebraic number field if its dimension over Q is finite. The dimension of K over Q is called the degree of K and is denoted [K : Q].
Let α be an algebraic number of degree n and define Q[α] = {f (α) | f (x) ∈ Q[x]}, a subring of C. Then Q[α] is an algebraic number filed of degree n over Q. From now on, we will denote Q[α]
by Q(α).
Let α and β be algebraic numbers. Q(α, β) is a field since it is the intersection of the subfields of C containing Q, α and β. The intersection of a finite number of subfields in a fixed field is again a field.
Proposition 3.3.1. If α and β are algebraic numbers, then there exists an algebraic number θ such that Q(α, β) = Q(θ).
Proposition 3.3.1 can be generalized quite easily using induction: for a finite set of algebraic numbers α1, . . . , αn, there exists an algebraic number θ such that Q(α1, . . . , αn) = Q(θ). Therefore, any algebraic number field K is Q(θ) for some algebraic number θ.
The roots of minimal polynomial p(x) of α are called the conjugates of α. Since, p(x) has no repeated roots, if n is the degree of p(x), then α has n conjugates. If θ = θ(1) and θ(2), . . . , θ(n)are the conjugates of θ, then Q(θ(i)) for i = 2, . . . , n is called a conjugate field to Q(θ). Further, the maps θ 7→ θ(i) are monomorphisms of Q(θ) → Q(θ(i)) which is referred to as embeddings of Q(θ) to C. we can partition the conjugates of θ into real roots and nonreal roots (called complex roots).
Exercise 11
K is called a normal extension of Q if all the conjugate fields of K are identical to K. We also define the normal closure of any field K as the extension eK of smallest degree containing all the conjugate fields of K.
Using Galois Theory, Liouville’s theorem also holds for α where α is a complex algebraic number of degree n ≥ 2.
The following theorem gives several characterizations of algebraic integers.
Theorem 3.3.2. The following statements are equivalent:
(1) α is an algebraic integer.
(2) The minimal polynomial of α is monic over Z (3) Z[α] is a finite generated Z-module.
(4) There exists a finitely generated Z-submodule M 6= {0} of C such that αM ⊆ M .
Part (c) and (d) of Theorem 3.3.2 are the most useful because they supply us with an immediate tool to test whether a given number is an algebraic integer or not.
Proposition 3.3.3. Let K be an algebraic number field. Let OK be the set of all algebraic integers in K. Then OK is a ring.
Exercise
(1) Show that if r ∈ Q is an algebraic integer, then r ∈ Z.
(2) Find the minimal polynomial of√
n where n is a squarefree integer.
(3) Show that
X∞ n=0
1 an!
is transcendental for a ∈ Z and a ≥ 2.
(4) Show that
X∞ n=0
1 af (n) is transcendental for a ∈ Z and a ≥ 2 when
n→∞lim
f (n + 1) f (n) > 2 .
(5) Show that there are only finitely many integral solutions to the equation x3+ 3x2y + xy2+ y3= m.
(6) Let α be an algebraic number. Show that there exists m ∈ Z such that mα is an algebraic integer.
(7) Let K = Q(θ) be of degree n over Q. Let ω1, . . . , ωn be a basis of K as a vector space over Q.
Show that the matrix (aij) is invertible, where aij = wi(j).
(8) Let f (x) = xn+ an−1xn−1+ · · · + a1x + a0 and assume that p | ai for i = 0, . . . , k − 1 and p26 |a0. Show that f (x) has an irreducible factor of degree at least k.
(9) Prove that f (x) = x6+ 7x5− 12x3+ 6x + 8 is irreducible over Q.
12 3. Algebraic Numbers and Integers
(10) Let ζm be a primitive m-th root of unity. Show that Y
0≤i6=j≤m−1
(ζmi − ζmj ) = (−1)m−1mm. (11) Let
φm(x) = Y
1≤i≤m, (i,m)=1
(x − ζmi ) denote the m-th cyclotomic polynomial.
(a) Prove that
xm− 1 =Y
d|m
φd(x).
(b) Let I be a subset of the positive integers smaller than m which are co-prime to m. Suppose that
f (x) =Y
i ∈I
(x − ζmi ) ∈ Z[x].
Show that if f (ζm) = 0 and f (ζmp) 6= 0 for some prime p, then p | m.
(c) Show that φm(x) ∈ Z[x] and is irreducible in Q[x].
(12) Let a be a squarefree and greater than 1 and let p be a prime. Show that the normal closure of Q(a1/p) is Q(a1/p, ζp).
Chapter 4
Integral Bases
In this chapter, we look more closely at the algebraic structure of OK, the ring of integers of an algebraic number field K. In particular, we show that OK is always a finitely Z-module admitting a Q-basis for K as a generating set (where K is viewed as a Q-vector space). We will also define an important invariant of a number field called the discriminant which arises in many calculations within the number field.
4.1. The Norm and the Trace
Recall that if K is an algebraic number field, then K can be viewed as a finite dimensional vector space over Q. If α ∈ K, the map from K to K defined by Φα : v 7→ αv gives a linear operator on K. We define the trace of α by TrK(α) := Tr(Φα) and the norm of α by NK(α) := det(Φα).
Thus, we choose any Q-basis ω1, ω2, . . . , ωn of K and write αωi =Pn
j=1aijωj for all 1 ≤ i ≤ n, so TrK(α) =Pn
i=1aii and NK(α) = det(aij).
Lemma 4.1.1. If K is an algebraic number field of degree n over Q and α ∈ OK its ring of integers, then TrK(α) and NK(α) are in Z.
Given an algebraic number field K and ω1, ω2, . . . , ωn a Q-basis for K, consider the corre- spondence from K to Mn(Q) (the space of n × n matrix over Q) given by α 7→ (aij) where αωi =Pn
j=1aijwj. This is readily seen to give a homomorphism from K to Mn(Q). From this we can deduce that TrK(·) is in fact a Q-linear map from K to Q.
Lemma 4.1.2. The bilinear pairing given by B(x, y) : K × K → Q such that (x, y) 7→ TrK(xy) is nondegenerate.
We remark that the definition of nondegeneracy above is independent of the choice of basis.
4.2. Existence of an Integral Basis
Let K be an algebraic number field of degree n over Q and OK its ring of integers. We say that ω1, ω2, . . . , ωn is an integral basis for OK if ωi ∈ OK for all i and OK = Zω1+ Zω2+ · · · + Zωn. In general, we say that a Z-module M has an integral basis if there exists α1, α2, . . . , αm ∈ M such that M = Zα1+ Zα2+ · · · + Zαm, for some positive integer m, and α1, . . . , αm are linearly independent over Z.
13
14 4. Integral Bases
For a module M with submodule N , we can define the index of N in M to be the number of elements in M/N and denote this by [M : N ].
Lemma 4.2.1. Let α1, α2, . . . , αn be a set of generators for a finitely generated Z-module M and let N be a submodule.
(1) There exist β1, β2, . . . , βm in N with m ≤ n such that N = Zβ1+ Zβ2 + · · · Zβm and βi =P
j≥ipijαj with 1 ≤ i ≤ m and pij ∈ Z.
(2) If m = n, then [M : N ] = p11p22· · · pnn.
In Exercise 2, we ask you to prove that there exist w1, w2. . . , wn ∈ K such that OK ⊆ Zw1+ Zw2+ · · · + Zwn. Applying Lemma 4.2.1, we have the following.
Theorem 4.2.2. Ok has an integral basis.
For example, let K = Q(√
D) with D a squarefree integer. If D ≡ 1 (mod 4), then we can choose 1, (1 +√
D)/2 as an integral basis. If D ≡ 2, 3 (mod 4), then we can choose 1, √
D as an integral basis.
We are justified now in making the following definition.
Definition 4.2.3. If K is an algebraic number field of degree n over Q and OK its ring of integers, define the discriminant of K as dK := det(ωi(j))2, where ω1, ω2, . . . , ωn is an integral basis for OK.
Note that the discriminant is independent of the choice of integral basis.
We can generalize the notion of a discriminant for arbitrary elements of K. Let K be an algebraic number field of degree n over Q. Let σ1, σ2, . . . , σn be the embeddings of K. For a1, a2, . . . , an∈ K, we can define dK/Q(a1, a2, . . . , an) = det(σi(aj))2.
We denote dK/Q(1, a, a2, . . . , an−1) by dK/Q(a). If α is an algebraic integer with minimal poly- nomial f (x). Then we have
dK/Q(α) = (−1)n(n−1)/2 Yn i=1
f0(α(i)).
Definition 4.2.4. Suppose α is an algebraic integer of degree n, generating a field K. We define the index of α to be the index of Z + Zα + · · · + Zαn−1 in OK.
Suppose that the minimal polynomial of α is an Eisenstein polynomial with respect to a prime number p. We can show that the index of α is not divisible by p.
4.3. Ideals in OK
At this point, we have shown that OK is indeed much like Z in its algebraic structure. It turns out that we are only halfway to the final step in our generalization of an integer in a number field. We may think of the ideals in OK as the most general integers in K, and we remark that when this set of ideals is endowed with the usual operations of ideal addition and multiplication, we recover an arithmetic most like that of Z.
Because for a nonzero ideal a of OK, we have a ∩ Z 6= {0}, it implies that the index of a in OK must be finite. By Lemma 4.2.1, a has an integral basis.
We define the norm of a nonzero ideal a in OK to be its index in OK. We will denote the norm of a by N (a).
Exercise 15
Exercise
(1) Determine the algebraic integers of K = Q(√
−5).
(2) Show that there exist w1, w2. . . , wn∈ K such that OK⊆ Zw1+ Zw2+ · · · + Zwn.
(3) Let K be an algebraic number field of finite degree over Q. Show that dK ≡ 0 or 1 (mod 4).
(4) Let σ1, σ2, . . . , σnbe the embeddings of K. For a1, a2, . . . , an∈ K, define dK/Q(a1, a2, . . . , an) = det(σi(aj))2.
(a) Show that
dK/Q(1, a, . . . , an−1) =Y
i>j
¡σi(a) − σj(a)¢2 .
(b) Suppose that ui =Pn
j=1aijvj with aij ∈ Q, vj ∈ K. Show that dK/Q(u1, u2, . . . , un) =¡
det(aij)¢2
dK/Q(v1, v2, . . . , vn).
(c) Let a1, a2, . . . , an be linearly independent over Q. Let N = Za1+ Za2+ · · · + Zan and m = [OK : N ]. Prove that
dK/Q(a1, a2, . . . , an) = m2dK. (d) Let a be an integral ideal with basis α1, . . . , αn. Show that
|det(α(j)i )|2 = N (a)2dK.
(e) Let K be an algebraic number field. Suppose that θ ∈ OK is such that dK/Q(θ) is square- free. Show that OK = Z[θ].
(f) Let f (x) = xn+ an−1xn−1+ · · · + a1x + a0 with ai ∈ Z be the minimal polynomial of θ.
Let K = Q(θ). If for each prime p such that p | dK/Q(θ) we have f (x) Eisensteinian with respect to p, show that OK = Z[θ].
(5) Let K = Q(α) where α = r1/3, r = ab2 ∈ Z, where ab is squarefree. Suppose that 36 | b. Find an integral basis for OK.
(6) Let K = Q(θ) where θ3− θ2− 2θ − 8 = 0
(a) Show that f (x) = x3− x2− 2x − 8 is irreducible over Q.
(b) Consider β = (θ2+ θ)/2. Show that β is an algebraic integer.
(c) Show that dK/Q(1, θ, β) = −503 and dK/Q(θ) = −4 · 503. Deduce that 1, θ, β is a Z-basis of OK.
(d) Show that every integer x of K has an even discriminant.
(e) Deduce that OK is not of the form Z[α].
(7) Let m = pa with p prime and let K = Q(ζm) where ζm is a primitive m-th root of 1.
(a) Show that (1 − ζm)φ(m)= p OK. (b) Show that
dK/Q(ζm) = (−1)φ(m)2 mφ(m) pmp .
(c) Show that {1, ζm, . . . , ζmφ(m)−1} is an integral basis for OK. Deduce that dK = dK/Q(ζm).
(d) Show that Z[ζm+ ζm−1] is the ring of integers of Q(ζm+ ζm−1).
(8) Suppose that K is a number field with r1 real embeddings and 2r2complex embeddings so that r1+ 2r2= [K : Q]. Show that dK has sign (−1)r2.
(9) Show that only finitely many imaginary quadratic fields are Euclidean.
16 4. Integral Bases
(10) Let K and L be algebraic number fields of degree m and n, respectively, over Q. Let d = gcd(dK, dL).
(a) Show that if [KL : Q] = mn, then dOKL⊆ OKOL. (b) Suppose d = 1. Show that dKL = dnKdmL.
(c) Let K = Q(√
2) and L = Q(√
−3). Find an integral basis for OKL.
(d) Let ζm be a primitive m-th root of 1 and let K = Q(ζm). Show that OK = Z[ζm] and dK = (−1)φ(m)2 mφ(m)
Q
p | mpφ(m)p−1 .
(11) Let K be an algebraic number field of degree n over Q. Let a1, a2, . . . , an ∈ OK be linearly independent over Q and set
4 = dK/Q(a1, a2, . . . , an).
(a) Show that if α ∈ OK, then 4α ∈ Z[a1, . . . , an].
(b) For each i ∈ {1, . . . , n}, choose the least natural number dii so that for some dij ∈ Z, the number
wi = 4−1 Xi j=1
dijaj ∈ OK.
Show that w1, . . . , wn is an integral basis of OK.
(c) Show that there is an integral basis w1, . . . , wn of OK such that for j = 1, . . . , n, aj = cj1w1+ · · · + cjjwj for some cij ∈ Z.
(d) If Q ⊆ K ⊆ L and K, L are algebraic number fields, show that dK| dL.
Chapter 5
Dedekind Domains
The notion of a Dedekind domain is the concept we need in order to establish the unique factoriza- tion of ideals as a product of prime ideals. We will also meet the fundamental idea of a Noetherian ring. It turns out that Dedekind domains can be studied in the wider context of Noetherian rings.
5.1. Characterizing Dedekind Domains
Let R be a commutative ring with identity. If p is a prime ideal containing the product a1a2· · · ar of r ideals of R, then ai ⊆ p for some i.
We know that a finite integral domain is a field. Let K be an algebraic number field and OK its ring of integers. Since every nonzero ideal in OK has finite index in OK, we have that OK/p is a finite field for every nonzero prime ideal p of OK. Hence, every nonzero prime ideal of OK is maximal.
Definition 5.1.1. For any field containing R, we say that α ∈ L is integral over R if α satisfies a monic polynomial equation f (α) = 0 with f (x) ∈ R[x]. R is said to be integrally closed if every element in the quotient field of R which is integral over R, already lies in R
The following theorem and its proof were exactly the same as Theorem 3.3.2, with OK replacing Z.
Theorem 5.1.2. Let K be an algebraic number field and OK its ring of integers. The following statements are equivalent:
(1) α ∈ OK
(2) OK[α] is a finite generated OK-module.
(3) There exists a finitely generated OK-submodule M 6= {0} of C such that αM ⊆ M . By Theorem 5.1.2, we have the following important result.
Theorem 5.1.3. Let K be an algebraic number field and OK its ring of integers. Then OK is integrally closed.
Definition 5.1.4. A ring is called Noetherian if every ascending chain a1 ⊆ a2 ⊆ a3⊆ · · · of ideals terminates, i.e., if there exists n such that an= an+k for all k ≥ 0.
Lemma 5.1.5. For any commutative ring R with identity, the following are equivalent:
17
18 5. Dedekind Domains
(1) R is Noetherian;
(2) every nonempty set of ideals contains a maximal element; and (3) every ideal of R is finitely generated.
If a ( b are ideals of OK, then we have that N (a) > N (b). Since N (a) is a positive integer for every ideal a in OK, this shows that OK is Noetherian.
Thus, we have proved that:
(1) OK is integrally closed;
(2) every nonzero prime ideal of OK is maximal; and (3) OK is Noetherian.
A commutative integral domain which satisfies these three conditions is called a Dedekind domain. We have thus seen that OK is a Dedekind domain.
5.2. Fractional Ideals and Unique Factorization
A fractional ideal A of OK is an OK-module contained in K such that there exists m ∈ Z with mA ⊆ OK. Of course, any ideal of OK is a fractional ideal by taking m = 1.
It is easy tocheck that any fractional ideal is finitely generated as an OK-module and the sum and product of two fractional ideals are again fractional ideals.
To show that every ideal can be written as a product of prime ideals uniquely, we need the following lemmas.
Lemma 5.2.1. Any proper ideal of OK contains a product of nonzero prime ideals.
Lemma 5.2.2. Let p be a prime ideal of OK. There exists z ∈ K \ OK, such that zp ⊆ OK. Let p be a prime ideal. Define
p−1= {x ∈ K | xp ⊆ OK}.
Lemma 5.2.2 implies, in particular, that p−16= OK.
Lemma 5.2.3. Let p be a prime ideal of OK. Then p−1 is a fractional ideal and pp−1 = OK. Theorem 5.2.4. Any ideal of OK can be written as a product of prime ideals uniquely.
For a and b ideals of OK, we say a divides b (denoted a | b), if a ⊇ b.
Given two ideals a and b, the ideal d satisfying:
(1) d | a and d | b; and (2) if e | a and e | b then e | d
is called the great common divisor of a and b and denoted by gcd(a, b).
Similarly, the ideal m satisfying:
(1) a | m and b | m; and (2) if a | n and b | n then m | n
is called the least common multiple of a and b and denoted by lcm(a, b).
Theorem 5.2.5 (Chinese Remainder Theorem). Let a and b be ideals so that gcd(a, b) = 1. Given a, b ∈ OK, we can solve x ≡ a (mod a) and x ≡ b (mod b). Furthermore, let p1, p2, . . . , pr be r distinct prime ideals in OK. Given ai ∈ OK, ei∈ Z+, there exist x such that x ≡ ai (mod pri1) for all i = 1, . . . , r.
5.2. Fractional Ideals and Unique Factorization 19
Definition 5.2.6. Given a prime ideal p. We define the order of a in p by ordp(a) = t if pt| a and pt+16 | a.
It easy to show that ordp(ab) = ordp(a) + ordp(b), by unique factorization of ideals.
Proposition 5.2.7. Let p, p1, · · · , pr be prime ideals in OK. We have that (1) OK/p ' pe−1/pe and N (pe) = (N (p))e, for every integer e ≥ 1.
(2) If a =Qr
i=1peii, then N (a) =Qr
i=1N (pei1).
