Chapter 1 Algebraic Numbers and Algebraic Integers 1.1 Rings of integers We start by introducing two essential notions: number ¯eld and algebraic inte- ger. De¯nition 1.1. A number ¯eld is a ¯nite ¯eld extension K of Q, i.e., a ¯eld which is a Q-vector space of ¯nite dimension. We note this dimension [K : Q] and call it the degree of K. Examples 1.1. 1. The ¯eld p p Q( 2) = fx + y 2 j x; y 2 Qg is a number ¯eld. It is of degree 2 over Q. Number ¯elds of degree 2 over Q are called quadratic ¯elds. More generally, Q[X]=f(X) is a number ¯eld if f is irreducible. It is of degree the degree of the polynomial f. 2. Let ³n be a primitive nth root of unity. The ¯eld Q(³n) is a number ¯eld called cyclotomic ¯eld. 3. The ¯elds C and R are not number ¯elds. Let K be a number ¯eld of degree n. If ® 2 K, there must be a Q-linear dependency among f1; ®; : : : ; ®ng, since K is a Q-vector space of dimension n. In other words, there exists a polynomial f(X) 2 Q[X] such that f(X) = 0. We call ® an algebraic number. De¯nition 1.2. An algebraic integer in a number ¯eld K is an element ® 2 K which is a root of a monic polynomial with coe±cients in Z. 3 4 CHAPTER 1. ALGEBRAIC NUMBERS AND ALGEBRAIC INTEGERS p p Example 1.2. Since X2 ¡2 = 0, 2 2 Q( 2) is an algebraic integer. Similarly, i 2 Q(i) is an algebraic integer, since X2 + 1 = 0. However, an element a=b 2 Q is not an algebraic integer, unless b divides a. Now that we have the concept of an algebraic integer in a number ¯eld, it is natural to wonder whether one can compute the set of all algebraic integers of a given number ¯eld. Let us start by determining the set of algebraic integers in Q. De¯nition 1.3. The minimal polynomial f of an algebraic number ® is the monic polynomial in Q[X] of smallest degree such that f(®) = 0. Proposition 1.1. The minimal polynomial of ® has integer coe±cients if and only if ® is an algebraic integer. Proof. If the minimal polynomial of ® has integer coe±cients, then by de¯nition (De¯nition 1.2) ® is algebraic. Now let us assume that ® is an algebraic integer. This means by de¯nition that there exists a monic polynomial f 2 Z[X] such that f(®) = 0. Let g 2 Q[X] be the minimal polyonial of ®. Then g(X) divides f(X), that is, there exists a monic polynomial h 2 Q[X] such that g(X)h(X) = f(X): (Note that h is monic because f and g are). We want to prove that g(X) actually belongs to Z[X]. Assume by contradiction that this is not true, that is, there exists at least one prime p which divides one of the denominators of the coe±cients of g. Let u > 0 be the smallest integer such that pug does not have anymore denominators divisible by p. Since h may or may not have denominators divisible by p, let v ¸ 0 be the smallest integer such that pvh has no denominator divisible by p. We then have pug(X)pvh(X) = pu+vf(X): The left hand side of this equation does not have denominators divisible by p anymore, thus we can look at this equation modulo p. This gives u v p g(X)p h(X) ´ 0 2 Fp[X]; where Fp denotes the ¯nite ¯eld with p elements. This give a contradiction, since the left hand side is a product of two non-zero polynomials (by minimality of u and v), and Fp[X] does not have zero divisor. Corollary 1.2. The set of algebraic integers of Q is Z. a a Proof. Let b 2 Q. Its minimal polynomial is X ¡ b . By the above proposition, a b is an algebraic integer if and only b = §1. De¯nition 1.4. The set of algebraic integers of a number ¯eld K is denoted by OK . It is usually called the ring of integers of K. 1.1. RINGS OF INTEGERS 5 The fact that OK is a ring is not obvious. In general, if one takes a; b two algebraic integers, it is not straightforward to ¯nd a monic polynomial in Z[X] which has a + b as a root. We now proceed to prove that OK is indeed a ring. Theorem 1.3. Let K be a number ¯eld, and take ® 2 K. The two statements are equivalent: 1. ® is an algebraic integer. 2. The Abelian group Z[®] is ¯nitely generated (a group G is ¯nitely generated if there exist ¯nitely many elements x1; :::; xs 2 G such that every x 2 G can be written in the form x = n1x1 + n2x2 + ::: + nsxs with integers n1; :::; ns). Proof. Let ® be an algebraic integer, and let m be the degree of its minimal polynomial, which is monic and with coe±cients in Z by Proposition 1.1. Since all ®u with u ¸ m can be written as Z-linear combination of 1; ®; : : : ; ®m¡1, we have that Z[®] = Z © Z® © ::: © Z®m¡1 and f1; ®; : : : ; ®m¡1g generate Z[®] as an Abelian group. Note that for this proof to work, we really need the minimal polynomial to have coe±cients in Z, and to be monic! Conversely, let us assume that Z[®] is ¯nitely generated, with generators a1; : : : ; am, where ai = fi(®) for some fi 2 Z[X]. In order to prove that ® is an algebraic integer, we need to ¯nd a monic polynomial f 2 Z[X] such that f(®) = 0. Let N be an integer such that N > deg fi for i = 1; : : : ; m. We have that Xm N ® = bjaj; bj 2 Z j=1 that is Xm N ® ¡ bjfj(®) = 0: j=1 Let us thus choose Xm N f(X) = X ¡ bjfj(X): j=1 Clearly f 2 Z[X], it is monic by the choice of N > deg fi for i = 1; : : : ; m, and ¯nally f(®) = 0. So ® is an algebraic integer. Example 1.3. We have that na o Z[1=2] = j b is a power of 2 b 1 is not ¯nitely generated, since 2 is not an algebraic integer. Its minimal poly- 1 nomial is X ¡ 2 . 6 CHAPTER 1. ALGEBRAIC NUMBERS AND ALGEBRAIC INTEGERS Corollary 1.4. Let K be a number ¯eld. Then OK is a ring. Proof. Let ®; ¯ 2 OK . The above theorem tells us that Z[®] and Z[¯] are ¯nitely generated, thus so is Z[®; ¯]. Now, Z[®; ¯] is a ring, thus in particular ®§¯ and ®¯ 2 Z[®; ¯]. Since Z[®§¯] and Z[®¯] are subgroups of Z[®; ¯], they are ¯nitely generated. By invoking again the above theorem, ® § ¯ and ®¯ 2 OK . Corollary 1.5. Let K be a number ¯eld, with ring of integers OK . Then QOK = K. Proof. It is clear that if x = b® 2 QOK , b 2 Q, ® 2 OK , then x 2 K. Now if ® 2 K, we show that there exists d 2 Z such that ®d 2 OK (that is ®d = ¯ 2 OK , or equivalently, ® = ¯=d). Let f(X) 2 Q[X] be the minimal polynomial of ®. Choose d to be the least common multiple of the denominators of the coe±cients of f(X), then (recall that f is monic!) µ ¶ X ddeg(f)f = g(X); d and g(X) 2 Z[X] is monic, with ®d as a root. Thus ®d 2 OK . 1.2 Norms and Traces De¯nition 1.5. Let L=K be a ¯nite extension of number ¯elds. Let ® 2 L. We consider the multiplication map by ®, denoted by ¹®, such that ¹® : L ! L x 7! ®x: This is a K-linear map of the K-vector space L into itself (or in other words, an endomorphism of the K-vector space L). We call the norm of ® the determinant of ¹®, that is NL=K (®) = det(¹®) 2 K; and the trace of ® the trace of ¹®, that is TrL=K (®) = Tr(¹®) 2 K: Note that the norm is multiplicative, since NL=K (®¯) = det(¹®¯) = det(¹® ± ¹¯) = det(¹®) det(¹¯) = NL=K (®)NL=K (¯) while the trace is additive: TrL=K (®+¯) = Tr(¹®+¯) = Tr(¹®+¹¯) = Tr(¹®)+Tr(¹¯) = TrL=K (®)+TrL=K (¯): In particular, if n denotes the degree of L=K, we have that n NL=K (a®) = a NL=K (®); TrL=K (a®) = aTrL=K (®); a 2 K: 1.2. NORMS AND TRACES 7 Indeed, the matrix of ¹a is given by a diagonal matrix whose coe±cients are all a when a 2 K. Recall that the characteristic polynomial of ® 2 L is the characteristic poly- nomial of ¹®, that is ÂL=K (X) = det(XI ¡ ¹®) 2 K[X]: This is a monic polynomial of degree n = [L : K], the coe±cient of Xn¡1 is ¡TrL=K (®) and its constant term is §NL=K (®). p Examplep 1.4. Let L be the quadratic ¯eld Q( 2), K = Qp, and take ® 2 Q( 2). In order to compute ¹®, we need to ¯x a basis of Q( 2) as Q-vector space, say p f1; 2g: p Thus, ® can be written ® = a + b 2, a; b 2 Q. By linearity, it is enough to compute ¹® on the basis elements: p p p p p ¹®(1) = a + b 2; ¹®( 2) = (a + b 2) 2 = a 2 + 2b: We now have that µ ¶ ¡ p ¢ a 2b ¡ p p ¢ 1; 2 = a + b 2; 2b + a 2 b a | {z } M and M is the matrix of ¹® in the chosen basis.