Explicit Methods in Algebraic Number Theory

Explicit Methods in Algebraic Number Theory Amalia Pizarro Madariaga Instituto de Matemáticas Universidad de Valparaíso, Chile amaliapizarro@uvcl 1 Lecture 1 11 Number fields and ring of integers Algebraic number theory studies the arithmetic aspects of the number fields Such fields are involved in the solution of many rational problems, as the following diophantine problems Pell Equation: Find integer numbers x, y such that x dy 1 = 0, with d > 1 squarefree Note that x dy = (x dy)(x + dy), if we consider the ring Z[ d] = {a + b d : a, b Z} So, to solve this diophantine equation is equivalent to look into Z[ d] Pythagorean triples: Find integers numbers without common factors x, y, z such that x + y = z x + y = (x + yi)(x iy) in Z[i] We have Z[i] in a unique factorization domain (exercise), so each element in Z[i] can be written in a unique way (unless order an multiplication by units) as the product of irreducible elements By using this fact, it is possible to prove that x + iy = uα, with α, u Z[i] and u a unit (ie u {±1} (exercise) If α = m + ni, with m, n Z, then x + iy = ±(m + ni) = ±(m + mni n ), ie, x = ±(m n ), y = ±mn Therefore, z = (m + n ) and z = ±(m + n ) m and n must be relatively primes and not both odd Is it possible to apply this idea to solve the general case x n + y n = z n, n >? Fermat 1 conjectured that there is no integer solution non zero for n > In order to study this problem, it is enough to consider the case n = p, with p an odd prime Suppose that for some odd prime p there is a solution x, y, z Z {0} with no common factors Let us consider the following cases: (a) p does not divide any x, y, z 1 Now it is known as the Last Fermat Theorem and was proved by Andrew Wiles in 94 1

(b) p divides exactly one of them We will only see the case (a) p = 3 If x, y, z are not multiples of 3, then x 3, y 3, z 3 ±1 (mod 9) and x 3 + y 3 z 3 (mod 9), so it cannot have non trivial solution p > 3 x p + y p = (x + y)(x + ζ p y)(x + ζ py) (x + ζ p 1 p y) = z p, where ζ p = e πi/p in Z[ζ p ] = {a p ζ p p + a 1 ζ p + a 0 } (exercise) Kummer asserted that this ring was a unique factorization domain and from here he obtained a proof of the Fermat Theorem However, only is valid if p < 3 Idea: If we assume that Z[ζ p ] is a UFD, it is possible to prove that x + ζ p y = uα p, for some α Z[ζ p ] and u Z[ζ p ] and also that if x, y are not divisible by p, then x y (mod) p Putting x p + ( z) p = ( y) p, we obtain that x z (mod) p This implies x p x x p + y p = z p x ( x) p (mod) p, so p 3x p, but p 3 and p does not divide x, which is a contradiction and then there is no solutions of the case (a) More general case: Dedekind discovered that although the elements of Z[ζ p ] may not factor in a unique way in irreducibles, the ideals of this ring always factors in product of prime ideals From here, it is possible to prove that the principal ideal generated by x + ζ p y may be written as (x + ζ p y) = I p, for some I ideal There are certain primes p (regular primes) for which I may be a principal ideal I = (α), then (x + ζ p y) = I p = (α) p = (α p ), and again (x + ζ p y) = uα p, for u a unit Then x y (mod) p, which is a contradiction 1 Number Fields Definition 1 A field K is an algebraic number field if is a finite extension of Q Their elements will be called algebraic numbers, that is, they are roots of nonzero polynomials with rational coefficients The monic polynomial P α (x) of lowest degree of which α K is a root is called de minimal polynomial of α If α is root of g(x) Q[x], then P α (x) g(x) Example 1 Quadratic fields Quadratic fields are extension K of Q of degree Q( d), where d is squarefree If d < 0 we say that Q( d) is an imaginary quadratic field and of d > 0 a real quadratic field

Example Cyclotomic fields Let n 1 and let ζ n be a primitive n-th root of unity in C The n-th cyclotomic field is the field Q(ζ n ) The degree of this field over Q is φ(n), where φ is the Euler s phi function The minimal polynomial of ζ n over Q is called the cyclotomic polynomial Φ n (x) and it verifies the following: (i) Let U n be the group of n-th roots of unity in C and let U n = {ζ a n : 0 a < n, gcd(a, n) = 1} Then Φ n (x) = ζ U (x ζ) (ii) Φ n (x) is a monic polynomial with integer coefficients and irreducible over Q Its degree is φ(n) (ii) d n Φ d(x) = x n 1 13 Algebraic Integers Definition An element α in a number field will be called algebraic integer if there exists a monic polynomial f(x) Z[x] such that f(α) = 0 Example 3 integer 3, + are algebraic integers 3 is algebraic, but it is no an algebraic Theorem 11 Let α be an algebraic integer Then, the minimal polynomial of α has integer coefficients Proof Let P α (x) Q[x] the minimal polynomial of α and g(x) Z[x] with g(α) = 0 Then g = P α h, for some h(x) Q[x] If P α (x) / Z[x], then there is a prime p dividing the denominator of some coefficient of P α Let p i the biggest power of p with this property and p j the biggest power dividing the coefficients of h Then: p i+j g = (p i P α )(p j h) 0 (mod) p As Z p [x] is an integral domain, p i P α or p j h are zero mod p, which is a contradiction From now, we will denote by O K the set of algebraic integers in the number field K Corollary 1 O Q = Z 3

14 Characterization of Algebraic Integers Theorem 1 The following assertions are equivalents: (i) α is algebraic integer (ii) Z[α] = {f(α) : f(x) Z[x]} is a finitely generated Z-module (iii) There exists a finitely generated Z-module M such that αm M and γm {0} for all γ Z[α] {0} Proof i) ii) Let f(x) = x n + a 1 x + a 0 Z[x] and f(α) = 0 Let us consider the following Z-module: M = Z+Zα+ Zα n 1 It is clear that M Z[α] By induction: suppose that α k M, then: α n+k = α k α n = α k [ (a n 1 α n 1 + + a 0 ] = ( α k a n 1 )α n 1 + + ( α k a 0 ) Because α k a i Z[α] for i = 0, 1,, n 1, we have that α n+k M, therefore M = Z[α] ii) iii) We take M = Z[α] As α M, then αm M and γ = γ 1 γm iii) i) Let {x 1, x x r } be a generators of M By hypothesis αx i M, then there exists a set of integers numbers c ij such that αx i = r j=1 c ijx j, for all i = 1,, r Let C = (c ij ) ij, then C x 1 x r = α x 1 x r (C αi d) x 1 x r = 0 There is at least one x i non zero, so det(c αi d ) = 0 and then det(c xi d ) Z[x] Theorem 13 Let K be a number field Then O K is a ring Proof If α, β are algebraic integers, then Z[α] and Z[β] are a finitely generated as Z- modules From here, we have that M = Z[α, β] also is a finitely generated Z-module Moreover, (α ± β)m M and (αβ)m M, and then α ± β and αβ belong to the set of algebraic integers 15 Discriminant of Number Fields Let K be a number field with [K : Q] = n and let σ 1,, σ n be the complex embeddings of K For α 1,, α n K we define the discriminant of α 1,, α n by D K (α 1,, α n ) = det(σ i (α j )) (11) 4

Theorem 14 D K (α 1,, α n ) = det(t K (α i α j ) Lemma 1 If γ i = n j=1 c ijα j, with c ij Q, then Proof γ k γ m = n i,j=1 c kic mj α i α j D K (γ 1,, γ n ) = det(c ij ) D K (α 1,, α n ) Theorem 15 D K (α 1,, α n ) 0 if and only if the set {α 1,, α n } is linearly independent over Q Proof If {α 1,, α n } is linearly dependent over Q then the columns if the matrix (σ i (α j )) are linearly dependent, so D K (α 1,, α n ) = 0 Reciprocally, if D K (α 1,, α n ) = 0 then the columns of (T K (α i α j )) ij are ld Let us suppose that α 1,, α n is li We fix rational numbers (not all zero) such that a 1 R 1 + + a n R n = 0, where R l are the columns of (T K (α i α j )) ij and let α = a 1 α 1 + + a n α n 0 Looking the j-th coordinate of each row, we see that T k (αα j ) = 0 for all j Note that {α 1, α,, α n } is in fact, a basis for K over Q and then {αα 1, αα,, αα n } is a also a basis, then T K (β) = 0 for all β K, which is a contradiction Theorem 16 Let K = Q(α), and α 1, α,, α n the conjugated of α over Q Then D K (1, α, α,, α n 1 ) = (α r α s ) = ±N K (f (α)), 1 r<s n where f is the irreducible monic polynomial of α over Q and the signe is + if and only if n 0 or 1 (mod 4) Proof It is not difficult to prove that 1 α 1 α1 α n 1 1 D K (1, α, α,, α n 1 1 α α α n 1 ) = det 1 α n αn αn n 1 = 1 r<s n By using that N K (f (α)) = n i=1 σ i(f (α)), we prove the second equality 16 Integral basis (α r α s ) By using discriminant, we can prove that the ring of integers O K of a number field K with [K : Q] = n is a free abelian group of rank n, that is, is the direct product of n subgroups, each of which is isomorphic to Z It is known that if A and B are free abelian groups of rank n, and A B C, then so is C If α K, then there exists an integer m Z such that mα is an algebraic integer Following this, we can find a basis of K over Q, say {α 1,, α n }, contained in O K So, the free abelian group of rank n A = Zα 1 + + Zα n is contained in O K 5

Theorem 17 Let {α 1,, α n } be a basis for K over Q consisting entirely of algebraic integers, and set D = D K (α 1,, α n ) Then, every α O K can be expressed in the form 1 D (m 1α 1 + + m n α n ) with m j Z and m j are divisible by D It follows that O K is containdes in the free abelian group B = Z α 1 D + + αn D, so we have the following corollary Corollary O K is a free abelian group of rank n It means that there exits β 1,, β n in O K such that every α O K has unique representation m 1 β 1 + + m n β n, where m i Z The set {β 1,, β n } is called integral basis for O K Although ring of integers has many integral basis, their discriminants are the same Theorem 18 Let {β 1,, β n } and {α 1,, α n } be two integral bases for O K Then Proof It is enough to apply lemma(1) D K (β 1,, β n ) = D K (α 1,, α n ) Definition 3 Let K be a number field of degree n over Q We define the disciminant of K by D K := D K (α 1,, α n ), where α 1,, α n is a integral basis of O K 17 Some explicit computations I 18 Ring of Integers of Quadratic Number Fields Let us consider a quadratic number field K = Q( d) with d square free Let α = a+b d O K, then its conjugate α = a b d is also in O K We have that α+α O K, but a Q, so a is in fact an integer so a = a, Z Looking the equation of α over Q 0 = (x α)(x α ) = x (α + α )x + αα, it follows that αα Z (because αα O K Q), that is, ( ) a αα = a b d = b d Z Then, (a ) 4b d 4Z Because a Z, 4b d Z and so 4b Z due to d is square free Now it follows that b Z and so b = b, with b Z Now, we can see that α has the following representation: 6

Also, because αα Z, α = a + b d ( ) a ( ) b d Z Note that d 0 (mod 4), so d 1,, 3 (mod 4) and (a ) (b ) d (mod) 4 Therefore, a and b have the same parity Let us see the two cases: If a and b are even, then α = ã + b d, with ã and b Z If a and b are odd, then (a ) (b ) 1 (mod) 4, so d 1 (mod 4) Finally, we have the following proposition: Proposition 1 If Q( d) with d squarefree, then { Z[ d], if d, 3 (mod 4) O K = [ ] Z 1+ d, if d 1 (mod 4) From the proposition it is clear that an integral basis, depending on d, is the following: { {1, d}, if d, 3 (mod 4) {1, 1+ d }, if d 1 (mod 4) 19 Discriminant of Quadratic Number Fields The complex embeddings of Q( d) are σ ( a+b d) = a+b d and τ(a+b d) = a b d So, T k (a + b d) = a Let {1, α} be a integral basis of O K Then, ( Tk (1) T D K (1, α) = det K (α) T K (α) T k (α ) Finally, we have: D K = ) ( a = det a (a + b d) { 4d, if d, 3 (mod 4) d, d 1 (mod 4) Moreover, this discriminant satisfies the following: ) = 4b d D K 0, 1 (mod 4), Stickelgerger s criterion 7

110 Ring of Integers and Discriminant of Cyclotomic Number Fields Proposition Let n = p l with p a prime number and ζ a primitive n-th root of unity in C Then {1, ζ,, ζ φ(n) 1 } is a Q- basis of K = Q(ζ) and D K (1, ζ,, ζ φ(n) 1 ) = ±r s, where s = p l 1 (lp l 1) Proof The main steps are the following: Φ n (x) = xpl 1 x pl 1 = x pl 1 (l 1) + + x pl 1 + 1 From (16), D K (1, ζ,, ζ φ(n) 1 ) = ±N K (φ (ζ)) Φ n (x) = pl ζ pl 1 ζ n/p N K (φ (ζ)) = N K(p l ζ pl 1 ) N K (ζ n/p ) = plφ(n) N K (ζpl 1 ) N K (ζ n/p ) Proposition 3 Let n = p, with p a prime number and let ζ be a n-th primitive root of unity If K = Q(ζ), then {1, ζ, ζ,, ζ p } is and integral basis for O K Proof The main steps are the following: Φ p (x) = xp 1 x 1 = xp + + x + 1 p 1 From (16), D K (1, ζ,, ζ p ) = ± (φ (ζ i )) Φ ( ζ i ) = pζ i ζ i 1 p 1 D K (1, ζ,, ζ p pζ i ) = ± ζ i 1 = ± pp 1 Φ p (1) = ±pp 0 i=1 i=1 ζ i are algebraic integers, so if {α 1,, α p } is an integral basis, then from (1), D K (1, ζ,, ζ p ) = c D K (α 1,, α p ), where c is the determinant of the matrix C = (c ij ) where ζ i = c ij α j it verifies that c = 1 Let us considere the following result: p j=1 8

Let a 1,, a n O K linearly independent over Q Let N = Za 1 + + Za n and m = [O K : N] Prove that D K (a 1,, a n ) = m D K If we fix N = Z 1 + Zζ + Zζ p 1, then [O K : N] = 1, so 1, ζ,, ζ p is an integral basis Theorem 19 Let ζ be a n-th primitive root of unity If K = Q(ζ), then {1, ζ, ζ,, ζ φ(n) } is and integral basis for O K, ie O K = Z[ζ] In particular, the discriminant of K is D K = ( 1)φ(n)/ n φ(n) p n pφ(n)/p 1 111 Exercises 1 (i) Let K be a number field with [K : Q] = n and β O K Prove that β O K if and only if N K (β) = 1 (ii) Prove that Z[ ] = {±(1 + ) k : k Z} and Z[ ] = {±1} If K is a number field, prove that its discriminant D K is an integer 3 Let a 1,, a n O K linearly independent over Q Let N = Za 1 + + Za n and m = [O K : N] Prove that D K (a 1,, a n ) = m D K (Hint: Use the following result: Let M = Zα 1 + + Zα n and N be a submodule Then there exists β 1,, β m N with m n such that N = Zβ 1 + + β m Z and β i = j i p ijα j with 1 i m and p ij Z) 4 Prove the Stickelgerger s criterion 5 Let f(x) = x n + a n 1 x n 1 + + a 1 x + a 0 Z[x] the minimal polynomial of θ Let K = Q(θ) If for each prime p such that p D K (θ) we have f(x) Eisensteinian (that is, f(x) satisfies the Eisenstein s criterion for irreducibility for p) with respect to p show that O K = Z[θ] (Hint: Use the problem 3) 6 If the minimal polynomial of α is f(x) = x n + ax + b, show that for K = Q(α), D K (α) = ( 1) (n ) (n n b n 1 + a n (1 n) n 1 ) 7 Determine an integral basis for K = Q(θ), where θ 3 + θ + 1 = 0 9