Hijos de Agar

and L(s, χ) is the Dirichlet L-function

∞ n=1

χ(n)/n^s=

p

(1 − χ(p)p^−s)⁻¹.

All the sums and products here converge for s> 1 and indeed for any complex variable s= σ + it with σ > 1. In the case of the L-function we have conver-gence for s= 1 and there are classical results of Dirichlet giving the values of L(1, χ) in terms of the class number of K ; see Section 15.6.

12.6 The cyclotomic field

We shall need Eisenstein’s criterion relating to a polynomial f(x) = xⁿ+ an−1xⁿ⁻¹+ · · · + a1x+ a0

with integer coefficients. It is stated as follows.

Theorem 12.6 (Eisenstein) If there exists a prime p such that p divides a0,a1, . . . ,an−1but p²does not divide a0then f(x) is irreducible.

Proof Suppose on the contrary that f(x) = g(x)h(x) where g(x) and h(x) are polynomials with integer coefficients and m < n is the degree of g(x). Let Z = Fpbe the field with p elements as in Section 12.5 and let f, g, h be the reductions of f, g, h (mod p). Then f (x) = g(x)h(x) and f = xⁿ. But, by hy-pothesis, p² a0, whence the constant terms in g(x), h(x) cannot both be 0, and this plainly gives a contradiction; indeed Z[x] is a unique factorization domain, whence we have g= x^m, h= x^n−m. This proves the theorem.

The qth cyclotomic field is defined as K= Q(ζ ) where ζ = e^2πi/q and q is an integer> 2. We shall discuss here only the case when q is a prime. Then, sinceζ^q= 1, we see that ζ is a zero of the cyclotomic polynomial given by

q(x) = x^q−1+ x^q−2+ · · · + 1 and that this is irreducible; for plainly Eisen-stein’s criterion applies to

q(x + 1) =(x + 1)^q− 1

x = x^q−1+

q 1



x^q−2+ · · · +

 q q− 1

 . Thusq(x) is the minimum polynomial for ζ , whence ζ has degree q − 1 and conjugatesζ, ζ², . . . , ζ^q−1. This gives at once the basic cyclotomic property ofζ , namely that, for any integer k,

TK/Q(ζ^k) =

 −1 if q  k, q− 1 if q | k.

132 Units and ideal classes

Furthermore the following holds.

Theorem 12.7 (Basis) An integral basis for K is given by 1, ζ, . . . , ζ^q−2.

Proof First we observe that from the factorization

q(x) = (x − ζ )(x − ζ²) . . . (x − ζ^q−1)

we obtain NK/Q(1−ζ )=q(1)=q. Further, we observe that (1−ζ^j)/(1−ζ ) is a unit in K for certainly it is inOK and we have

NK/Q(1 − ζ ) = NK/Q(1 − ζ^j) (1 ≤ j ≤ q − 1).

Taking the product over all j , it follows that q/(1 − ζ )^q−1is a unit in K . We now suppose thatα ∈ OK. Then, for some rationals a0, . . . , aq−2,

α = a0+ a1ζ + · · · + aq−2ζ^q−2. By the cyclotomic property ofζ recorded above, this gives

TK/Q(αζ^{− j}− αζ ) = qaj (0 ≤ j ≤ q − 2), whence qaj is a rational integer. Hence we have

qα = b0+ b1(1 − ζ ) + · · · + bq−2(1 − ζ )^q−2 (^∗∗) for some integers b0, . . . , bq−2. We proceed to show that q divides bjfor all j and this will suffice to establish the result.

Now if we assume that b0, . . . , br−1are divisible by q, where 1≤ r ≤ q − 2, then(^∗∗) shows that br/(1 − ζ ) ∈ OK; thus q divides NK/Q(br) = b^qr⁻¹and so, since q is a prime, it divides br. Plainly, by(^∗∗), b0/(1 − ζ ) ∈ OK, whence q divides b0and the assertion follows by induction.

Theorem 12.8 (Discriminant) The discriminant of K is(−1)^12(q−1)q^q−2. Proof We shall use the integral basisζ, ζ², . . . , ζ^q−1. By the cyclotomic prop-erty ofζ referred to above we obtain, for 1 ≤ i, j ≤ q − 1,

det(TK/Q(ζ^{i+ j})) =

−1 −1 . . . −1 q− 1

−1 −1 . . . q − 1 −1

...

−1 q− 1 . . . −1 −1

q− 1 −1 . . . −1 −1

.

12.6 The cyclotomic field 133

Subtracting the first row from the others and then adding the first q−2 columns to the last we get

Now adding the last column to the others we obtain zeros except on the diago-nal and the result follows.

Theorem 12.9 (Factorization) Let p be a prime. If p q then p = p1. . . pl

where p1, . . . , pl are distinct prime ideals. If p= q then p = p^q−1where p = [1− ζ ].

Proof We haveOK= Z(ζ ) and so the hypothesis of Dedekind’s theorem is satisfied. Suppose first that p q. Then the polynomial x^q− 1 in Z[x], where Z = Fp, is relatively prime to its derivative and hence the cyclotomic poly-nomial q(x) has no repeated factor (mod p). This gives the desired result.

When p= q we have x^q− 1 ≡ (x − 1)^q(mod q) and, since q/(1 − ζ )^q−1is a unit in K , we see that [q, 1 − ζ ] = [1 − ζ ]. The assertion follows.

We note now that K is a normal field; indeed its Galois group is isomorphic to the multiplicative subgroup ofFq. Hence in the case when p q all prime divisors p of p have the same degree, say f , and we have l = (q − 1)/ f (cf.

Section 11.6). In fact we can calculate f explicitly.

Theorem 12.10 We have N p = p^f for each prime divisor p of p where f is the least positive integer such that p^f≡ 1 (mod q).

Proof It is easily seen that 1, ζ , . . . , ζ^q−1 are in distinct residue classes (mod p). Indeed otherwise p would divide 1 − ζ^j for some j with 1≤ j < q, whence it would divide NK/Q(1 − ζ ) = q contrary to p  q. The classes form a subgroup of the multiplicative group of the fieldOK/p and this has order N p − 1. Hence N p ≡ 1 (mod q) and so N p = p^f where f^≥ f . But since p^f ≡ 1 (mod q) we have ζ^pf = ζ , whence α^pf ≡ α (mod p) for all α in OK. Thus each of the N p elements of OK/p is a zero of x^pf − x and this gives

p^f≤ p^f. We conclude that f^= f .

134 Units and ideal classes

It follows as a corollary to Theorem 12.10 that p splits completely in K if and only if p≡ 1 (mod q); the result has connections with class field theory and it is sometimes referred to as the cyclotomic reciprocity law.

Theorem 12.11 (Units) There are ¹₂(q − 3) elements in any fundamental set of units of K . The roots of unity are given by(−ζ )^j with 0≤ j < 2q.

Proof The first part follows at once from Dirichlet’s theorem since here s= 0, t=¹₂(q − 1) and so r = s + t − 1 =¹₂(q − 3). The roots of unity form a finite cyclic group which includes the elements indicated above as a subgroup.

Thus 2q divides the order of the group, say m. Then m= 2q follows since the group is generated by a primitive mth root of unity and its degreeφ(m) cannot exceed the degreeφ(q) of K . Alternatively, one can argue that if there were a primitive nth root of unity in the field with(n, q) = 1 and n > 2, then the product withζ would be a primitive nqth root of unity in K ; but this has degree φ(nq) = φ(n)φ(q) which exceeds the degree φ(q) of K . Plainly if (n, q) > 1, then the degreeφ(n) of a primitive nth root of unity exceeds φ(q) unless n = q or n= 2q, in which case the root is just a power of ζ or −ζ . This completes the proof except for a verification that a primitive mth root of unity with composite m has degreeφ(m), and for this we refer to the recommended books.

It was already observed in the proof of Theorem 12.7 that(1 − ζ^j)/(1 − ζ ) is a unit in K for j= 2, . . . ,¹₂(q − 1). Now we have

1− ζ^j

1− ζ = ζ^{( j−1)/2}ζ^j/2− ζ^{− j/2}

ζ^1/2− ζ^−1/2= ζ^{( j−1)/2}sin( jπ/q) sin(π/q) .

Further, sinceζ^1/2= ±ζ^(q+1)/2and q is odd, we see thatζ^{( j−1)/2}is a root of unity in K . Hence the numbers sin( jπ/q)/ sin(π/q) are real positive units in K ; they can be verified as being multiplicatively independent but they do not necessarily comprise a fundamental system.

We note that the units referred to above all lie in the field K⁺= Q(ζ + ζ⁻¹).

Indeedζ is a zero of the polynomial x²− (ζ + ζ⁻¹)x + 1, whence [K : K⁺]= 2. Thus [K⁺:Q] =¹₂(q − 1) and, since ζ + ζ⁻¹= 2 cos(2π/q), it follows that K⁺is the maximal real subfield of K . It plays a significant role in studies on the class number of K . One result relating to K⁺can be obtained easily.

Theorem 12.12 An integral basis for K⁺ is 1, η, . . . , η^m withη = ζ + ζ⁻¹ and m=¹₂(q − 1) − 1.

Proof Letα be in the ring of integers of K⁺. Plainly we haveα = a0+ a1η + · · · + amη^m for some rationals a0, a1, . . . , am. Suppose that these are not all

12.6 The cyclotomic field 135 integers. Then there exists a suffix k such that the aj with j > k are integers and akis not. But we have

ζ^k(α − ak+1η^k+1− · · · − amη^m) = ζ^k(a0+ a1η + · · · + akη^k) and this is in the ring of integers of K . Thus, since 1, ζ, . . . , ζ^q−2is an integral basis for K and 2k< q − 2, the coefficient akofζ^2kis an integer and this is a contradiction.

The Dedekind zeta-functionζK(s) for K , where s = σ + it and σ > 1, is



a

(Na)^−s=

p

(1 − (Np)^−s)⁻¹

and from Theorems 12.9 and 12.10 we deduce the following.

Theorem 12.13 We have

ζK(s) = ζ(s)

χχ0

L(s, χ),

where

L(s, χ) =^∞

n=1

χ(n)/n^−s=

p

(1 − χ(p)p^−s)⁻¹.

Here the charactersχ = χj with 0≤ j ≤ q − 2 are defined by χj(n) =

 e²πi jν/(q−1) if q  n,

0 if q| n,

whereν denotes the index of n with respect to a primitive root (mod q); see Section 15.3. The characterχ0is called the principal character and it satisfies χ0(n) = 0 if q|n and χ0(n) = 1 otherwise.

Proof of Theorem 12.13 We have (1 − q^−s)ζK(s) = 

pq, p|p

(1 − (Np)^−s)⁻¹=

pq

(1 − p^{− f s})^−l

and

(1 − q^−s)ζ(s)

χχ0

L(s, χ) =

pq q−2

j=0

(1 − χj(p)p^−s)⁻¹.

Now q− 1 = f l and if p  q then p^f≡ 1 (mod q) with f minimal. Thus ν = kl for some integer k when n= p and here (k, f ) = 1; for p^{f/(k, f )}≡ 1 (mod q)

136 Units and ideal classes

We remark finally that the L-functions converge for s= 1 and we have



We now give some examples to show how the preceding theories can be used to compute class numbers and to solve certain related Diophantine equations.

Example 12.1 Consider the cyclotomic field K= Q(ζ ) where ζ = e^2πi/5. By Theorem 12.7 an integral basis for K is ζ, ζ², ζ³, ζ⁴ and by Theorem 12.8 the discriminant of K is 125. Further, the Minkowski constant of K is 3/(2π²). Hence every ideal class of K contains an ideal with norm at most 3√

125/(2π²) < 2. This implies that every ideal in K is equivalent to a prin-cipal ideal and so is prinprin-cipal. Thus the class number of K is 1 and the field has unique factorization.

Example 12.2 Consider the cubic field K = Q(√³

2). By Section 10.10 Exercise (ix), an integral basis for K is 1,√³

2,√³

4. Thus by Section 10.7 the discriminant of K is the same as the discriminant of x³− 2 and this is −108.

The Minkowski constant for K is 8/(9π), whence every ideal class of K con-tains an ideal with norm at most 8√

108/(9π) < 3. Now we have 2 = [√³ 2]³ where [√³

2] has norm 2 and so is a prime ideal as well as being principal. It follows that K has class number 1. Similarly we can treat K= Q(√³

3). An integral basis for K is 1,√³

3,√³

9 and the discriminant of K is−243. Thus every ideal class contains an ideal with norm at most 8√

243/(9π) < 5. We

9] and, by Dedekind’s theorem, this

In document Reos y espías. La Monarquía hispánica y los renegados (1550-1630) (página 68-74)