Number fields
Definition 10.1 The discriminant of the polynomial
10.8 Calculation of bases
ji(αi− αj), where f(x) denotes the derivative of f (x), and hence disc( f ) = (−1)12n(n−1)a0n−2f(α1) . . . f(αn).
Further, in the special case when f(x) = xn+ px + q (q 0), we have f(x) = nxn−1+ p, whence αif(αi) = −nq + (1 − n)pαi; this gives
q f(α1) . . . f(αn) = (1 − n)npnf(nq/((1 − n)p)) and thus
disc( f ) = (−1)12n(n−1)
nnqn−1+ (1 − n)n−1pn
.
10.8 Calculation of bases
The quadratic field K= Q(√
d), where d is a square-free integer, has already been discussed in Chapter 7. In particular it was shown in Section 7.2 that if d≡ 2 or 3 (mod 4) then an integral basis for K is 1,√
d and the discriminant of K is 4d; if d≡ 1 (mod 4) then an integral basis for K is 1,12(1 +√
d) and the discriminant of K is d. We verify this again briefly with the current notation. The elements of K have the formα = u + v√d with u, v rational and, on putting a= 2u, b = 2v, the field polynomial (minimum if v 0) for α is x2− ax + c where c = Nα = u2− dv2. Thus if α ∈ OK then a, c ∈ Z and
10.8 Calculation of bases 107 assertion about the bases and then the discriminants are given in the two cases respectively by
We now give some examples to show how one can find integral bases for fields of higher degree. We begin with a simple instance.
Example 10.1 Letα denote a zero of the polynomial x3+ px + q, where p, q are integers and q 0. By Section 10.7 we have disc( f ) = −(27q2+ 4p3) and if the latter is square-free then it is the discriminant of the fieldQ(α) and 1, α, α2 is an integral basis. In particular this applies when p= q = −1 and
p= −2, q = −1; the discriminants are then −23 and 5 respectively.
Example 10.2 Consider the cubic field K= Q(α), where α =√3
These must be rational integers, say u, w, v. The field polynomial of v + wα is (x − v)3− 10w3, whence N(v + wα) = v3+ 10w3. Since N(α) = 10 this gives N(3α(b + cα)) =1001 (v3+ 10w3) and, since 3(θ − a) is in OK, it follows that the latter must be a rational integer; thus 10 dividesv and w. Hence we have θ =13(r + sα) + tβ with integers r = u −101w, s =101(v − w), t =101w. The field polynomial of 13(r + sα) is (x −13r)3− 10(13s)3and, sinceθ − tβ is in OK, this must have integer coefficients. The coefficient of x is 13r2, whence 3 divides r and from the constant coefficient we see that then 3 divides s.
Hence 1, α, β form an integral basis for K as asserted. On denoting the basis byω1, ω2, ω3we see that the discriminant of K is given by
108 Number fields is the only element of the Galois group of K apart from the identity that fixes k. Now suppose that θ ∈ OK and letθ = a + bα + cα2+ dα3 with rational a, b, c, d. Then TK/k(θ) = 2a + 2c√
2 and similarly TK/k(αθ) = 2b√ 2+ 4d.
Since these are inOK and since also 1,√
2 is an integral basis for k it follows that 2a, 2b, 2c and 4d are rational integers, say p, q,r, s. Further, we have NK/k(θ) = θθ = (a + c√ shows successively that p, s,r, q are even and furthermore that 8 divides s2, whence 4 divides s. Hence a, b, c, d are rational integers and so 1, α, α2, α3is an integral basis for K as asserted. We write the basis asω1, ω2, ω3, ω4 and note that the conjugates ofα are α, iα, −α, −iα; then the discriminant of K is given by
Example 10.4 Consider the biquadratic field K= Q(√2,√7). This has Ga-lois group 1, α, β, αβ where α takes√
7 is an integral basis for k, it follows that 2a and 2b are rational integers, say p and q. Similarly, from the subfields Q(√ s, that is, 2c and 2d, have the same parity. It follows that an integral basis for K is 1,√
2,√
7,12(1 +√ 7)√
2. For certainly the last element is inOK since the square of it is 4+√
7 and it has minimum polynomial x4− 8x2+ 9. To calculate the discriminant of K we denote the integral basis just obtained by ω1, ω2, ω3, ω4. Then on recalling that the Galois group of K is 1, α, β, αβ as above we see that the discriminant of K is given by
10.10 Exercises 109
An introduction to the literature was given in Section 7.7 and many of the works included under further reading in other chapters contain valuable expositions relating to the subject. Here we mention in addition the books by J. Esmonde and M. R. Murty, Problems in Algebraic Number Theory (Springer, 2004), and by A. Fröhlich and M. J. Taylor, Algebraic Number Theory (Cambridge University Press, 1991), both of which cover the topic well. The book Algebraic Number Theory by S. Lang (Springer, 1994) has been a stan-dard reference for many years, and Number Fields (Springer, 1995) by D. A.
Marcus is another accessible work. The volume Basic Number Theory (Springer, 1974) by A. Weil covers similar ground but it is written on a very sophisticated level. In connection with Section 10.8, the classic text is Berwick’s Integral Bases (Cambridge University Press, 1927).
10.10 Exercises
(i) Find the minimum polynomials overQ of (1 + i)√3, i +√3, i + eiπ/3. (ii) Find the field polynomials of i and√3
5 inQ(i +√3 5).
(iii) By the symmetric function theorem, or otherwise, prove that any zero of a monic polynomial with algebraic integer coefficients is an algebraic integer.
(iv) Which of the following are algebraic integers?
1/2, (√ the kernel is(X2+ 1)Z[X] and deduce that the above map induces an isomorphism of ringsZ[X]/(X2+ 1) → Z[i].
(vii) Show that, for a, b ∈ Q∗, whereQ∗denotes the multiplicative group of non-zero elements ofQ, the degree of Q(√
a,√
b) is equal to the order of the subgroup ofQ∗/Q∗2 generated by a, b. Determine whether the fieldQ(√
(2 +√
2)) is of the form Q(√ a,√
b) with a, b ∈ Q.
110 Number fields
(viii) Show that if a number field K has degree d= s + 2t, where s is the num-ber of real conjugate fields and 2t is the numnum-ber of complex conjugate fields, then the discriminant D of K satisfies(−1)tD> 0.
(ix) Let K= Q(α) where α =√3
d for a square-free integer d 0, ±1. Show that the ring of algebraic integersOK of K satisfies
Z[√3
d]⊆ OK⊆13Z[√3 d].
Verify that the field polynomial of a+ bα + cα2for rational a, b, c is x3− 3ax + 3(a2− bcd)x − (a3+ b3d+ c3d2− 3abcd).
By considering the cases when 3a, 3b, 3c are 0, ±1, prove that an inte-gral basis for K is given by 1, α, α2when d ±1 (mod 9) and by 1, α, β otherwise whereβ =13(1 ± α + α2) with corresponding ± signs.
(x) Let k⊂ K be number fields. Show that, for α ∈ OK, the trace TK/k(α) and norm NK/k(α) are in Ok. Let now K= Q(√3,√5). By computing traces and norms for the three quadratic subfields k of K , show that an integral basis for K is 1,√
3, 12(1 +√
5),12(1 +√ 5)√
3.
(xi) Prove that an integral basis forQ(i,√2) is 1, i,√2, 12(1 + i)√ 2. Prove further that an integral basis forQ(√
2,√
p), where p is a prime with p≡ 3 (mod 4), is 1,√2,√p, 12(1 +√p)√2. Calculate the discrimi-nants of the fields.
11
Ideals
11.1 Origins
The introduction of ideals was motivated by a desire to restore the property of unique factorization in number fields; certainly, as already observed in Section 7.4, the property is not universal. Consider, as there, the quadratic field K= Q(√(−5)) with basis 1,√(−5). We have
21= 3 × 7 = (1 + 2√
(−5))(1 − 2√ (−5)).
Now 3, 7, 1 + 2√
(−5), 1 − 2√
(−5) cannot be further factorized in OK. Sup-pose, for instance, 3= αβ with α, β in OK. Then NK/Q(α)NK/Q(β) = 9 and so if neither α nor β were ±1 (thus if neither has norm 1) we would have NK/Q(α) = 3, which is impossible since x2+ 5y2= 3 has no solution in inte-gers x, y. Similarly 7, 1 + 2√
(−5) and 1 − 2√
(−5) cannot be factorized. The situation was restored by Kummer; he proceeded by way of a mapping of the ringOKinto a multiplicative semigroup of elements which he called ‘ideals’.
11.2 Definitions
Let K be an algebraic number field with ring of integersOK. By an ideal in K we mean a non-empty subset a of OKsuch that
(i) ifα, α∈ a then α − α∈ a, (ii) ifα ∈ a and β ∈ OK thenαβ ∈ a.
For any elementsα1, . . . , αminOK the set of all numbersα1β1+ · · · + αmβm
withβ1, . . . , βminOKis plainly an ideal, say a, and we write a = [α1, . . . , αm].
Thenα1, . . . , αm are called generators for a. Conversely, for any ideal a, we 111
112 Ideals
have a = [α1, . . . , αm] for some generatorsα1, . . . , αm. Indeed we shall prove the following.
Theorem 11.1 There exists a basis for a; that is, there exist elements γ1, . . . , γn
in a such that, if α ∈ a, then α = u1γ1+ · · · + unγnwith u1, . . . , uninZ.
Proof Let ω1, . . . , ωn be an integral basis for K . If α ∈ a and α 0 then αω1, . . . , αωn is an a-basis for K over Q, that is, a basis for K as a vector space overQ with elements in a. The theorem follows on arguing as in Sec-tion 10.7; indeed we can take forγ1, . . . , γnany set of elements of a such that
|(γ1, . . . , γn)| assumes its smallest value.
Note that, if γ1, . . . , γn is a basis for a, then a = [γ1, . . . , γn] and so we have exhibited a set of generators for the ideal in accordance with the assertion above.
We define the product ab of ideals a and b as the ideal consisting of all ele-ments a1b1+ · · · + ajbjwith the a in a and the b in b. Hence if a = [α1, . . . , αl] and b = [β1, . . . , βm] then ab = [α1β1, . . . , αlβm], that is, the ideal with genera-torsαrβs(1 ≤r ≤l, 1 ≤ s ≤ m). Plainly multiplication of ideals is commutative (ab = ba) and associative ((ab)c = a(bc)). As for integers, we say that a divides b (written a|b) if b = ac for some ideal c. We write am= a . . . a (m factors) and a0= e = [1], that is, the ideal OK.