Math 129: Algebraic Number Theory Lecture 8

William Stein

Tuesday, March 2, 2004

1 The Discriminant

Suppose w_{1}, …, w_{n} are a basis for a number field K, which we view as a Q-vector space. Let σ : K ↪ C^{n} be the embedding σ(a) = (σ_{1}(a), …, σ_{n}(a)), where σ_{1}, …, σ_{n} are the distinct embeddings of K into C. Let A be the matrix whose rows are σ(w_{1}), …, σ(w_{n}). The quantity Det(A) depends on the ordering of the w_{i}, and need not be an integer.

If we consider Det(A)^{2} instead, we obtain a number that is a well-defined integer which can be either positive or negative. Note that

Det(A)^{2} = Det(AA) = Det(AA^{t}) = Det ( ∑_{k=1,...,n} σ_{k}(w_{i})σ_{k}(w_{j})) = Det(Tr(w_{i}w_{j})_{1≤i,j≤n}),

so Det(A)^{2} can be defined purely in terms of the trace without mentioning the embeddings σ_{i}. Also, changing the basis for O_{K} is the same as left multiplying A by an integer matrix U of determinant ±1, which does not change the squared determinant, since Det(UA)^{2} = Det(U)^{2} Det(A)^{2} = Det(A)^{2}. Thus Det(A)^{2} is well defined, and does not depend on the choice of basis.

If we view K as a Q-vector space, then (x, y) ↦ Tr(xy) defines a bilinear pairing K × K → Q on K, which we call the trace pairing. The following lemma asserts that this pairing is nondegenerate, so Det(Tr(w_{i}w_{j})) ≠ 0 hence Det(A) ≠ 0.

Lemma 1.1. The trace pairing is nondegenerate.

Proof. If the trace pairing is degenerate, then there exists a ∈ K such that for every b ∈ K we have Tr(ab) = 0. In particularly, taking b = a^{−1} we see that 0 = Tr(aa^{−1}) = Tr(1) = [K : Q] > 0, which is absurd.

Definition 1.2 (Discriminant). Suppose a_{1}, …, a_{n} is any Q-basis of K. The discriminant of a_{1}, …, a_{n} is

Disc(a_{1}, …, a_{n}) = Det(Tr(a_{i}a_{j})_{1≤i,j≤n}) ∈ Q.

The discriminant Disc(O) of an order O in O_{K} is the discriminant of any basis for O. The discriminant d_{K} = Disc(K) of the number field K is the discrimimant of O_{K}.

Note that the discriminants defined above are all nonzero by Lemma 1.1.

Warning: In MAGMA Disc(K) is defined to be the discriminant of the polynomial you happened to use to define K, which is (in my opinion) a poor choice and goes against most of the literature.

The following proposition asserts that the discriminant of an order O in O_{K} is bigger than disc(O_{K}) by a factor of the square of the index.

Proposition 1.3. Suppose O is an order in O_{K}. Then

Disc(O) = Disc(O_{K}) · [O_{K} : O]^{2}.

Proof. Let A be a matrix whose rows are the images via σ of a basis for O_{K}, and let B be a matrix whose rows are the images via σ of a basis for O. Since O ⊂ O_{K} has finite index, there is an integer matrix C such that CA = B, and | Det(C)| = [O_{K} : O]. Then

Disc(O) = Det(B)^{2} = Det(CA)^{2} = Det(C)^{2} Det(A)^{2} = [O_{K} : O]^{2} · Disc(O_{K}).

This result is enough to give an algorithm for computing O_{K}, albeit a potentially slow one. Given K, find some order O ⊂ K, and compute d = Disc(O). Factor d, and use the factorization to write d = s · f^{2}, where f^{2} is the largest square that divides d. Then the index of O in O_{K} is a divisor of f, and we (tediously) can enumerate all rings R with O ⊂ R ⊂ K and [R : O] | f, until we find the largest one all of whose elements are integral.

Example 1.4. Consider the ring O_{K} = Z[(1 + \sqrt{5})/2] of integers of K = Q(\sqrt{5}). The discriminant of the basis 1, a = (1 + \sqrt{5})/2 is

Disc(O_{K}) = |(\begin{matrix}2 & 1 \\ 1 & 3 \end{matrix})| = 5.

Let O = Z[\sqrt{5}] be the order generated by \sqrt{5}. Then O has basis 1, \sqrt{5}, so

Disc(O) = |(\begin{matrix}2 & 0 \\ 0 & 10 \end{matrix})| = 20 = [O_{K} : O]^{2} · 5.

2 Norms of Ideals

In this section we extend the notion of norm to ideals. This will be helpful in proving of class groups in the next section. For example, we will prove that the group of fractional ideals modulo principal fractional ideals of a number field is finite by showing that every ideal is equivalent to an ideal with norm at most some a priori bound.

Definition 2.1 (Lattice Index). If L and M are two lattices in vector space V, then the lattice index [L : M] is by definition the absolute value of the determinant of any linear automorphism A of V such that A(L) = M.

The lattice index has the following properties:

• If M ⊂ L, then [L : M] = #(L/M).

• If M, L, N are lattices then [L : N] = [L : M] · [M : N].

Definition 2.2 (Norm of Fractional Ideal). Suppose I is a fractional ideal of O_{K}. The norm of I is the lattice index

Norm(I) = [O_{K} : I] ∈ Q_{≥0},

or 0 if I = 0.

Note that if I is an integral ideal, then Norm(I) = #(O_{K}/I).

Lemma 2.3. Suppose a ∈ K and I is an integral ideal. Then

Norm(aI) = | Norm_{K/Q}(a)| Norm(I).

Proof. By properties of the lattice index mentioned above we have

[O_{K} : aI] = [O_{K} : I] · [I : aI] = Norm(I) · | Norm_{K/Q}(a)|.

Here we have used that [I : aI] = | Norm_{K/Q}(a)|, which is because left multiplication ℓ_{a} is an automorphism of K that sends I onto aI, so [I : aI] = | Det(ℓ_{a})| = | Norm_{K/Q}(a)|.

Proposition 2.4. If I and J are fractional ideals, then

Norm(IJ) = Norm(I) · Norm(J).

Proof. By Lemma 2.3, it suffices to prove this when I and J are integral ideals. If I and J are coprime, then Theorem ?? (Chinese Remainder Theorem) implies that Norm(IJ) = Norm(I) · Norm(J). Thus we reduce to the case when I = 𝔭^{m} and J = 𝔭^{k} for some prime ideal 𝔭 and integers m, k. By Proposition ?? (consequence of CRT that O_{K}/𝔭 ≅ 𝔭^{n}/𝔭^{n+1}), the filtration of O_{K}/𝔭^{n} given by powers of 𝔭 has successive quotients isomorphic to O_{K}/𝔭, so we see that #(O_{K}/𝔭^{n}) = #(O_{K}/𝔭)^{n}, which proves that Norm(𝔭^{n}) = Norm(𝔭)^{n}.

Lemma 2.5. Fix a number field K. Let B be a positive integer. There are only finitely many integral ideals I of O_{K} with norm at most B.

Proof. An integral ideal I is a subgroup of O_{K} of index equal to the norm of I. If G is any finitely generated abelian group, then there are only finitely many subgroups of G of index at most B, since the subgroups of index dividing an integer n are all subgroups of G that contain nG, and the group G/nG is finite. This proves the lemma.

3 Finiteness of the Class Group via Geometry of Numbers

We have seen examples in which O_{K} is not a unique factorization domain. If O_{K} is a principal ideal domain, then it is a unique factorization domain, so it is of interest to understand how badly O_{K} fails to be a principal ideal domain. The class group of O_{K} measures this failure. As one sees in a course on Class Field Theory, the class group and its generalizations also yield deep insight into the possible abelian Galois extensions of K.

Definition 3.1 (Class Group). Let O_{K} be the ring of integers of a number field K. The class group C_{K} of K is the group of nonzero fractional ideals modulo the sugroup of principal fractional ideals (a), for a ∈ K.

Note that if we let Div(K) denote the group of nonzero fractional ideals, then there is an exact sequence

0 → O_{K}^{∗} → K^{∗} → Div(K) → C_{K} → 0.

A basic theorem in algebraic number theory is that the class group C_{K} is finite, which follows from the first part of the following theorem and the fact that there are only finitely many ideals of norm less than a given integer.

Theorem 3.2 (Finiteness of the Class Group). Let K be a number field. There is a constant C_{r,s} that depends only on the number r, s of real and pairs of complex conjugate embeddings of K such that every ideal class of O_{K} contains an integral ideal of norm at most C_{r,s}\sqrt{|d_{K}|}, where d_{K} = Disc(O_{K}). Thus by Lemma 2.5 the class group C_{K} of K is finite. One can choose C_{r,s} such that every ideal class in C_{K} contains an integral ideal of norm at most

\sqrt{|d_{K}|} · (\frac{4}{π} )^{s} \frac{n!}{n^{n}} .

The explicit bound in the theorem is called the Minkowski bound, and I think it is the best known unconditional general bound (though there are better bounds in certain special cases).

Before proving Theorem 3.2, we prove a few lemmas. The strategy of the proof will be to start with any nonzero ideal I, and prove that there is some nonzero a ∈ K, with very small norm, such that aI is an integral ideal. Then Norm(aI) = Norm_{K/Q}(a) Norm(I) will be small, since Norm_{K/Q}(a) is small. The trick is to determine precisely how small an a we can choose subject to the condition that aI be an integral ideal, i.e., that a ∈ I^{−1}.

Let S be a subset of V = R^{n}. Then S is convex if whenever x, y ∈ S then the line connecting x and y lies entirely in S. We say that S is symmetric about the origin if whenever x ∈ S then −x ∈ S also. If L is a lattice in V, then the volume of V/L is the volume of the compact real manifold V/L, which is the same thing as the absolute value of the determinant of any matrix whose rows form a basis for L.

Lemma 3.3. Let L be a lattice in V = R^{n}, and let S be a bounded closed convex subset of V that is symmetric about the origin. Assume that Vol(S) ≥ 2^{n} Vol(V/L). Then S contains a nonzero element of L.

Proof. First assume that Vol(S) > 2^{n} · Vol(V/L). If the map π : \frac{1}{2}S → V/L is injective, then

\frac{1}{2^{n}} Vol(S) = Vol (\frac{1}{2}S) ≤ Vol(V/L),

a contradiction. Thus π is not injective, so there exist P_{1} ≠ P_{2} ∈ \frac{1}{2}S such that P_{1} − P_{2} ∈ L. By symmetry −P_{2} ∈ \frac{1}{2}S. By convexity, the average \frac{1}{2}(P_{1} − P_{2}) of P_{1} and −P_{2} is also in \frac{1}{2}S. Thus 0 ≠ P_{1} − P_{2} ∈ S ∩ L, as claimed.

Next assume that Vol(S) = 2^{n} · Vol(V/L). Then for all ε > 0 there is 0 ≠ Q_{ε} ∈ L ∩ (1 + ε)S, since Vol((1 + ε)S) > Vol(S) = 2^{n} · Vol(V/L). If ε < 1 then the Q_{ε} are all in L ∩ 2S, which is finite since 2S is bounded and L is discrete. Hence there exists Q = Q_{ε} ∈ L∩(1+ε)S for arbitrarily small ε. Since S is closed, Q ∈ L∩S.

Lemma 3.4. If L_{1} and L_{2} are lattices in V, then

Vol(V/L_{2}) = Vol(V/L_{1}) · [L_{1} : L_{2}].

Proof. Let A be an automorphism of V such that A(L_{1}) = L_{2}. Then A defines an isomorphism of real manifolds V/L_{1} → V/L_{2} that changes volume by a factor of | Det(A)| = [L_{1} : L_{2}]. The claimed formula then follows.

Fix a number field K with ring of integers O_{K}. Let σ : K → V = R^{n} be the embedding

σ(x) = (σ_{1}(x), σ_{2}(x), …, σ_{r}(x),
Re(σ_{r+1}(x)), …, Re(σ_{r+s}(x)), Im(σ_{r+1}(x)), …, Im(σ_{r+s}(x))),

where σ_{1}, …, σ_{r} are the real embeddings of K and σ_{r+1}, …, σ_{r+s} are half the complex embeddings of K, with one representative of each pair of complex conjugate embeddings. Note that this σ is not exactly the same as the one at the beginning of Section 1.

Lemma 3.5.

Vol(V/σ(O_{K})) = 2^{−s}\sqrt{|d_{K}|}.

Proof. Let L = σ(O_{K}). From a basis w_{1}, …, w_{n} for O_{K} we obtain a matrix A whose ith row is

(σ_{1}(w_{i}), ⋯, σ_{r}(w_{i}), Re(σ_{r+1}(w_{i})), …, Re(σ_{r+s}(w_{1})), Im(σ_{r+1}(w_{i})), …, Im(σ_{r+s}(w_{1})))

and whose determinant has absolute value equal to the volume of V/L. By doing the following three column operations, we obtain a matrix whose rows are exactly the images of the w_{i} under all embeddings of K into C, which is the matrix that came up when we defined d_{K}.

1. Add i = \sqrt{−1} times each column with entries Im(σ_{r+j}(w_{i})) to the column with entries Re(σ_{r+j}(w_{i})).

2. Multiply all columns Im(σ_{r+j}(w_{i})) by −2i, thus changing the determinant by (−2i)^{s}.

3. Add each columns with entries Re(σ_{r+j}(w_{i})) to the the column with entries −2iIm(σ_{r+j}(w_{i})).

Recalling the definition of discriminant, we see that if B is the matrix constructed by the above three operations, then Det(B)^{2} = d_{K}. Thus

Vol(V/L) = | Det(A)| = |(−2i)^{−s} · Det(B)| = 2^{−s}\sqrt{|d_{K}|}.

Lemma 3.6. If I is a nonzero fractional ideal for O_{K}, then σ(I) is a lattice in V, and

Vol(V/σ(I)) = 2^{−s}\sqrt{|d_{K}|} · Norm(I).

Proof. We know that [O_{K} : I] = Norm(I) is a nonzero rational number. Lemma 3.5 implies that σ(O_{K}) is a lattice in V, since σ(O_{K}) has rank n as abelian group and spans V, so σ(I) is also a lattice in V. For the volume formula, combine Lemmas 3.4– 3.5 to get

Vol(V/σ(I)) = Vol(V/σ(O_{K})) · [O_{K} : I] = 2^{−s}\sqrt{|d_{K}|} Norm(I).

Proof of Theorem 3.2. Let K be a number field with ring of integers O_{K}, let σ : K ↪ V ≅ R^{n} be as above, and let f : V → R be the function defined by

f(x_{1}, …, x_{n}) = |x_{1} ⋯ x_{r} · (x_{r+1}^{2} + x_{(r+1)+s}^{2}) ⋯ (x_{r+s}^{2} + x_{n}^{2}).

Notice that if x ∈ K then f(σ(x)) = | Norm_{K/Q}(x)|.

Let S ⊂ V be any closed, bounded, convex, subset that is symmetric with respect to the origin and has positive volume. Since S is closed and bounded,

M = max{f(x) : x ∈ S}

exists.

Suppose I is any nonzero fractional ideal of O_{K}. Our goal is to prove there is an integral ideal aI with small norm. We will do this by finding an appropriate a ∈ I^{−1}. By Lemma 3.6,

c = Vol(V/I^{−1}) = \frac{2^{−s}\sqrt{|d_{K}|}}{Norm(I)}.

Let λ = 2 · (\frac{c}{v})^{1/n}, where v = Vol(S). Then

Vol(λS) = λ^{n} Vol(S) = 2^{n} \frac{c}{v} · v = 2^{n} · c = 2^{n} Vol(V/I^{−1}),

so by Lemma 3.3 there exists 0 ≠ a ∈ I^{−1} ∩ λS. Since M is the largest norm of an element of S, the largest norm of an element of I^{−1} ∩ λS is at most λ^{n}M, so

| Norm_{K/Q}(a)| ≤ λ^{n}M.

Since a ∈ I^{−1}, we have aI ⊂ O_{K}, so aI is an integral ideal of O_{K} that is equivalent to I, and

Norm(aI) = | Norm_{K/Q}(a)| · Norm(I)

≤ λ^{n}M · Norm(I)

≤ 2^{n} \frac{c}{v} M · Norm(I)

≤ 2^{n} · 2^{−s}\sqrt{|d_{K}|} · M · v^{−1}

= 2^{r+s}\sqrt{|d_{K}|} · M · v^{−1}.

Notice that the right hand side is independent of I. It depends only on r, s, |d_{K}|, and our choice of S. This completes the proof of the theorem, except for the assertion that S can be chosen to give the claim at the end of the theorem, which we leave as an exercise.

Corollary 3.7. Suppose that K ≠ Q is a number field. Then |d_{K}| > 1.

Proof. Applying Theorem 3.2 to the unit ideal, we get the bound

1 ≤ \sqrt{|d_{K}|} · (\frac{4}{π} )^{s} \frac{n!}{n^{n}} .

Thus

\sqrt{|d_{K}|} ≥ (\frac{π}{4} )^{s} \frac{n^{n}}{n!} ,

and the right hand quantity is strictly bigger than 1 for any s ≤ n/2 and any n > 1 (exercise).

3.1 An Open Problem

Conjecture 3.8. There are infinitely many number fields K such that the class group of K has order 1.

For example, if we consider real quadratic fields K = Q(\sqrt{d}), with d positive and square free, many class numbers are probably 1, as suggested by the MAGMA output below. It looks like 1’s will keep appearing infinitely often, and indeed Cohen and Lenstra conjecture that they do. Nobody has found a way to prove this yet.

> for d in [2..1000] do
     if d eq SquareFree(d) then
        h := ClassNumber(NumberField(x^2-d));
        if h eq 1 then
           printf "%o, ", d;
        end if;
     end if;
   end for;
2, 3, 5, 6, 7, 11, 13, 14, 17, 19, 21, 22, 23, 29, 31, 33, 37,
38, 41, 43, 46, 47, 53, 57, 59, 61, 62, 67, 69, 71, 73, 77, 83,
86, 89, 93, 94, 97, 101, 103, 107, 109, 113, 118, 127, 129, 131,
133, 134, 137, 139, 141, 149, 151, 157, 158, 161, 163, 166, 167,
173, 177, 179, 181, 191, 193, 197, 199, 201, 206, 209, 211, 213,
214, 217, 227, 233, 237, 239, 241, 249, 251, 253, 262, 263, 269,
271, 277, 278, 281, 283, 293, 301, 302, 307, 309, 311, 313, 317,
329, 331, 334, 337, 341, 347, 349, 353, 358, 367, 373, 379, 381,
382, 383, 389, 393, 397, 398, 409, 413, 417, 419, 421, 422, 431,
433, 437, 446, 449, 453, 454, 457, 461, 463, 467, 478, 479, 487,
489, 491, 497, 501, 502, 503, 509, 517, 521, 523, 526, 537, 541,
542, 547, 553, 557, 563, 566, 569, 571, 573, 581, 587, 589, 593,
597, 599, 601, 607, 613, 614, 617, 619, 622, 631, 633, 641, 643,
647, 649, 653, 661, 662, 669, 673, 677, 681, 683, 691, 694, 701,
709, 713, 717, 718, 719, 721, 734, 737, 739, 743, 749, 751, 753,
757, 758, 766, 769, 773, 781, 787, 789, 797, 809, 811, 813, 821,
823, 827, 829, 838, 849, 853, 857, 859, 862, 863, 869, 877, 878,
881, 883, 886, 887, 889, 893, 907, 911, 913, 917, 919, 921, 926,
929, 933, 937, 941, 947, 953, 958, 967, 971, 973, 974, 977, 983,
989, 991, 997, 998,

In contrast, if we look at class numbers of quadratic imaginary fields, only a few at the beginning have class number 1.

> for d in [1..1000] do
     if d eq SquareFree(d) then
        h := ClassNumber(NumberField(x^2+d));
        if h eq 1 then
           printf "%o, ", d;
        end if;
     end if;
  end for;
1, 2, 3, 7, 11, 19, 43, 67, 163

It is a theorem that the above list of 9 fields is the complete list with class number 1. More generally, it is possible (in theory), using deep work of Gross, Zagier, and Goldfeld involving zeta functions and elliptic curves, to enumerate all quadratic number fields with a given class number.