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We will use Minkowski theory, which belongs to the general area of geometry of numbers,

to gain insight into the ideal class group of a number field. We have already mentioned

the ideal class group briefly in (3.4.5); it measures how close a Dedekind domain is to a

principal ideal domain.

**5.1 Lattices**

**5.1.1 Definitions and Comments**

Let e1, . . . , en ∈ Rn, with the ei linearly independent over R. Thus the ei form a basis

for Rn as a vector space over R. The ei also form a basis for a free Z-module of rank n,

namely

H = Ze1 + · · · + Zen.

A set H constructed in this way is said to be a lattice in Rn. The fundamental domain

of H is given by

T = {x ∈ Rn : x =

n

i=1

aiei, 0 ≤ ai < 1}.

In the most familiar case, e1 and e2 are linearly independent vectors in the plane, and T is

the parallelogram generated by the ei. In general, every point of Rn is congruent modulo

H to a unique point of T, so Rn is the disjoint union of the sets h + T, h ∈ H. If μ is

Lebesgue measure, then the volume μ(T) of the fundamental domain T will be denoted by

v(H). If we generate H using a different Z-basis, the volume of the fundamental domain

is unchanged. (The change of variables matrix between Z-bases is unimodular, hence has

determinant ±1. The result follows from the change of variables formula for multiple

integrals.)

**5.1.2 Lemma**

Let S be a Lebesgue measurable subset of Rn with μ(S) > v(H). Then there exist distinct

points x, y ∈ S such that x − y ∈ H.

Proof. As we observed in (5.1.1), the sets h + T, h ∈ H, are (pairwise) disjoint and cover

Rn. Thus the sets S ∩ (h + T), h ∈ H, are disjoint and cover S. Consequently,

μ(S) =

h∈H

μ(S ∩ (h + T)).

By translation-invariance of Lebesgue measure, μ(S ∩ (h + T)) = μ((−h + S) ∩ T). Now

if S ∩ (h1 + T) and S ∩ (h2 + T) are disjoint, it does not follow that (−h1 + S) ∩ T and

(−h2 + S) ∩ T are disjoint, as we are not subtracting the same vector from each set. In

fact, if the sets (−h + S) ∩ T, h ∈ H, were disjoint, we would reach a contradiction via

v(H) = μ(T) ≥

h∈H

μ((−h + S) ∩ T) = μ(S).

Thus there are distinct elements h1, h2 ∈ H such that (−h1+S)∩(−h2+S)∩T

= ∅. Choose

(necessarily distinct) x, y ∈ S such that −h1 + x = −h2 + y. Then x − y = h1 − h2 ∈ H,

as desired. ♣

**5.1.3 Minkowski’s Convex Body Theorem**

Let H be a lattice in Rn, and assume that S is a Lebesgue measurable subset of Rn that

is symmetric about the origin and convex. If

(a) μ(S) > 2nv(H), or

(b) μ(S) ≥ 2nv(H) and S is compact,

then S ∩ (H \ {0})

= ∅.

Proof.

(a) Let S = 1

2S. Then μ(S ) = 2−nμ(S) > v(H) by hypothesis, so by (5.1.2), there exist

distinct elements y, z ∈ S such that y − z ∈ H. But y − z = 1

2 (2y + (−2z)), a convex

combination of 2y and −2z. But y ∈ S ⇒ 2y ∈ S, and z ∈ S ⇒ 2z ∈ S ⇒ −2z ∈ S by

symmetry about the origin. Thus y−z ∈ S and since y and z are distinct, y−z ∈ H\{0}.

(b)We apply (a) to (1+1/m)S,m = 1, 2, . . . . Since S, hence (1+1/m)S, is a bounded set,

it contains only finitely many points of the lattice H. Consequently, for every positive

integer m, Sm = (1+1/m)S ∩ (H \ {0}) is a nonempty finite, hence compact, subset

of Rn. Since Sm+1 ⊆ Sm for all m, the sets Sm form a nested sequence, and therefore

∩∞

m=1Sm

= ∅. If x ∈ ∩∞

m=1Sm, then x ∈ H \ {0} and x/(1+1/m) ∈ S for every m. Since

S is closed, we may let m→∞ to conclude that x ∈ S. ♣

**5.1.4 Example**

With n = 2, take e1 = (1, 0) and e2 = (0, 1). The fundamental domain is the unit square,

closed at the bottom and on the left, and open at the top and on the right. Let S be the

set of all a1e1 + a2e2 with −1 < ai < 1, i = 1, 2. Then μ(S) = 4v(H), but S contains no

nonzero lattice points. Thus compactness is a necessary hypothesis in part (b).

5.2 A Volume Calculation

We will use n-dimensional integration technique to derive a result that will be needed in

the proof that the ideal class group is finite. We will work in Rn, realized as the product

of r1 copies of R and r2 copies of C, where r1 + 2r2 = n. Our interest is in the set

Bt = {(y1, . . . , yr1, z1, . . . , zr2 ) ∈ Rr1 × Cr2 :

r1

i=1

|yi| + 2

r2

j=1

|zj| ≤ t}, t ≥ 0.

We will show that the volume of Bt is given by

V (r1, r2, t) = 2r1 (π

2

)r2 tn

n! .

The proof is by double induction on r1 and r2. If r1 = 1 and r2 = 0, hence n = 1, we

are calculating the length of the interval [−t, t], which is 2t, as predicted. If r1 = 0 and

r2 = 1, hence n = 2, we are calculating the area of {z1 : 2|z1| ≤ t}, a disk of radius t/2.

The result is πt2/4, again as predicted. Now assume that the formula holds for r1, r2, and

all t. Then V (r1 + 1, r2, t) is the volume of the set described by

|y| +

r1

i=1

|yi| + 2

r2

j=1

|zj| ≤ t

or equivalently by

r1

i=1

|yi| + 2

r2

j=1

|zj| ≤ t − |y|.

Now if |y| > t, then Bt is empty. For smaller values of |y|, suppose we change |y| to

|y| + dy. This creates a box in (n + 1)-space with dy as one of the dimensions. The

volume of the box is V (r1, r2, t − |y|)dy. Thus

V (r1 + 1, r2, t) =

t

−t

V (r1, r2, t − |y|)dy

which by the induction hypothesis is 2

t

0 2r1 (π/2)r2 [(t − y)n/n!] dy. Evaluating the integral,

we obtain 2r1+1(π/2)r2 tn+1/(n + 1)!, as desired.

Finally, V (r1, r2 + 1, t) is the volume of the set described by

r1

i=1

|yi| + 2

r2

j=1

|zj | + 2|z| ≤ t.

As above,

V (r1, r2 + 1, t) =

|z|≤t/2

V (r1, r2, t − 2|z|)dμ(z)

where μ is Lebesgue measure on C. In polar coordinates, the integral becomes

2π

θ=0

t/2

r=0

2r1 (π

2

)r2 (t − 2r)n

n! r dr dθ

which reduces to 2r1 (π/2)r2(2π/n!)

t/2

r=0(t − 2r)n r dr. We may write the integrand as

(t − 2r)n r dr = −rd(t − 2r)n+1/2(n + 1). Integration by parts yields (for the moment

ignoring the constant factors preceding the integral)

t/2

0

(t − 2r)n+1dr/2(n + 1) =

−(t − 2r)n+2

2(n + 1)2(n + 2)

t/2

0

= tn+2

4(n + 1)(n + 2).

Therefore V (r1, r2 + 1, t) = 2r1 (π/2)r2(2π/n!)tn+2/4(n + 1)(n + 2), which simplifies to

2r1 (π/2)r2+1tn+2/(n+2)!, completing the induction. Note that n+2 (rather than n+1)

is correct, because r1 + 2(r2 + 1) = r1 + 2r2 + 2 = n + 2.

**5.3 The Canonical Embedding**

**5.3.1 Definitions and Comments**

Let L be a number field of degree n over Q, and let σ1, . . . , σn be the Q-monomorphisms

of L into C. If σi maps entirely into R, we say that σi is a real embedding; otherwise it

is a complex embedding. Since the complex conjugate of a Q-monomorphism is also a Qmonomorphism,

we can renumber the σi so that the real embeddings are σ1, . . . ,σr1 and

the complex embeddings are σr1+1, . . . , σn, with σr1+j paired with its complex conjugate

σr1+r2+j, j = 1, . . . , r2. Thus there are 2r2 complex embeddings, and r1 + 2r2 = n.

The canonical embedding σ : L → Rr1 × Cr2 = Rn is the injective ring homomorhism

given by

σ(x) = (σ1(x), . . . , σr1+r2 (x)).

**5.3.2 Some Matrix Manipulations**

Let x1, . . . , xn ∈ L be linearly dependent over Z (hence the xi form a basis for L over Q).

Let C be the matrix whose kth column (k = 1, . . . , n) is

σ1(xk), . . . , σr1 (xk), Re σr1+1(xk), Im σr1+1(xk), . . . , Re σr1+r2 (xk), Im σr1+r2 (xk).

The determinant of C looks something like a discriminant, and we can be more precise

with the aid of elementary row operations. Suppose that

σj(xk)

σj(xk)

=

x + iy

x − iy

.

We are fixing j and allowing k to range from 1 to n, so we have two rows of an n by

n matrix. Add the second row to the first, so that the entries on the right become 2xout 2 and −i, we get

−2i

x

y

= −2i

Re σj(xk)

Im σj(xk)

.

Do this for each j = 1, . . . , r2. In the above calculation, σj appears immediately under

σj , but in the original ordering they are separated by r2, which introduces a factor of

(−1)r2 when we calculate a determinant. To summarize, we have

detC = (2i)−r2 det(σj(xk))

Note that j and k range from 1 to n; no operations are needed for the first r1 rows.

Now let M be the free Z-module generated by the xi, so that σ(M) is a free Z-module

with basis σ(xi), i = 1, . . . , n, hence a lattice in Rn. The fundamental domain is a

parallelotope whose sides are the σ(xi), and the volume of the fundamental domain is the

absolute value of the determinant whose rows (or columns) are the σ(xi). Consequently

[see (5.1.1) for notation],

v(σ(M)) = | detC| = 2−r2 | det σj(xk)|.

We apply this result in an algebraic number theory setting.

**5.3.3 Proposition**

Let B be the ring of algebraic integers of a number field L, and let I be a nonzero integral

ideal of B, so that by (4.2.4) and (5.3.2), σ(I) is a lattice in Rn. Then the volume of the

fundamental domain of this lattice is

v(σ(I)) = 2−r2 |d|1/2N(I);

in particular, v(σ(B)) = 2−r2 |d|1/2, where d is the field discriminant.

Proof. The result for I = B follows from (5.3.2) and (2.3.3), taking the xk as an integral

basis for B. To establish the general result, observe that the fundamental domain for σ(I)

can be assembled by taking the disjoint union of N(I) copies of the fundamental domain

of σ(B). To convince yourself of this, let e1 and e2 be basis vectors in the plane. The

lattice H generated by 2e1 and 3e2 is a subgroup of the lattice H generated by e1 and

e2, but the fundamental domain T of H is larger than the fundamental domain T of H.

In fact, exactly 6 copies of T will fit inside T . ♣

**5.3.4 Minkowski Bound on Element Norms**

If I is a nonzero integral ideal of B, then I contains a nonzero element x such that

|NL/Q(x)| ≤ (4/π)r2 (n!/nn)|d|1/2N(I).

Proof. The set Bt of Section 5.2 is compact, convex and symmetric about the origin.

The volume of Bt is μ(Bt) = 2r1 (π/2)r2 tn/n!, with μ indicating Lebesgue measure. We choose t so that μ(Bt) = 2nv(σ(I)), which by (5.3.3) is 2n−r2 |d|1/2N(I). Equating the

two expressions for μ(Bt), we get

tn = 2n−r1π

−r2 n! |d|1/2N(I).

Apply (5.1.3b) with H = σ(I) and S = Bt. By our choice of t, the hypothesis of (5.1.3b)

is satisfied, and we have S ∩ (H \ {0})

= ∅. Thus there is a nonzero element x ∈ I such

that σ(x) ∈ Bt. Now by (2.1.6), the absolute value of the norm of x is the product of the

positive numbers ai = |σi(x)|, i = 1, . . . ,n. To estimate N(x), we invoke the inequality of

the arithmetic and geometric means, which states that (a1 · · · an)1/n ≤ (a1 +· · ·+an)/n.

It follows that a1 · · · an ≤ (

n

i=1 ai/n)n. With our ai’s, we have

|N(x)| ≤ [

1

n

r1

i=1

|σi(x)| +

2

n

r 1+r2

i=r1+1

|σi(x)| ]n.

Since σ(x) ∈ Bt, we have |N(x)| ≤ tn/nn. By choice of t,

|N(x)| ≤ (1/nn)2n−r1π

−r2 n! |d|1/2N(I).

But n − r1 = 2r2, so 2n−r1π−r2 = 22r2π−r2 = (4/π)r2 , and the result follows. ♣

**5.3.5 Minkowski Bound on Ideal Norms**

Every ideal class [see (3.4.5)] of L contains an integral ideal I such that

N(I) ≤ (4/π)r2 (n!/nn) |d|1/2.

Proof. Let J be a fractional ideal in the given class. We can multiply by a principal

ideal of B without changing the ideal class, so we can assume with loss of generality that

J = (J )−1 is an integral ideal. Choose a nonzero element x ∈ J such that x satisfies the

norm inequality of (5.3.4). Our candidate is I = xJ .

First note that I is an integral ideal because x ∈ J and JJ = B. Now (x) = IJ, so

by (4.2.6) and (5.3.4),

N(I)N(J) = N(x) ≤ (4/π)r2 (n!/nn) |d|1/2N(J).

Cancel N(J) to get the desired result. ♣

**5.3.6 Corollary**

The ideal class group of a number field is finite.

Proof. By (4.2.13), there are only finitely many integral ideals with a given norm. By

(5.3.5), we can associate with each ideal class an integral ideal whose norm is bounded

above by a fixed constant. If the ideal class group were infinite, we would eventually use

the same integral ideal in two different ideal classes, which is impossible. ♣

**5.3.7 Applications**

Suppose that a number field L has a Minkowski bound on ideal norms that is less than 2.

Since the only ideal of norm 1 is the trivial ideal (1) = B, every ideal class must contain

(1). Thus there can be only one ideal class, and the class number of L, that is, the order

of the ideal class group, is hL = 1. By (3.4.5), B is a PID, equivalently, by (3.2.8), a

UFD.

If the Minkowski bound is greater than 2 but less than 3, we must examine ideals

whose norm is 2. If I is such an ideal, then by (4.2.9), I divides (2). Thus the prime

factorization of (2) will give useful information about the class number.

In the exercises, we will look at several explicit examples.

Problems For Section 5.3

1. Calculate the Minkowski bound on ideal norms for an imaginary quadratic field, in

terms of the field discriminant d. Use the result to show that Q(

√

m) has class number 1

for m = −1,−2,−3,−7.

2. Calculate the Minkowski bound on ideal norms or a real quadratic field, in terms

of the field discriminant d. Use the result to show that Q(

√

m) has class number 1 for

m = 2, 3, 5, 13.

3. Show that in the ring of algebraic integers of Q(

√

−5), there is only one ideal whose

norm is 2. Then use the Minkowski bound to prove that the class number is 2.

4. Repeat Problem 3 for Q(

√

6).

5. Show that the only prime ideals of norm 2 in the ring of algebraic integers of Q(

√

17)

are principal. Conclude that the class number is 1.

6. Find the class number of Q(

√

14). (It will be necessary to determine the number of

ideals of norm 3 as well as norm 2.)

Problems 7-10 consider bounds on the field discriminant.

7. Let L be a number field of degree n over Q, with field discriminant d. Show that

|d| ≥ an = (π/4)n n2n/(n!)2.

8. Show that a2 = π2/4and an+1/an ≥ 3π/4. From this, derive the lower bound

|d| ≥ (π/3)(3π/4)n−1 for n ≥ 2.

9. Show that n/ log |d| is bounded above by a constant that is independent of the

particular number field.

10. Show that if L

= Q, then |d| > 1, hence in any nontrivial extension of Q, at least one

prime must ramify.

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