Algebraic Number Theory — Chapter 4 Geometry of numbers

Chapter 4
Geometry of numbers

4.1. Lattices

Definition 4.1.1. §

A lattice in an finite-dimensional $ℝ$ -vector space $V$ is an abelian subgroup of $V$ that is generated by a finite set of $ℝ$ -linearly independent vectors.

Definition 4.1.2. §

A lattice in a finite-dimensional $ℝ$ -vector space $V$ is said to be complete, or full, if its elements span $V$ .

The following is an immediate consequence of the definitions.

Lemma 4.1.3. §

If $Λ$ is a complete lattice in a finite-dimensional $ℝ$ -vector space $V$ , then $Λ$ is generated by a basis of $V$ .

Definition 4.1.4. §

Let $Λ$ be a complete lattice in a finite-dimensional vector space $V$ . The fundamental domain of $Λ$ relative to a $ℤ$ -basis ${v_{1}, \dots, v_{n}}$ of $Λ$ is the set

D = {\sum_{i = 1}^{n} c_{i} v_{i} | c_{i} \in [0, 1) for all 1 \leq i \leq n} .

The following is essentially immediate.

Lemma 4.1.5. §

If $D$ is a fundamental domain of a complete lattice $Λ$ in a vector space $V$ , then every element of $V$ may be written uniquely in the form $x + y$ for some $x \in Λ$ and some $y \in D$ .

Of course, a finite-dimensional $ℝ$ -vector space $V$ has a Euclidean metric with respect to any basis of $V$ (as such a basis defines an isomorphism $𝑉≅ ℝ^{n}$ for some $n$ ). Though the metric depends upon the choice of basis, the resulting topology is independent of the choice of basis.

Definition 4.1.6. §

We say that a subgroup of an finite-dimensional $ℝ$ -vector space $V$ is discrete if it has the discrete topology as a subspace of $V$ .

In other words, a subgroup $Λ$ of a finite-dimensional real vector space $V$ is discrete if for every $v \in Λ$ there exists an open neighborhood $U$ of $v$ in $V$ such that $U \cap Λ = {v}$ . The following is elementary.

Lemma 4.1.7. §

A discrete subgroup of a finite-dimensional $ℝ$ -vector space $V$ is a closed subset of $V$ .

Proposition 4.1.8. §

A subgroup of a finite-dimensional $ℝ$ -vector space is discrete if and only if it is a lattice.

Proof.

Let $V$ be an $n$ -dimensional $ℝ$ -vector space for some $n \geq 0$ , and let $Λ$ be a subgroup. If $Λ$ is a lattice, then $Λ = \sum_{i = 1}^{m} ℤ v_{i}$ for some linearly independent $v_{i} \in Λ$ and $m \leq n$ . Extend these to an ordered basis $v_{1}, \dots, v_{n}$ of $V$ . Let $v = \sum_{i = 1}^{m} a_{i} v_{i} \in Λ$ for some $a_{i} \in ℤ$ . Then

U = {v + \sum_{i = 1}^{n} c_{i} v_{i} | c_{i} \in (- 1, 1) for all 1 \leq i \leq n}

is an open neighborhood in $V$ containing $v$ but no other elements of $Λ$ . Thus $Λ$ is discrete.

Conversely, suppose $Λ$ is discrete. Let $W$ be the subspace of $V$ spanned by $Λ$ , and let $v_{1}, \dots, v_{m} \in Λ$ be linearly independent with $m$ maximal, so that the $v_{i}$ necessarily span $W$ . Let $Σ = \sum_{i = 1}^{m} ℤ v_{i} \leq Λ$ . We claim that $Σ$ is of finite index in $Λ$ . By definition, $Σ$ is a complete lattice in $W$ , so so we may choose a system $S$ of representatives of the cosets $Λ ∕ Σ$ inside the fundamental domain $D$ of $Σ$ in $W$ . That is, $S \subset Λ \cap D$ . However, $Λ$ is discrete and closed and $D$ is a bounded set, so the intersection of $Λ$ with the closure of $D$ is discrete and compact, hence finite. Since $Λ \cap D$ is then finite, $S$ has only finitely many elements.

Now set $d = [Λ : Σ]$ . Then $Λ \subseteq \frac{1}{d} Σ$ , and $\frac{1}{d} Σ ≅ Σ$ is a free abelian group of rank $m$ , so $Λ$ is free of rank (at most) $m$ . In other words, $Λ$ is a lattice. □

Lemma 4.1.9. §

A lattice $Λ$ in $V$ is complete if and only if there exists a bounded subset $B$ of $V$ such that every element of $V$ is the sum of an element of $B$ and an element of $Λ$ .

Proof.

If $Λ$ is a complete lattice, we may let $B$ be the fundamental domain of $V$ relative to a basis, and in fact every element of $V$ may in that case be written uniquely as an element of $B + Λ$ .

Suppose that there exists a $B$ as in the statement of the lemma. Let $W$ be the $ℝ$ -span of $Λ$ , and let $v \in V$ . For each $k \geq 1$ , write $𝑘𝑣 = b_{k} + x_{k}$ with $b_{k} \in B$ and $x_{k} \in Λ$ . As $B$ is bounded, we have

\lim_{k \to \infty} \frac{1}{k} b_{k} = 0,

v = \lim_{k \to \infty} \frac{1}{k} x_{k} .

But $\frac{1}{k} x_{k} \in W$ for all $k$ and $W$ is closed, so $v \in W$ . □

We will suppose now that our finite-dimensional vector space $V$ comes equipped with a symmetric, positive definite inner product

⟨, ⟩ : V \times V \to ℝ .

In other words, $V$ is an finite-dimensional real inner product space. The Euclidean metric defined by a choice of an orthonormal basis of $V$ is independent of the choice. The resulting Lebesgue measure $μ_{V}$ on $V$ has the property that the cube defined by an orthonormal basis of $V$ has volume (i.e., measure) one.

Definition 4.1.10. §

Let $V$ be an finite-dimensional real vector space with a symmetric, positive definite inner product $⟨, ⟩$ . Let $Λ$ be a complete lattice in $V$ . The volume $Vol (Λ)$ of lattice is the volume of the fundamental domain of $Λ$ relative to a $ℤ$ -basis of $Λ$ .

Remark 4.1.11. §

That the volume of $Λ$ is indepedendent of the choice of basis is a consequence of the following lemma.

Lemma 4.1.12. §

Let $V$ be an $n$ -dimensional real inner product space. Let $e_{1}, \dots, e_{n}$ be an orthonormal basis of $V$ . Let $Λ$ be a complete lattice in $V$ with basis $v_{1}, \dots, v_{n}$ , write

v_{i} = \sum_{j = 1}^{n} a_{𝑖𝑗} e_{j},

and let $A = (a_{𝑖𝑗})$ . Then

Vol (Λ) = | \det A | = | \det (⟨ v_{i}, v_{j} ⟩) |^{1 ∕ 2} .

Proof.

The first equality is a standard statement of linear algebra. The second is the fact that

⟨ v_{i}, v_{j} ⟩ = \sum_{k = 1}^{n} \sum_{l = 1}^{n} a_{𝑖𝑘} a_{𝑗𝑙} ⟨ e_{k}, e_{l} ⟩ = \sum_{k = 1}^{n} a_{𝑖𝑘} a_{𝑗𝑘} = {(A A^{T})}_{𝑖𝑗},

since $\det (A A^{T}) = \det {(A)}^{2}$ . □

Definition 4.1.13. §

Let $T$ be a subset of a finite-dimensional vector space $V$ .

a.

We say that $T$ is convex if it contains the line segment

ℓ = {(1 - c) v + 𝑐𝑤 ∣ c \in [0, 1]}

between any two vectors $v, w \in T$ .

b.

We say that $T$ is symmetric about the origin if $- v \in T$ for all $v \in T$ .

Now we come to the main theorem of this subsection.

Theorem 4.1.14 (Minkowski’s theorem). §

Let $V$ be an $n$ -dimensional real inner product space, and let $Λ$ be a complete lattice in $V$ . Let $X$ be a convex, measurable subset of $V$ that is symmetric about the origin, and suppose that

μ_{K \otimes_{ℚ} ℝ} (X) > 2^{n} Vol (Λ) .

Then $X \cap Λ \neq {0}$ .

Proof.

Let

Y = \frac{1}{2} X = {\frac{1}{2} x ∣ x \in X} .

If $y \in Y$ , then $2 y \in X$ , so $- 2 y \in X$ as $X$ is symmetric about the origin. If $y^{'} \in Y$ as well, then

y^{'} - y = \frac{1}{2} (2 y^{'}) + \frac{1}{2} (- 2 y) \in X,

since $X$ is convex. We therefore have that the difference of any two points of $Y$ is in $X$ . Note also that $μ_{V} (Y) = \frac{1}{2^{n}} μ_{V} (X) > Vol (Λ)$ .

Let $D$ a the fundamental domain for $Λ$ . Suppose that the sets $v + Y$ for $v \in Λ$ were pairwise disjoint. Then we would have

Vol (Λ) \geq \sum_{v \in Λ} μ_{V} (D \cap (v + Y)) = \sum_{v \in Λ} μ_{V} ((D - v) \cap Y) = μ_{V} (Y),

the latter equality following since the sets $D - v$ cover $V$ . This does not hold by assumption, so there exist distinct $v, w \in Λ$ such that $(v + Y) \cap (w + Y)$ is nonempty. But then $v - w \in X$ , so $X \cap Λ \neq {0}$ . □

4.2. Real and complex embeddings

Let $K$ be an number field of degree $n$ over $ℚ$ . Then, since $K ∕ ℚ$ is separable and $ℂ$ is algebraically closed, then we obtain the product

K ↪ K \otimes_{ℚ} ℂ \to_{}^{\sim} \prod_{σ : K ↪ ℂ} ℂ

of embeddings of $K$ in $ℂ$ .

Definition 4.2.1. §

Let $K$ be a number field.

a.: A real embedding (or real prime) of $K$ is a field embedding of $K$ in $ℝ$ .
b.: A complex embedding of $K$ is a field embedding of $K$ in $ℂ$ that does not have image contained in $ℝ$ .
c.: A complex prime of $K$ is an unordered pair ${σ, \bar{σ}}$ of complex embeddings such that $\bar{σ} (α) = \bar{σ (α)}$ for $α \in K$ , where $\bar{z}$ denotes the complex conjugate of $z \in ℂ$ .
d.: An archimedean prime of $K$ is either a real prime or a complex prime.

Notation 4.2.2. §

The number of real (resp., complex) primes of a number field $K$ is denoted $r_{1} (K)$ (resp., $r_{2} (K)$ ).

We can also apply Proposition 1.1.1 to obtain the following.

Theorem 4.2.3. §

Let $K$ be a number field. Then $r_{1} (K) + 2 r_{2} (K) = [K : ℚ]$ . In fact, we have an isomorphism of $ℝ$ -algebras

κ_{ℝ} : K \otimes_{ℚ} ℝ \to_{}^{\sim} ℝ^{r_{1} (K)} \times ℂ^{r_{2} (K)}

where $κ_{ℝ} (α \otimes 1)$ for $α \in K$ is the product of the real embeddings of $K$ applied to $α$ and one complex embedding from each complex prime of $ℚ$ applied to $α$ .

Proof.

Let $K = ℚ (𝜃)$ and $f \in K [x]$ be the minimal polynomial of $𝜃$ . In $ℝ [x]$ , we may write

f = \prod_{i = 1}^{r_{1}} (x - α_{i}) \cdot \prod_{j = 1}^{r_{2}} f_{j},

for some $r_{1}, r_{2} \geq 0$ with $r_{1} + 2 r_{2} = [K : ℚ]$ , where $α_{i} \in ℝ$ and $f_{j}$ is irreducible quadratic. Note that $ℝ [x] ∕ (f_{j}) ≅ℂ$ for each $1 \leq j \leq r_{2}$ via maps that take $x + (f_{j})$ to a chosen root of $f_{j}$ in $ℂ$ . By Proposition 1.1.1, we then have

K \otimes_{ℚ} ℝ≅ \prod_{i = 1}^{r_{1}} ℝ \times \prod_{i = 1}^{r_{2}} ℂ

and the composition of the natural inclusion of $K$ in $K \otimes_{ℚ} ℝ$ is the product of the real embeddings of $K$ and one choice of complex embedding of $K$ among pairs of complex conjugate embeddings (i.e., complex primes). We therefore have $r_{1} = r_{1} (K)$ and $r_{2} = r_{2} (K)$ . □

Examples 4.2.4. §

a.: The field $ℚ (i)$ has one complex prime, corresponding to the two complex embeddings of $ℚ (i)$ taking $i$ to $\pm i$ .
b.: The field $ℚ (\sqrt[3]{2})$ has one real prime and that takes $\sqrt[3]{2}$ to $\sqrt[3]{2}$ one complex prime corresponding to the two complex embeddings taking $\sqrt[3]{2}$ to $ω^{\pm 1} \sqrt[3]{2}$ .

Remark 4.2.5. §

Given a number field $K$ , Theorem 4.2.3 provides us with a product of embeddings

ι_{K} : K \to \prod_{i = 1}^{r_{1}} ℝ \times \prod_{i = 1}^{r_{2}} ℂ

corresponding to the real and complex primes of $K$ .

4.3. Finiteness of the class group

Throughout this section, we let $K$ be a number field of degree $n$ over $ℚ$ . We set $r_{1} = r_{1} (K)$ and $r_{2} = r_{2} (K)$ . We let $σ_{i}$ for $1 \leq i \leq r_{1}$ be the real embeddings of $K$ , and we let $τ_{i}$ for $1 \leq i \leq r_{2}$ be a choice of complex embedding from each complex prime of $K$ . We identify $K \otimes_{ℚ} ℝ$ with $ℝ^{r_{1}} \times ℂ^{r_{2}}$ as in Theorem 4.2.3.

We endow $K \otimes_{ℚ} ℝ$ with the Lebesgue measure $μ_{K \otimes_{ℚ} ℝ}$ on $ℝ^{n} ≅ ℝ^{r_{1}} \times ℂ^{r_{2}}$ , where the inverse of said isomorphism is defined by

(x_{1}, \dots, x_{r_{1}}, z_{1}, \dots, z_{r_{2}}) \mapsto (x_{1}, \dots, x_{r_{1}}, Re (z_{1}), Im (z_{1}), \dots, Re (z_{r_{2}}), Im (z_{r_{2}})) .

Proposition 4.3.1. §

Let $𝔞$ be a nonzero ideal of $𝒪_{K}$ . Then $ι_{K} (𝔞)$ is a complete lattice in $K \otimes_{ℚ} ℝ$ and

Vol (ι_{K} (𝔞)) = 2^{- r_{2}} \cdot 𝑁𝔞 \cdot | disc (K) |^{1 ∕ 2} .

Proof.

Let $α_{1}, \dots, α_{n}$ be a $ℤ$ -basis of $𝔞$ . Consider the matrix $A \in M_{n} (ℂ)$ with $i$ th row

(σ_{1} (α_{i}), \dots, σ_{r_{1}} (α_{i}), τ_{1} (α_{i}), \bar{τ_{1} (α_{i})}, \dots, τ_{r_{2}} (α_{i}), \bar{τ_{r_{2}} (α_{i})})

and the matrix $B \in M_{n} (ℝ)$ with $i$ th row

(σ_{1} (α_{i}), \dots, σ_{r_{1}} (α_{i}), Re (τ_{1} (α_{i})), Im (τ_{1} (α_{i})), \dots, Re (τ_{r_{2}} (α_{i})), Im (τ_{r_{2}} (α_{i}))) .

Simple column operations yield that $\det A = {(- 2 i)}^{r_{2}} \det B$ , and Lemmas 1.4.10 and 1.4.6 imply that

| \det A | = | D (α_{1}, \dots, α_{n}) |^{1 ∕ 2} = 𝑁𝔞 \cdot | disc (K) |^{1 ∕ 2} .

In particular, $\det B \neq 0$ , so the rows of images of the $α_{i}$ in $K \otimes_{ℚ} ℝ$ are $ℝ$ -linearly independent, which is to say that $ι_{K} (𝔞)$ is complete. Since Lemma 4.1.12 implies that $Vol (ι_{K} (𝔞)) = | \det B |$ , we have the result. □

Notation 4.3.2. §

We define a norm on $K \otimes_{ℚ} ℝ≅ ℝ^{r_{1}} \times ℂ^{r_{2}}$ by

∥ v ∥_{K} = \sum_{i = 1}^{r_{1}} | x_{i} | + 2 \sum_{j = 1}^{r_{2}} | z_{j} |

for $v = (x_{1}, \dots, x_{r_{1}}, z_{1}, \dots, z_{r_{2}})$ . For $t > 0$ , let

D_{t} = {v \in K \otimes_{ℚ} ℝ ∣ ∥ v ∥_{K} < t} .

We compute the volume of $D_{t}$ .

Lemma 4.3.3. §

We have

μ_{K \otimes_{ℚ} ℝ} (D_{t}) = 2^{r_{1} - r_{2}} π^{r_{2}} \frac{t^{n}}{n!}

Proof.

We treat this as a problem about the real inner product space $V = R^{r_{1}} \times ℂ^{r_{2}}$ of dimension $n = r_{1} + 2 r_{2}$ . Let $B_{t}^{(r_{1}, r_{2})}$ be the subset of $D_{t} = D_{t}^{(r_{1}, r_{2})}$ inside $V$ of elements with nonnegative real coordinates. Then $μ_{V} (B_{t}^{(r_{1}, r_{2})}) = 2^{- r_{1}} μ_{V} (D_{t}^{(r_{1}, r_{2})})$ . Suppose first that $r_{2} = 0$ and the result holds for $(r_{1} - 1, 0)$ . Then we have

μ_{V} (B_{t}^{(r_{1}, 0)}) = \int_{x = 0}^{t} μ_{V} (B_{t - x}^{(r_{1} - 1, 0)}) 𝑑𝑥 = \int_{x = 0}^{t} \frac{x^{n - 1}}{(n - 1)!} 𝑑𝑥 = \frac{t^{n}}{n!} .

Next, suppose that $r_{2} > 0$ and the result holds for $(r_{1}, r_{2} - 1)$ . By elementary calculation, we obtain

μ_{V} (B_{t}^{(r_{1}, r_{2})}) = \int_{r = 0}^{t ∕ 2} \int_{𝜃 = 0}^{2 π} μ_{V} (B_{t - 2 r}^{(r_{1}, r_{2} - 1)}) 𝑟𝑑𝑟𝑑𝜃 = 2 π {(\frac{π}{2})}^{r_{2} - 1} \int_{r = 0}^{t ∕ 2} \frac{{(2 r)}^{n - 2}}{(n - 2)!} (t - 2 r) 𝑑𝑟 = {(\frac{π}{2})}^{r_{2}} \frac{t^{n}}{n!} .

□

Proposition 4.3.4. §

For any nonzero ideal $𝔞$ of $𝒪_{K}$ , there exists $α \in 𝔞 - {0}$ such that

| N_{K ∕ ℚ} (α) | \leq {(\frac{4}{π})}^{r_{2}} \frac{n!}{n^{n}} \cdot 𝑁𝔞 \cdot | disc (K) |^{1 ∕ 2} .

Proof.

Let $t$ be such that

μ_{K \otimes_{ℚ} ℝ} (D_{t}) > 2^{n} Vol (ι_{K} (𝔞)) .

(4.3.1)

Since $ι_{K} (𝔞)$ is a lattice, Minkowski’s theorem ensures that $D_{t}$ contains $ι_{K} (α)$ for some nonzero $α \in 𝔞$ . Note that $N_{K ∕ ℚ} (α)$ is the product of images of $α$ under the distinct field embeddings of $K$ in $ℂ$ . Hence, we have

| N_{K ∕ ℚ} (α) | = | σ_{1} (α) | \dots | σ_{r_{1}} (α) | | τ_{1} (α) |^{2} \dots | τ_{r_{2}} (α) |^{2},

and since the arithmetic mean bounds the geometric mean, this is at most the $n$ th power of

\frac{1}{n} (\sum_{i = 1}^{r_{1}} | σ_{i} (α) | + 2 \sum_{j = 1}^{r_{2}} | τ_{j} (α) |) .

On the other hand, since $α \in D_{t}$ , the latter quantity is less than $\frac{t}{n}$ , so we have $| N_{K ∕ ℚ} (α) | < {(\frac{t}{n})}^{n}$ . Now, Lemma 4.3.3 and Proposition 4.3.1 allow us to rewrite (4.3.1) as

2^{r_{1} - r_{2}} π^{r_{2}} \frac{t^{n}}{n!} > 2^{n - r_{2}} \cdot 𝑁𝔞 \cdot | disc (K) |^{1 ∕ 2} .

Set

C = {(\frac{4}{π})}^{r_{2}} \frac{n!}{n^{n}} \cdot 𝑁𝔞 \cdot | disc (K) |^{1 ∕ 2},

so ${(\frac{t}{n})}^{n} > C$ . We can and do choose ${(\frac{t}{n})}^{n}$ to be less than the smallest integer greater than $C$ . Since $| N_{K ∕ ℚ} (α) |$ is an integer, we must then have that $| N_{K ∕ ℚ} (α) | \leq C$ . □

Definition 4.3.5. §

Let $K$ be a number field of degree $n$ over $ℚ$ . The Minkowski bound $B_{K}$ is the quantity

B_{K} = \frac{n!}{n^{n}} {(\frac{4}{π})}^{r_{2}} | disc (K) |^{1 ∕ 2} .

Proposition 4.3.4 has the following interesting corollary

Corollary 4.3.6. §

The discriminant of a number field $K$ satisfies $| disc (K) |^{1 ∕ 2} \geq {(\frac{π}{4})}^{r_{2}} \frac{n^{n}}{n!}$ .

Proof.

Take $𝔞 = 𝒪_{K}$ in Proposition 4.3.4, and note that $| N_{K ∕ ℚ} (α) | \geq 1$ for all $α \in 𝒪_{K}$ . □

Theorem 4.3.7 (Minkowski). §

Let $K$ be a number field of degree $n$ over $ℚ$ . Then there exists a set of representatives for the ideal class group of $K$ consisting of integral ideals $𝔞$ with $𝑁𝔞 \leq B_{K}$ .

Proof.

Let $𝔞$ be a fractional ideal of $𝒪_{K}$ . Let $d \in K^{\times}$ be such that $𝔟 = d 𝔞^{- 1}$ is an integral ideal. By Proposition 4.3.4, there exists $β \in 𝔟 - {0}$ with $| N_{K ∕ ℚ} (β) | \leq B_{K} \cdot 𝑁𝔟$ . Now, note that $(β) = 𝔟𝔠$ for some ideal $𝔠$ of $𝒪_{K}$ , and the ideal class of $𝔠$ is the ideal class of $𝔟^{- 1}$ , which is the ideal class of $𝔞$ . Furthermore, we have that $| N_{K ∕ ℚ} (β) | = 𝑁𝔟 \cdot 𝑁𝔠$ , so $𝑁𝔠 \leq B_{K}$ , as desired. □

As a consequence, we have the following theorem.

Theorem 4.3.8. §

The class group of a number field is finite.

Proof.

In fact, the set of nonzero integral ideals $𝔞$ with $𝑁𝔞 \leq B_{K}$ is finite. To see this, write $𝔞 = 𝔭_{1}^{r_{1}} \dots 𝔭_{k}^{r_{k}}$ for distinct prime ideals $𝔭_{i}$ and $r_{i} \geq 1$ for some $1 \leq i \leq k$ . Then

𝑁𝔞 = p_{1}^{r_{1} f_{1}} \dots p_{k}^{r_{k} f_{k}},

where $N 𝔭_{i} = p_{i}^{f_{i}}$ with $p_{i}$ prime. Since there are only finitely many positive integers less than $B_{K}$ , there are only finitely many primes that could divide $𝑁𝔞$ , and the exponents of these primes are bounded (e.g., by $\log_{2} (B_{K})$ ). Since each prime $(p)$ of $ℤ$ has only finitely many primes of $𝒪_{K}$ lying over it, we are done. □

Definition 4.3.9. §

The class number $h_{K}$ of a number field $K$ is the order of the class group ${Cl}_{K}$ .

The Minkowski bound allows us to actually give bounds on the class number of a number field and sometimes, to actually compute the class group.

Example 4.3.10. §

Consider $K = ℚ (\sqrt{- 5})$ . Note that $disc (K) = - 20$ , so the Minkowski bound is

B_{K} = \frac{2}{π} \sqrt{20} < 3 .

Since $ℤ [\sqrt{- 5}]$ is not a UFD, it is not a PID, so $h_{K} \geq 2$ . Since $2$ ramifies in $ℤ [\sqrt{- 5}]$ , the class of the prime over it in $K$ is the only nontrivial element in ${Cl}_{K}$ . We therefore have $h_{K} = 2$ .

Example 4.3.11. §

Consider $K = ℚ (\sqrt{17})$ . In this case $disc (K) = 17$ , and the Minkowski bound is

B_{K} = \frac{1}{2} \sqrt{17} < 3 .

Set $α = \frac{\sqrt{17} + 1}{2}$ so that $𝒪_{K} = ℤ [α]$ . Since the norm of any prime $𝔭$ of $ℤ [α]$ is at least the prime $p$ with $𝔭 \cap ℤ = (p)$ , there exists a set of representatives for ${Cl}_{K}$ dividing $2$ . The minimal polynomial of $α$ is $x^{2} - x - 4$ , which splits modulo $2$ . Therefore, the class group ${Cl}_{K}$ is generated by the two primes over $2$ . Their classes are inverse to each other, so ${Cl}_{K}$ is generated any one of them, which we denote $𝔭$ . For $α \in 𝒪_{K}$ , we have $𝔭 = (α)$ if and only if $N (α) = \pm 2$ . Note that

N_{K ∕ ℚ} (\frac{\sqrt{17} + 5}{2}) = \frac{25 - 17}{4} = 2,

so $𝔭$ is principal and $K$ is a UFD.

4.4. Dirichlet’s unit theorem

Lemma 4.4.1. §

Let $n, N \geq 1$ . The set of algebraic integers $α$ such that $[ℚ (α) : ℚ] \leq n$ and $| σ (α) | \leq N$ for all archimedean embeddings $σ$ of $ℚ (α)$ is finite.

Proof.

Let $α$ be such an integer, and let $f = \sum_{i = 0}^{n} a_{i} x^{i}$ be its minimal polynomial in $ℤ [x]$ . In $ℂ [x]$ , we have $f = \prod_{σ} (x - σ (α))$ , the product being taken over the archimedean embeddings. Then $| a_{i} |$ is bounded: e.g., it satisfies $| a_{i} | \leq N^{n - i} (\binom{n}{i})$ for all $i$ . As each $a_{i}$ is an integer, the number of minimal polynomials of elements in the stated set, and hence the order of the set, is finite. □

Corollary 4.4.2. §

Let $K$ be a number field. Then $μ (K)$ is a finite group, equal to the set of all $α \in 𝒪_{K}$ such that $| σ (α) | = 1$ for all archimedean embeddings $σ$ of $K$ .

Proof.

If $α \in 𝒪_{K}$ satisfies $| σ (α) | = 1$ for all $σ$ , then so does $α^{n}$ for all $n$ . Since the set $T$ of such $α$ is finite by Lemma 4.4.1, the group $⟨ α ⟩$ is finite, and hence $α \in μ (K)$ . In particular, since a root of unity has complex absolute value $1$ under any archimedean embeddings, the set $μ (K) = T$ is finite. □

Definition 4.4.3. §

The unit group of a number field $K$ is the group $𝒪_{K}^{\times}$ of units in $𝒪_{K}$ .

Definition 4.4.4. §

For a number field $K$ , we let $ℓ_{K} : K^{\times} \to ℝ^{r_{1} + r_{2}}$ be defined on $α \in K^{\times}$ as

ℓ_{K} (α) = (\log | σ_{1} (α) |, \dots, \log | σ_{r_{1}} (α) |, \log | τ_{1} (α) |, \dots, \log | τ_{r_{2}} (α) |),

where $σ_{1}, \dots, σ_{r_{1}}$ are the real embeddings of $K$ and $τ_{1}, \dots, τ_{r_{2}}$ are the complex embeddings of $K$ .

Proposition 4.4.5. §

For a number field $K$ , the set $ℓ_{K} (𝒪_{K}^{\times})$ is a lattice in $ℝ^{r_{1} + r_{2}}$ that sits in the hyperplane

H = {(x_{1}, \dots, x_{r_{1} + r_{2}}) | \sum_{i = 1}^{r_{1}} x_{i} + 2 \sum_{j = 1}^{r_{2}} x_{j + r_{1}} = 0},

and $\ker ℓ_{K} = μ (K)$ .

Proof.

That $\ker ℓ_{K} = μ (K)$ is just a rewording of Corollary 4.4.2. For $α \in 𝒪_{K}^{\times}$ , we have that

\sum_{i = 0}^{r_{1}} \log | σ_{i} (α) | + 2 \sum_{j = 0}^{r_{2}} \log | τ_{j} (α) | = \log | N_{K ∕ ℚ} (α) | = 0,

the latter equality by Corollary 2.5.18, so $ℓ_{K} (α) \in H$ . For $N \geq 0$ , consider the bounded subset

D_{N} = {(x_{1}, \dots, x_{r_{1} + r_{2}}) \in H ∣ | x_{i} | \leq N for all i}

of $H$ . By Lemma 4.4.1, the set of elements of $𝒪_{K}^{\times}$ contained in $D_{N}$ is finite, which means that there exists an open neighborhood $U$ in $ℝ^{r_{1} + r_{2}}$ such that the intersection $U \cap ℓ_{K} (𝒪_{K}^{\times})$ is ${0}$ . This implies that $ℓ_{K} (𝒪_{K}^{\times})$ is a discrete group: the set $ℓ_{K} (α) + U$ is also open and its intersection with $ℓ_{K} (𝒪_{K}^{\times})$ is ${ℓ_{K} (α)}$ . Finally, recall that a discrete subgroup of $ℝ^{r_{1} + r_{2}}$ is a lattice. □

Proposition 4.4.5 implies that the sequence

1 \to μ (K) \to 𝒪_{K}^{\times} \to_{}^{ℓ_{K}} ℓ_{K} (𝒪_{K}^{\times}) \to 0

is exact and that the set $ℓ_{K} (𝒪_{K}^{\times})$ is a free abelian group of finite rank. In particular, $𝒪_{K}^{\times}$ is a finitely generated abelian group of rank that of $ℓ_{K} (𝒪_{K}^{\times})$ , and the sequence is split. If $ℓ_{K} (𝒪_{K}^{\times})$ is a complete lattice in $H$ , then we know that the rank of $𝒪_{K}^{\times}$ is $r_{1} + r_{2} - 1$ . Dirichlet’s unit theorem tells us that this is in fact the case. For this, we require the following lemma.

Lemma 4.4.6. §

Let $A = (a_{𝑖𝑗}) \in M_{k} (ℝ)$ for some $k \geq 1$ satisfy $a_{𝑖𝑗} < 0$ for all $i \neq j$ and $\sum_{j = 1}^{k} a_{𝑖𝑗} > 0$ for all $i$ . Then $A$ is invertible.

Proof.

Suppose by way of contradication that $v = (v_{1}, \dots, v_{k}) \in ℝ^{k}$ is a solution to $𝐴𝑣 = 0$ , scaled so that $v_{j} = 1$ for some $j$ and $| v_{i} | \leq 1$ for all $i \neq j$ . Then

0 = \sum_{i = 1}^{k} a_{𝑗𝑖} v_{i} = a_{𝑗𝑗} + \sum_{\begin{array}{c} i = 1 \\ i \neq j \end{array}}^{k} a_{𝑗𝑖} v_{i} .

(4.4.1)

Note that $a_{𝑗𝑖} v_{i} \geq a_{𝑗𝑖}$ for $i \neq j$ since $a_{𝑗𝑖}$ is negative and $v_{i} \leq 1$ . Thus, the right-hand side of (4.4.1) is at least $\sum_{i = 1}^{k} a_{𝑖𝑗} > 0$ , which is the desired contradiction. □

For a number field $K$ , let us use $κ_{ℝ}$ to identify $K \otimes_{ℚ} ℝ$ and $ℝ^{r_{1}} \times ℂ^{r_{2}}$ as rings. We can then make the following definition.

Notation 4.4.7. §

For a number field $K$ , we let

[\cdot] : K \otimes_{ℚ} ℝ \to ℝ

denote the multiplicative function given by

[(x_{1}, \dots, x_{r_{1}}, z_{1}, \dots, z_{r_{2}})] = | x_{1} | \dots | x_{r_{1}} | | z_{1} |^{2} \dots | z_{r_{2}} |^{2},

for $x_{i} \in ℝ$ with $1 \leq i \leq r_{1}$ and $z_{j} \in ℂ$ with $1 \leq j \leq r_{2}$ .

The following rather complicated lemma is useful in showing the completeness of $ℓ_{K} (𝒪_{K}^{\times})$ in $H$ .

Lemma 4.4.8. §

Let $K$ be a number field of degree $n$ . Set

D = {v \in K \otimes_{ℚ} ℝ ∣ 1 ∕ 2 \leq [v] \leq 1} .

Let $X$ be a bounded, convex subset of $K \otimes_{ℚ} ℝ$ that is symmetric about the origin and has volume

μ_{V} (X) > 2^{r_{1} + r_{2}} | disc (K) |^{1 ∕ 2} .

There exist $α_{1}, \dots, α_{s} \in 𝒪_{K}$ for some $s \geq 1$ such that for each $v \in D$ , there exists some $𝜖 \in 𝒪_{K}^{\times}$ and $1 \leq i \leq s$ such that $v ι_{K} (𝜖) \in ι_{K} (α_{i}^{- 1}) X$ .

Proof.

For $w \in K \otimes_{ℚ} ℝ$ , the set

w \cdot ι_{K} (𝒪_{K}) = {w \cdot ι_{K} (α) ∣ α \in 𝒪_{K}}

is a complete lattice with volume $2^{- r_{2}} | disc (K) |^{1 ∕ 2} \cdot [w]$ . Suppose $w \in D$ . Since $[w] \leq 1$ , we have

2^{n} Vol (w \cdot ι_{K} (𝒪_{K})) \leq 2^{n} Vol (ι_{K} (𝒪_{K})) = 2^{r_{1} + r_{2}} | disc (K) |^{1 ∕ 2},

the latter equality by Proposition 4.3.1. Minkowski’s theorem then tells us that there exists $α \in 𝒪_{K} - {0}$ such that $w \cdot ι_{K} (α) \in X$ .

Since $X$ is bounded, there exists $M > 0$ such that $[x] \leq M$ for all $x \in X$ . In particular, we have $[w \cdot ι_{K} (α)] \leq M$ , which since $[w] \geq \frac{1}{2}$ forces $[ι_{K} (α)] \leq 2 M$ and then $N (α 𝒪_{K}) \leq 2 M$ . As there are only finitely many ideals of absolute norm at most $2 M$ , there are only finitely many ideals $α 𝒪_{K}$ such that $w \cdot ι_{K} (α 𝒪_{K}) \cap X \neq {0}$ for some $w \in D$ . Let $α_{1} 𝒪_{K}, \dots, α_{s} 𝒪_{K}$ be these ideals.

Finally, given $v \in D$ , let $β \in 𝒪_{K} - {0}$ have the property that $v \cdot ι_{K} (β) \in X$ . Then $β 𝒪_{K} = α_{i} 𝒪_{K}$ for some $1 \leq i \leq s$ , so $𝜖 = β α_{i}^{- 1} \in 𝒪_{K}^{\times}$ and $v \cdot ι_{K} (𝜖) \in ι_{K} (α_{i}^{- 1}) X$ . □

Theorem 4.4.9 (Dirichlet’s unit theorem). §

Let $K$ be a number field. Then we have an isomorphism

𝒪_{K}^{\times} ≅ ℤ^{r_{1} (K) + r_{2} (K) - 1} \times μ (K) .

Proof.

Let $D$ , $X$ , and $α_{1}, \dots, α_{s} \in 𝒪_{K}$ be as in Lemma 4.4.8. Set

Y = ⋃_{i = 1}^{s} ι_{K} (α_{i}^{- 1}) X .

Since $Y$ is bounded, we may let $N$ be such that if $(λ_{1}, \dots, λ_{r_{1} + r_{2}}) \in Y$ , then $| λ_{i} | \leq N$ for all $1 \leq i \leq r_{1} + r_{2}$ . For $1 \leq i \leq r_{1} + r_{2}$ , let

v^{(i)} = (v_{1}^{(i)}, \dots, v_{r_{1} + r_{2}}^{(i)}) \in K \otimes_{ℚ} ℝ

be an element such that $| v_{j}^{(i)} | > N$ for $j \neq i$ but $[v^{(i)}] = 1$ . Then $v^{(i)} \in D$ , so there exists $𝜖^{(i)} \in 𝒪_{K}^{\times}$ such that $v^{(i)} \cdot ι_{K} (𝜖^{(i)}) \in Y$ . In particular, we must have

ι_{K} (𝜖^{(i)}) = (𝜖_{1}^{(i)}, \dots, 𝜖_{r_{1} + r_{2}}^{(i)}) = (σ_{1} (𝜖^{(i)}), \dots, σ_{r_{1}} (𝜖^{(i)}), τ_{1} (𝜖^{(i)}), \dots, τ_{r_{2}} (𝜖^{(i)}))

with $| 𝜖_{j}^{(i)} | < 1$ for all $j \neq i$ . In other words, $ℓ_{K} (𝜖^{(i)})$ has negative coordinates for $j \neq i$ .

Assume without loss of generality that $r_{1} + r_{2} > 1$ . Set

a_{i, j} = {\begin{matrix} \log | σ_{j} (𝜖^{(i)}) | & for 1 \leq j \leq r_{1} \\ 2 \log | τ_{j - r_{1}} (𝜖^{(i)}) | & for r_{1} + 1 \leq j \leq r_{1} + r_{2} . \end{matrix}

Consider the matrix $A = {(a_{i, j})}_{i, j} \in M_{r_{1} + r_{2} - 1} (ℝ)$ . For $i < r_{1} + r_{2}$ , we have that the sum of the $i$ th row is

\sum_{j = 1}^{r_{1} + r_{2} - 1} a_{i, j} = - a_{i, r_{1} + r_{2}} > 0

as $ℓ_{K} (𝜖^{(i)}) \in H$ and $i \neq r_{1} + r_{2}$ . Lemma 4.4.6 then tells us that $A$ is invertible, so the $ℓ_{K} (𝜖^{(i)})$ for $1 \leq i < r_{1} + r_{2}$ form a system of $r_{1} + r_{2} - 1$ linearly independent vectors in $H$ . In other words, $ℓ_{K} (𝒪_{K}^{\times})$ is a complete lattice. □

Example 4.4.10. §

Let $K = ℚ (\sqrt{d})$ for some square-free integer $d \in ℤ$ , $d \neq 1$ . If $K$ is a real quadratic field, which is to say that $d > 0$ , then $r_{1} (K) = 2$ and $r_{2} (K) = 0$ . Since the complex conjugate of a root of unity $ζ$ equals $ζ$ if and only if $ζ^{2} = 1$ , we have in this case that $μ (K) = {\pm 1}$ and

𝒪_{K}^{\times} ≅ℤ ∕ 2 ℤ \times ℤ .

If $K$ is an imaginary quadratic field, which is to say that $d < 0$ , then $r_{1} (K) = 0$ and $r_{2} (K) = 1$ . In this case, since $[K : ℚ] = 2$ , the field $K$ cannot contain any roots of unity aside from $4$ th and $6$ th roots of $1$ . But $ℚ (i) = ℚ (\sqrt{- 1})$ and $ℚ (ω) = ℚ (\sqrt{- 3})$ for $ω$ a primitive $3$ rd root of $1$ , so for $d < 0$ , we have $𝒪_{K}^{\times} ≅ℤ ∕ 2 ℤ$ unless $d = - 1$ or $d = - 3$ , in which case $𝒪_{K}^{\times} ≅ℤ ∕ 4 ℤ$ and $ℤ ∕ 6 ℤ$ , respectively.

Example 4.4.11. §

Let $K = ℚ (\sqrt{2})$ . In this case, we know that $𝒪_{K}$ is generated by $\pm 𝜖$ for some $𝜖 \in 𝒪_{K}^{\times}$ of infinite order. To find $a, b \in ℤ$ such that $𝜖 = a + b \sqrt{2}$ , we note that $N_{K ∕ ℚ} (a + b \sqrt{2}) = a^{2} - 2 b^{2} = \pm 1$ . Suppose we choose $𝜖$ with $a, b > 0$ , which can be done since

{\pm 𝜖, \pm 𝜖^{- 1}} = {\pm a \pm b \sqrt{2}} .

The $𝜖$ is known as the fundamental unit for $K$ .

If we consider $𝜖^{n} = a_{n} + b_{n} \sqrt{2}$ for $n > 1$ , it is not hard to check that $a_{n} > a$ and $b_{n} > b$ . Therefore, we must find a solution to $a^{2} - 2 b^{2} = \pm 1$ with $a > 0$ or $b > 0$ minimal (and therefore both), e.g., $a = b = 1$ . That is, $1 + \sqrt{2}$ is a fundamental unit for $K$ .

Example 4.4.12. §

Let $n \geq 3$ be a positive integer, and let $m$ be $n$ or $2 n$ according to whether $n$ is even or odd, respectively. We have

ℤ {[μ_{n}]}^{\times} ≅ ℤ^{\frac{φ (n)}{2} - 1} \times μ_{m} .