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Minkowski Theory and Number Field Class Groups

1 · Prerequisites

2 · Summary

The unscaled Minkowski embedding turns $O_{K}$ and its nonzero fractional ideals into full Euclidean lattices of covolume $2^{- r_{2}} \sqrt{∣ d_{K} ∣} N a$ in $R^{r_{1} + 2 r_{2}}$ . No $\sqrt{2}$ scaling is used, so conjugate pairs contribute the factor $2^{- r_{2}}$ to covolumes and $4 / π$ to the arithmetic constant; mixing this convention with the scaled one is a false shortcut.

The geometry of numbers is proved in full on this page: the fundamental parallelotope and its bounded intersections, Blichfeldt's principle, the Minkowski convex-body theorem in strict and equality form, and the two-sided second theorem for the successive minima via an adapted flag and a triangular volume deformation.

The arithmetic conclusions are finiteness results. Every ideal class contains an integral ideal of norm at most $M_{K} = (4 / π)^{r_{2}} (n! / n^{n}) \sqrt{∣ d_{K} ∣}$ ; only finitely many integral ideals have bounded norm; the class group is finite and generated by the prime ideals of norm at most $M_{K}$ . The same bound gives $∣ d_{K} ∣ > 1$ for every degree greater than one and hence, through the ramification-discriminant criterion, a ramified rational prime. Hermite-Minkowski finiteness closes the page with the bounded-conjugate polynomial lemma and a bounded primitive integral element.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Literature-sourcedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Unscaled Minkowski embedding

Definition

Let $K$ be a number field (Number field) of degree $n = [K : Q]$ and signature $(r_{1}, r_{2})$ , so that $n = r_{1} + 2 r_{2}$ and the field has $r_{1}$ real embeddings $σ_{1}, \dots, σ_{r_{1}} : K \to R$ and $r_{2}$ complex-conjugate pairs of nonreal embeddings from which one representative $τ_{1}, \dots, τ_{r_{2}}$ is chosen; the notation and the count $r_{1} + 2 r_{2} = n$ are those of Archimedean embeddings and signature. The unscaled Minkowski embedding of $K$ is the injective map

$σ : K ⟶ R^{r_{1}} \times C^{r_{2}}, σ (x) = (σ_{1} (x), \dots, σ_{r_{1}} (x), τ_{1} (x), \dots, τ_{r_{2}} (x)),$

composed with the identification $C^{r_{2}} ≅ R^{2 r_{2}}$ that sends $z$ to the pair $(Re z, Im z)$ of real coordinates. This exhibits $σ$ as a map into $R^{r_{1}} \times R^{2 r_{2}} = R^{n}$ .

No factor $\sqrt{2}$ is inserted in the complex coordinates: each complex coordinate contributes the two coordinates $Re τ_{j} (x)$ and $Im τ_{j} (x)$ with equal weight. Thus the Euclidean norm of a complex block $(z_{1}, \dots, z_{r_{2}})$ is $\sqrt{\sum_{j} ∣ z_{j} ∣^{2}}$ , and the complex block of $σ (x)$ has norm $\sqrt{\sum_{j} ∣ τ_{j} (x) ∣^{2}}$ . All volumes, covolumes and determinants on this item use this unscaled convention.

Remarks

Injectivity and linearity. Since $n = r_{1} + 2 r_{2} \geq 1$ , at least one of the listed embeddings exists. If $σ (x) = 0$ , every listed embedding sends $x$ to zero; any one of them is injective, so $x = 0$ . This also covers the case $r_{2} = 0$ , when there are no complex representatives. Each embedding is $Q$ -linear, as is the real-coordinate identification, so $σ$ is $Q$ -linear. Injectivity alone does not imply that the images of a $Q$ -basis are linearly independent over $R$ ; that fact follows from the determinant calculation below.

Relation to the all-complex embedding determinant. Let $α_{1}, \dots, α_{n}$ be a $Q$ -basis of $K$ and let $M = (ψ_{i} (α_{j}))$ be the $n \times n$ matrix of all complex embeddings. By Embedding determinant formula, $\det M \neq 0$ and $\det (M)^{2} = disc (α_{1}, \dots, α_{n})$ . Reorder its rows so that each complex-conjugate pair is adjacent. For a pair $τ, \overset{ˉ}{τ}$ , the old rows are obtained from the real rows $(Re τ, Im τ)$ by the transition matrix whose determinant is $- 2 i$ , of modulus $2$ . Replacing all such pairs by their real and imaginary rows therefore gives the real $n \times n$ matrix $A$ with columns $σ (α_{j})$ and

$∣ \det A ∣ = 2^{- r_{2}} ∣ \det M ∣ = 2^{- r_{2}} \sqrt{∣ disc (α_{1}, \dots, α_{n}) ∣} .$

In particular, the images of every $Q$ -basis form a real basis of $R^{n}$ . This factor $2^{- r_{2}}$ is responsible for the covolume formula $covol (σ (a)) = 2^{- r_{2}} \sqrt{∣ d_{K} ∣} N a$ proved later in this development.

DefinitionDefinition: Literature-sourcedProof: Literature-sourcedjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Full Euclidean lattice and covolume

Definition

Fix $n \geq 1$ . A full lattice in $R^{n}$ is a subgroup $Λ \subseteq R^{n}$ of the form

$Λ = Z b_{1} \oplus \dots \oplus Z b_{n} = {\sum_{i = 1}^{n} m_{i} b_{i} : m_{i} \in Z},$

where $b_{1}, \dots, b_{n} \in R^{n}$ are linearly independent over $R$ , that is, they form a real basis of $R^{n}$ . Write $B = (b_{1} \dots b_{n})$ for the $n \times n$ matrix whose $i$ -th column is $b_{i}$ ; determinants of square matrices are as in For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix. The covolume of $Λ$ is

$covol (Λ) : = ∣ \det B ∣ .$

The value is positive: linear independence of the columns makes $B$ invertible, so $\det B \neq 0$ .

Remarks

Basis independence. Suppose $c_{1}, \dots, c_{n}$ is a second integer basis of $Λ$ , with matrix $C = (c_{1} \dots c_{n})$ . Each $c_{j}$ is an integer combination of the $b_{i}$ and each $b_{i}$ is an integer combination of the $c_{j}$ , so there are matrices $A, A^{'} \in M_{n} (Z)$ with $C = B A$ and $B = C A^{'}$ . Substituting gives $B = B A A^{'}$ , hence $A A^{'} = I_{n}$ because $B$ is invertible, and symmetrically $A^{'} A = I_{n}$ . Taking determinants, $(\det A) (\det A^{'}) = 1$ with both factors integers, so $\det A = \pm 1$ and therefore $∣ \det C ∣ = ∣ \det B ∣ ∣ \det A ∣ = ∣ \det B ∣$ . Thus the covolume does not depend on the chosen basis, and the definition above is unambiguous.

Discrete subgroups. A subgroup $Λ \subseteq R^{n}$ is a full lattice in the sense above exactly when it is discrete in the Euclidean topology and spans $R^{n}$ ; this equivalence is Milne's Lemma 4.14 together with the identification of full lattices with discrete spanning subgroups (Milne, Ch. 4, pp.73-75). The half-open fundamental parallelotope $P = {\sum_{i} t_{i} b_{i} : 0 \leq t_{i} < 1}$ of a full lattice tiles $R^{n}$ by $Λ$ -translates with volume $covol (Λ)$ , and bounded sets meet $Λ$ in finitely many points; both facts are proved in this batch and used below.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-10-02Open item page →

Fundamental parallelotope and finite bounded intersections

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ and let $Λ = Z b_{1} \oplus \dots \oplus Z b_{n} \subseteq R^{n}$ be a full lattice with covolume $covol (Λ) = ∣ \det B ∣$ , $B = (b_{1} \dots b_{n})$ (Full Euclidean lattice and covolume). Let

$P = {\sum_{i = 1}^{n} t_{i} b_{i} : 0 < t_{i} \leq 1}$

be its half-open fundamental parallelotope. Then:

every $x \in R^{n}$ has a unique representation $x = λ + p$ with $λ \in Λ$ and $p \in P$ ;
$P$ is Lebesgue measurable and $λ_{n} (P) = covol (Λ)$ ;
every bounded subset $S \subseteq R^{n}$ meets $Λ$ in finitely many points: $S \cap Λ$ is finite.

The convention $0 < t_{i} \leq 1$ is the published one for half-open boxes, $(a, b]$ -faces (Half-open boxes in $R^{n}$ and their volume); it is a translate of the equally common $0 \leq t_{i} < 1$ parallelotope and carries the same volume.

Facts & Assumptions

Given: The Axiom of Choice, an integer $n \geq 1$ , the full lattice $Λ = Z b_{1} \oplus \dots \oplus Z b_{n}$ with matrix $B = (b_{1} \dots b_{n})$ , and its half-open fundamental parallelotope $P$ of the statement.

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), which is the choice hypothesis of [F3] and [F4], invoked in step 2.2; no further choice is used.

[F1]

The vectors $b_{1}, \dots, b_{n}$ are linearly independent over $R$ and form a real basis of $R^{n}$ , and $covol (Λ) = ∣ \det B ∣$ (Full Euclidean lattice and covolume).

[F2]

A square real matrix $B$ is invertible if and only if $\det B \neq 0$ (A finite square real matrix is invertible if and only if its determinant is nonzero).

[F3]

Invertible linear maps and Lebesgue measure: if $T : R^{n} \to R^{n}$ is linear with $\det T \neq 0$ , then $T [E]$ is Lebesgue measurable for every Lebesgue measurable $E$ and $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F4]

For real endpoints $a_{i} \leq b_{i}$ , the half-open box $B (a, b) = \prod_{i < n} (a_{i}, b_{i}]$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ ; in particular the half-open unit cube $(0, 1]^{n}$ has $λ_{n} ((0, 1]^{n}) = 1$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F5]

Integer part: for every real $y$ there is exactly one integer $k$ with $k \leq y < k + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[F6]

For every linear map $L : R^{m} \to R^{n}$ there is a real $K \geq 0$ with $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for every $h$ (Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ ).

[F7]

A subset $S$ of a metric space is bounded exactly when $S = \emptyset$ or $S \subseteq B (x_{0}, r)$ for some point $x_{0}$ and real $r > 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space); in $R^{n}$ with the Euclidean metric the triangle inequality gives $∥ s ∥ \leq ∥ s - x_{0} ∥ + ∥ x_{0} ∥$ for $s \in B (x_{0}, r)$ .

Proof

1.1F1given

By [F1] the vectors $b_{1}, \dots, b_{n}$ form a real basis of $R^{n}$ ; therefore every $x \in R^{n}$ has a unique coefficient vector $y = (y_{1}, \dots, y_{n}) \in R^{n}$ with $x = \sum_{i} y_{i} b_{i}$ .

1.2F5algebra

For every real $y$ there is exactly one pair $(m, t) \in Z \times (0, 1]$ with $y = m + t$ : apply [F5] to $- y$ to get the unique integer $k$ with $k \leq - y < k + 1$ , and put $m : = - k - 1$ , $t : = y - m$ ; then $m < y \leq m + 1$ and hence $t \in (0, 1]$ . Conversely if $m + t = m^{'} + t^{'}$ with $t, t^{'} \in (0, 1]$ , then $m - m^{'} = t^{'} - t$ has absolute value $< 1$ and is an integer, hence $m = m^{'}$ and $t = t^{'}$ .

1.3F7given

Let $S \subseteq R^{n}$ be bounded and suppose first $S \neq \emptyset$ . By [F7] there are $x_{0} \in R^{n}$ and $r > 0$ with $S \subseteq B (x_{0}, r)$ , so every $s \in S$ satisfies $∥ s ∥ \leq ∥ s - x_{0} ∥ + ∥ x_{0} ∥ \leq r + ∥ x_{0} ∥ = : M$ .

2.1F1step 1.1step 1.2

(Tiling.) Let $x \in R^{n}$ have coefficient vector $y$ as in step 1.1 and write $y_{i} = m_{i} + t_{i}$ as in step 1.2. Put $λ = \sum_{i} m_{i} b_{i} \in Λ$ and $p = \sum_{i} t_{i} b_{i} \in P$ ; then $x = λ + p$ . For uniqueness, suppose $λ + p = λ^{'} + p^{'}$ with $λ = \sum_{i} m_{i} b_{i}$ , $λ^{'} = \sum_{i} m_{i}^{'} b_{i} \in Λ$ and $p = \sum_{i} t_{i} b_{i}$ , $p^{'} = \sum_{i} t_{i}^{'} b_{i} \in P$ . Then $\sum_{i} ((m_{i} - m_{i}^{'}) + (t_{i} - t_{i}^{'})) b_{i} = 0$ , and linear independence of the $b_{i}$ forces $(m_{i} - m_{i}^{'}) + (t_{i} - t_{i}^{'}) = 0$ for every $i$ . Here $m_{i} - m_{i}^{'}$ is an integer and $t_{i}^{'} - t_{i} \in (- 1, 1)$ , so $m_{i} - m_{i}^{'} = t_{i}^{'} - t_{i} \in Z \cap (- 1, 1) = {0}$ ; thus $m_{i} = m_{i}^{'}$ and $t_{i} = t_{i}^{'}$ for all $i$ , that is $λ = λ^{'}$ and $p = p^{'}$ .

2.2F1F2F3F4A1step 1.1

Let $T : R^{n} \to R^{n}$ be the linear map $T (t) = \sum_{i} t_{i} b_{i}$ , with matrix $B$ . By step 1.1 the map $T$ is a bijection, so the square matrix $B$ is invertible and [F2] gives $\det B \neq 0$ . Since $T [(0, 1]^{n}] = P$ and $(0, 1]^{n}$ is Lebesgue measurable of measure $1$ by [F4], [F3] and [F1] give $λ_{n} (P) = ∣ \det B ∣ λ_{n} ((0, 1]^{n}) = covol (Λ)$ , with the Countable Choice hypotheses of [F3] and [F4] supplied by [A1].

2.3F6step 1.3

For each $i$ the $i$ -th coordinate functional $f_{i} (z) : = (B^{- 1} z)_{i}$ is linear, so by [F6] there is $K_{i} \geq 0$ with $∣ f_{i} (z) ∣ \leq K_{i} ∥ z ∥_{2}$ for every $z$ . If $λ = \sum_{i} m_{i} b_{i} \in S$ , then $B^{- 1} λ = (m_{1}, \dots, m_{n})$ , so $f_{i} (λ) = m_{i}$ and step 1.3 gives $∣ m_{i} ∣ \leq K_{i} M = : C_{i}$ .

3.1F1step 2.3

Every integer $m_{i}$ with $∣ m_{i} ∣ \leq C_{i}$ satisfies $- ⌈ C_{i} ⌉ \leq m_{i} \leq ⌈ C_{i} ⌉$ ; the set ${m \in Z : ∣ m ∣ \leq C_{i}}$ is therefore a subset of the finite set ${- ⌈ C_{i} ⌉, \dots, ⌈ C_{i} ⌉}$ and is finite. Hence $S \cap Λ$ is contained in the image under $(m_{1}, \dots, m_{n}) \mapsto \sum_{i} m_{i} b_{i}$ of the finite set $\prod_{i = 1}^{n} {m \in Z : ∣ m ∣ \leq C_{i}}$ , so $S \cap Λ$ is finite; for $S = \emptyset$ it is empty.

4.1step 2.1step 2.2step 3.1∎

Step 2.1 proves the unique tiling, step 2.2 the volume $λ_{n} (P) = covol (Λ)$ , and step 3.1 the finiteness of $S \cap Λ$ for bounded $S$ ; these are the three claims of the statement.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-10-02Open item page →

Blichfeldt lattice-point principle

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ and let $Λ \subseteq R^{n}$ be a full lattice with $covol (Λ) > 0$ (Full Euclidean lattice and covolume). If $S \subseteq R^{n}$ is Lebesgue measurable with $λ_{n} (S) > covol (Λ)$ , then there are distinct points $x, y \in S$ with $x - y \in Λ$ .

Facts & Assumptions

Given: The Axiom of Choice, a full lattice $Λ = Z b_{1} \oplus \dots \oplus Z b_{n}$ in $R^{n}$ with covolume $covol (Λ) > 0$ , and a Lebesgue measurable set $S \subseteq R^{n}$ with $λ_{n} (S) > covol (Λ)$ .

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), the choice hypothesis of the complete-measure fact [F2], invoked in step 2.1; the translation-invariance fact [F3], the defining properties of a measure [F4] and the countability facts [F5] use no choice principle, and no further choice is used.

[F1]

The half-open fundamental parallelotope $P = {\sum_{i} t_{i} b_{i} : 0 < t_{i} \leq 1}$ tiles $R^{n}$ uniquely by $Λ$ -translates, $P$ is Lebesgue measurable with $λ_{n} (P) = covol (Λ)$ , and every bounded subset of $R^{n}$ meets $Λ$ in finitely many points (Fundamental parallelotope and finite bounded intersections).

[F2]

Assuming countable choice, $L (R^{n})$ is a sigma-algebra and $λ_{n}$ is a measure on it (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F3]

Translations preserve measurability and measure: $E$ is Lebesgue measurable if and only if $E + h$ is, and then $λ_{n} (E + h) = λ_{n} (E)$ (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F4]

A measure is countably additive: for pairwise disjoint measurable sets $E_{k}$ one has $μ (⋃_{k} E_{k}) = \sum_{k} μ (E_{k})$ , extended nonnegative sums included (Measures on sigma-algebras).

[F5]

$Z$ is at most countable: the quotient map $N \times N \to Z$ of The integers as equivalence classes of pairs of naturals is surjective, $N \times N$ is at most countable ( $N \times N \approx N$ , A product of two at most countable sets is at most countable), and an at most countable set that is a surjective image of an at most countable set is at most countable (A nonempty set is at most countable iff it is a surjective image of $N$ ). Hence $Z^{n}$ is at most countable, and so is $Λ$ , being a surjective image of $Z^{n}$ under $(m_{1}, \dots, m_{n}) \mapsto \sum_{i} m_{i} b_{i}$ (Finite, countably infinite, countable, uncountable).

Proof

1.1assume-contra

Assume for contradiction that there are no distinct $x, y \in S$ with $x - y \in Λ$ .

1.2F5A1

$Λ$ is at most countable by [F5], and it is infinite because $b_{1} \neq 0$ gives the distinct multiples $k b_{1}$ ; being at most countable and infinite, it is countably infinite, so fix a bijection $λ : N \to Λ$ , $k \mapsto λ_{k}$ (Finite, countably infinite, countable, uncountable). By [A1] the Axiom of Countable Choice holds; it discharges the choice hypothesis of the complete-measure fact [F2] applied below.

2.1F1F2F3step 1.2

For each $k$ put $S_{k} : = S \cap (P + λ_{k})$ and $T_{k} : = S_{k} - λ_{k}$ . Each $S_{k}$ is measurable: $S$ is measurable by hypothesis, the translate $P + λ_{k}$ is measurable by [F3] applied to the measurable tile $P$ of [F1], and the intersection is measurable because [F2] makes $L (R^{n})$ a sigma-algebra, its Countable Choice hypothesis having been discharged in step 1.2.

2.2F1F4step 1.2

The $S_{k}$ are pairwise disjoint with union $S$ , since the translates $P + λ$ tile $R^{n}$ by [F1]; hence $λ_{n} (S) = \sum_{k} λ_{n} (S_{k})$ by countable additivity [F4].

3.1F1F3step 2.1

By [F3] applied to the translation by $- λ_{k}$ , $T_{k}$ is measurable with $λ_{n} (T_{k}) = λ_{n} (S_{k})$ ; and $T_{k} \subseteq P$ , since $P + λ_{k}$ translated by $- λ_{k}$ is $P$ .

4.1step 1.1step 3.1

The sets $T_{k}$ are pairwise disjoint: if $z \in T_{j} \cap T_{k}$ with $j \neq k$ , then $z = x - λ_{j} = y - λ_{k}$ with $x \in S_{j} \subseteq S$ and $y \in S_{k} \subseteq S$ , so $x - y = λ_{j} - λ_{k} \in Λ$ while $x \neq y$ because $λ_{j} \neq λ_{k}$ ; this contradicts step 1.1.

5.1F4step 2.2step 3.1step 4.1

By countable additivity [F4] applied to the pairwise disjoint measurable sets $T_{k}$ , $λ_{n} (⋃_{k} T_{k}) = \sum_{k} λ_{n} (T_{k}) = \sum_{k} λ_{n} (S_{k}) = λ_{n} (S)$ , using steps 2.2, 3.1 and 4.1.

6.1F1F4step 5.1

Since $⋃_{k} T_{k} \subseteq P$ , monotonicity of a measure (additivity [F4] applied to $P = (⋃_{k} T_{k}) \cup (P ∖ ⋃_{k} T_{k})$ ) gives $λ_{n} (S) = λ_{n} (⋃_{k} T_{k}) \leq λ_{n} (P) = covol (Λ)$ .

7.1step 1.1step 6.1discharge-contradiction∎

Step 6.1 contradicts the hypothesis $λ_{n} (S) > covol (Λ)$ ; therefore the assumption of step 1.1 is false, and there exist distinct $x, y \in S$ with $x - y \in Λ$ .

Remarks

The proof works for unbounded $S$ and even for $λ_{n} (S) = + \infty$ : $S_{k} \subseteq P + λ_{k}$ and $T_{k} \subseteq P$ , so each piece has finite measure, but their measure sum may be infinite. Countable additivity permits extended nonnegative sums; under the no-pair assumption, the $T_{k}$ are disjoint in $P$ , which bounds that sum by $λ_{n} (P)$ and gives the contradiction. The hypothesis is strict: for $S = (0, 1]^{2}$ and $Λ = Z^{2}$ one has $λ_{2} (S) = covol (Λ) = 1$ and no two distinct points of $S$ differ by a lattice vector.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Minkowski convex-body theorem, strict form

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , let $Λ \subseteq R^{n}$ be a full lattice with $covol (Λ) > 0$ (Full Euclidean lattice and covolume), and let $C \subseteq R^{n}$ be Lebesgue measurable, convex and centrally symmetric (A convex subset of $R^{m}$ contains every line segment between two of its points). If

$λ_{n} (C) > 2^{n} covol (Λ),$

then $C$ contains a nonzero point of $Λ$ .

Facts & Assumptions

Given: The Axiom of Choice, a full lattice $Λ$ with $covol (Λ) > 0$ , and a Lebesgue measurable convex centrally symmetric set $C$ with $λ_{n} (C) > 2^{n} covol (Λ)$ .

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), the choice hypothesis of the scaling fact [F2], invoked in step 1.1; Blichfeldt's principle [F1] is applied under the Axiom of Choice assumed in the statement, and no further choice is used.

[F1]

Blichfeldt's principle: for a Lebesgue measurable $S \subseteq R^{n}$ with $λ_{n} (S) > covol (Λ)$ there are distinct $x, y \in S$ with $x - y \in Λ$ (Blichfeldt lattice-point principle).

[F2]

For a nonzero real $c$ , a set $E$ is Lebesgue measurable if and only if $c E$ is, and then $λ_{n} (c E) = ∣ c ∣^{n} λ_{n} (E)$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

[F3]

$C$ convex means $(1 - t) x + t y \in C$ for all $x, y \in C$ and $t \in [0, 1]$ ; central symmetry means $- C = C$ , so $- y \in C$ whenever $y \in C$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

Proof

1.1F2A1given

Put $C^{'} : = \frac{1}{2} C = {x / 2 : x \in C}$ . By [F2] with $c = 1 / 2$ , $C^{'}$ is Lebesgue measurable with $λ_{n} (C^{'}) = 2^{- n} λ_{n} (C) > covol (Λ)$ , the Countable Choice hypothesis of [F2] being supplied by [A1].

1.2F3algebra

$C^{'}$ is convex and centrally symmetric: for $u, v \in C^{'}$ write $u = x / 2$ , $v = y / 2$ with $x, y \in C$ ; then $(1 - t) u + t v = ((1 - t) x + t y) / 2 \in C^{'}$ by convexity of $C$ , and $- u = (- x) / 2 \in C^{'}$ by symmetry of $C$ .

2.1F1step 1.1

By [F1] applied to the measurable set $C^{'}$ of step 1.1 there are distinct $u, v \in C^{'}$ with $u - v \in Λ$ .

3.1F3step 2.1

The difference $u - v$ is nonzero because $u \neq v$ , and it lies in $C$ : $2 u \in C$ and $2 v \in C$ by definition of $C^{'}$ , so $- 2 v \in C$ by central symmetry, and convexity of $C$ gives $u - v = \frac{1}{2} (2 u) + \frac{1}{2} (- 2 v) \in C$ .

4.1step 2.1step 3.1∎

Thus $u - v$ is a nonzero point of $Λ$ lying in $C$ , as required.

Remarks

The factor $2^{n}$ is optimal for centrally symmetric convex bodies: for the open cube $C = (- 1, 1)^{n}$ and $Λ = Z^{n}$ one has $λ_{n} (C) = 2^{n} = 2^{n} covol (Λ)$ while $C \cap Z^{n} = {0}$ , so the strict inequality cannot be weakened to $\geq$ . The equality case for compact bodies is treated in the next item, where the strict form is applied to the dilates $(1 + 1 / m) C$ .

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Minkowski convex-body theorem at equality

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , let $Λ \subseteq R^{n}$ be a full lattice with $covol (Λ) > 0$ (Full Euclidean lattice and covolume), and let $C \subseteq R^{n}$ be compact, convex and centrally symmetric (A convex subset of $R^{m}$ contains every line segment between two of its points). If

$λ_{n} (C) \geq 2^{n} covol (Λ),$

then $C$ contains a nonzero point of $Λ$ .

Facts & Assumptions

Given: The Axiom of Choice, a full lattice $Λ$ with $covol (Λ) > 0$ , and a compact convex centrally symmetric $C$ with $λ_{n} (C) \geq 2^{n} covol (Λ)$ .

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), which licenses the countable selection of one nonzero lattice point $v_{m}$ from each of the sets $S_{m} \cap Λ$ in step 2.1; the strict theorem [F1] and the tiling lemma [F3] are applied under the Axiom of Choice assumed in the statement, and no further choice is used.

[F1]

Strict Minkowski: if a Lebesgue measurable convex centrally symmetric set $S$ satisfies $λ_{n} (S) > 2^{n} covol (Λ)$ , then $S$ contains a nonzero point of $Λ$ (Minkowski convex-body theorem, strict form).

[F2]

For nonzero real $t$ , $λ_{n} (t S) = ∣ t ∣^{n} λ_{n} (S)$ for Lebesgue measurable $S$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

[F3]

Every bounded subset of $R^{n}$ meets $Λ$ in finitely many points (Fundamental parallelotope and finite bounded intersections).

[F4]

$C$ convex: $(1 - t) x + t y \in C$ for $x, y \in C$ , $t \in [0, 1]$ ; $C$ compact, hence closed; central symmetry gives $- C = C$ , so $0 \in C$ since $C$ is nonempty; and for $c \in C$ the convexity relation $c / 2 = \frac{1}{2} c + \frac{1}{2} \cdot 0$ gives $c / 2 \in C$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

Proof

1.1F2F4given

For every $m \geq 1$ the dilate $S_{m} : = (1 + 1 / m) C$ is compact, convex and centrally symmetric, and by [F2] $λ_{n} (S_{m}) = (1 + 1 / m)^{n} λ_{n} (C) \geq (1 + 1 / m)^{n} 2^{n} covol (Λ) > 2^{n} covol (Λ)$ because $covol (Λ) > 0$ and $(1 + 1 / m)^{n} > 1$ .

2.1A1F1step 1.1

By [F1] each $S_{m}$ contains a nonzero lattice point; using [A1] choose one, say $0 \neq v_{m} \in S_{m} \cap Λ$ , for every $m \geq 1$ .

3.1F3F4step 2.1

Since $1 + 1 / m \leq 2$ for $m \geq 1$ and $0 \in C$ , convexity of $C$ gives $S_{m} \subseteq 2 C$ , so every $v_{m}$ lies in the bounded set $2 C$ ; by [F3] the set $2 C \cap Λ$ is finite, so some $v \in 2 C \cap Λ$ equals $v_{m}$ for infinitely many $m$ .

4.1F4step 3.1

Fix such an infinite set of indices $m$ . For each of them $v \in (1 + 1 / m) C$ , hence $v / (1 + 1 / m) \in C$ ; as $m \to \infty$ through those indices $v / (1 + 1 / m) \to v$ , and $C$ is closed by [F4], so $v \in C$ .

5.1step 4.1∎

The point $v$ is nonzero by step 2.1 and lies in $C \cap Λ$ , so $C$ contains a nonzero lattice point.

Remarks

Compactness is used twice: it makes $2 C \cap Λ$ finite and it makes $C$ closed, so that the limit of the points $v / (1 + 1 / m)$ stays in $C$ . For non-closed bodies the conclusion can fail: the open cube $(- 1, 1)^{n}$ has $λ_{n} = 2^{n} covol (Z^{n})$ and meets $Z^{n}$ only in the origin.

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Successive minima of a convex body

Definition

Let $Λ \subseteq R^{n}$ be a full lattice (Full Euclidean lattice and covolume) and let $C \subseteq R^{n}$ be centrally symmetric (that is, $- C = C$ ), convex in the sense of A convex subset of $R^{m}$ contains every line segment between two of its points, compact, and with nonempty interior. For a real $t > 0$ write

$t C : = {t x : x \in C} .$

For $1 \leq i \leq n$ the $i$ -th successive minimum of $C$ with respect to $Λ$ is

$λ_{i} (C, Λ) : = \inf {t > 0 : \dim_{R} {span}_{R} (t C \cap Λ) \geq i},$

the infimum ranging over the nonempty set of real numbers $t > 0$ for which the linear span of the finite set $t C \cap Λ$ has dimension at least $i$ . When no such $t$ exists the infimum is $+ \infty$ ; for the bodies considered here it is finite, as recorded in the remarks below. We write $λ_{i}$ for $λ_{i} (C, Λ)$ when $C$ and $Λ$ are fixed in context, and we write $λ_{0} : = 0$ by convention.

The scaling convention is that the body is enlarged and the lattice is held fixed: $t C \cap Λ$ is the set of lattice points lying in the $t$ -dilate of $C$ . Equivalently, because $C = - C$ , one may think of the shortest vectors of $Λ$ in the norm whose unit ball is $C$ .

Remarks

The sets involved are finite. Each $t C$ is bounded because $C$ is compact, and a bounded subset of $R^{n}$ meets a full lattice in finitely many points. Indeed, write $Λ = B Z^{n}$ using a basis matrix $B$ . If $S \subseteq R^{n}$ is bounded, choose $M > 0$ with $∣ x ∣ \leq M$ for all $x \in S$ . The inverse linear map is bounded by Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ , say $∣ B^{- 1} x ∣ \leq K ∣ x ∣$ . Thus if $B m \in S$ for $m \in Z^{n}$ , then $∣ m ∣ \leq K M$ , so every integer coordinate of $m$ lies in the finite interval $[- K M, K M]$ ; only finitely many such integer vectors occur. Consequently $span (t C \cap Λ)$ is a finite-dimensional real subspace and the dimension in the definition is a genuine nonnegative integer, never an undecided quantity.

Monotonicity. If $0 < s < t$ then $s C \subseteq t C$ : for $x \in C$ one has $0 \in C$ and $(s / t) x + (1 - s / t) \cdot 0 = (s / t) x \in C$ by convexity, so $s x \in t C$ . Hence $s C \cap Λ \subseteq t C \cap Λ$ and the dimension function is nondecreasing in $t$ ; the sets inside the infimum are therefore upward-closed, and $0 < λ_{1} \leq λ_{2} \leq \dots \leq λ_{n}$ .

Finiteness and positivity. Central symmetry and nonempty interior put the origin in the interior: if $v$ is an interior point then so is $- v$ , and $0 = (v + (- v)) / 2$ is an interior point of $C$ by convexity. So $C$ contains a Euclidean ball $ρ B^{n}$ about the origin with $ρ > 0$ , and compactness of $C$ bounds it by some $R = \sup_{x \in C} ∣ x ∣ < \infty$ , with $R > 0$ . Write $Λ = Z b_{1} \oplus \dots \oplus Z b_{n}$ and let $B : R^{n} \to R^{n}$ be the linear isomorphism $B (m) = \sum_{i} m_{i} b_{i}$ . Its inverse is a Euclidean linear map, so the bounded- linear-map result Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ gives a constant $K \geq 1$ such that $∣ B^{- 1} v ∣ \leq K ∣ v ∣$ for every $v \in R^{n}$ . For every nonzero integer vector $m \in Z^{n}$ , $∣ m ∣ \geq 1$ , and therefore $∣ B m ∣ \geq 1 / K$ . If $B m \in t C$ , then $∣ B m ∣ \leq t R$ , so $t \geq 1 / (K R)$ and hence $λ_{1} \geq 1 / (K R) > 0$ . Conversely $λ_{n} < \infty$ : for each $i$ choose $t_{i} \geq ∣ b_{i} ∣ / ρ$ , so that $b_{i} / t_{i}$ has norm at most $ρ$ and therefore lies in $C$ ; for $t = \max_{i} t_{i}$ the set $t C \cap Λ$ contains $b_{1}, \dots, b_{n}$ and spans $R^{n}$ . The infima defining the $λ_{i}$ are thus positive and finite, and the next items prove that they are attained and control the lattice vectors at the attained levels.

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Attained successive minima and adapted flag

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $C \subseteq R^{n}$ be compact, convex, centrally symmetric and of nonempty interior (A convex subset of $R^{m}$ contains every line segment between two of its points), let $Λ \subseteq R^{n}$ be a full lattice with $covol (Λ) > 0$ (Full Euclidean lattice and covolume), and let $λ_{1} \leq \dots \leq λ_{n}$ be the successive minima of $C$ with respect to $Λ$ (Successive minima of a convex body). Then:

(attainment) for every $i$ the space $span (λ_{i} C \cap Λ)$ has dimension at least $i$ ;
(adapted basis) there are linearly independent vectors $a_{1}, \dots, a_{n} \in Λ$ with $a_{i} \in λ_{i} C$ for every $i$ and $span (λ_{i} C \cap Λ) = span {a_{j} : λ_{j} \leq λ_{i}}$ for $1 \leq i \leq n$ ;
(interior flag) for every $i$ , every lattice point in the interior of $λ_{i} C$ lies in $span {a_{j} : λ_{j} < λ_{i}}$ provided by clause 2, that is $span (int (λ_{i} C) \cap Λ) \subseteq span {a_{j} : λ_{j} < λ_{i}}$ .

Facts & Assumptions

Given: The Axiom of Choice, a compact convex centrally symmetric body $C \subseteq R^{n}$ with nonempty interior, a full lattice $Λ = Z b_{1} \oplus \dots \oplus Z b_{n}$ , and the successive minima $λ_{1} \leq \dots \leq λ_{n}$ of the definition.

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), used in step 1.1 to select one real $t_{m}$ from each of the countably many nonempty sets $E_{i} \cap [λ_{i}, λ_{i} + 1 / m]$ ; the Axiom of Choice assumed in the statement already meets the hypothesis of [F2], and every other selection below is a least index in a fixed finite enumeration, requiring no further choice.

[F1]

The successive minima are defined by $λ_{i} = \inf {t > 0 : \dim_{R} span (t C \cap Λ) \geq i}$ ; the span of the finite set $t C \cap Λ$ is a genuine finite-dimensional space; $s < t$ implies $s C \subseteq t C$ because $0 \in C$ and $C$ is convex, so $i \mapsto λ_{i}$ is nondecreasing; $0 < λ_{1} \leq \dots \leq λ_{n} < \infty$ ; and there is a real $t_{0} > 0$ with $\dim span (t_{0} C \cap Λ) = n$ (Successive minima of a convex body).

[F2]

Every bounded subset of $R^{n}$ meets the full lattice $Λ$ in finitely many points (Fundamental parallelotope and finite bounded intersections).

[F3]

$C$ is compact, hence closed; it is convex, so $(1 - θ) x + θ y \in C$ for all $x, y \in C$ and $0 \leq θ \leq 1$ ; it is centrally symmetric, $- C = C$ , and $0 \in C$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[F4]

Terminology of linear algebra: a finite family that spans a space and is linearly independent is a basis, and the span of a set consists of its finite linear combinations (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ); a linear space spanned by a finite set with $m$ elements has a basis of at most $m$ elements, obtained by discarding one element at a time that lies in the span of the others.

Proof

1.1F1A1

(Attainment.) Fix $i$ and let $t_{0}$ be as in [F1]. Since $λ_{i} \leq t_{0} < \infty$ and $λ_{i} = \inf E_{i}$ for $E_{i} = {t > 0 : \dim span (t C \cap Λ) \geq i}$ , [A1] lets us select for each $m \geq 1$ a real $t_{m} \in E_{i}$ with $λ_{i} \leq t_{m} < λ_{i} + 1 / m$ . Thus $t_{m} \to λ_{i}$ and $t_{m} < t_{0} + 1$ , because $λ_{i} \leq t_{0}$ and $1 / m \leq 1$ .

1.2F3algebra

For any $v \in int (λ_{i} C)$ , since $λ_{i} > 0$ , the point $c : = v / λ_{i}$ lies in $int (C)$ . The map $ε \mapsto v / (λ_{i} - ε)$ is continuous at $0$ with value $c$ ; because $C$ contains an open ball about $c$ , there is $ε_{0} \in (0, λ_{i})$ such that $v / (λ_{i} - ε) \in C$ for every $0 < ε < ε_{0}$ . Thus $v \in (λ_{i} - ε) C$ for every such $ε$ .

2.1F2step 1.1

By [F2] the sets $t_{m} C \cap Λ$ are contained in the finite set $F : = (t_{0} + 1) C \cap Λ$ , since $t_{m} < t_{0} + 1$ implies $t_{m} C \subseteq (t_{0} + 1) C$ by [F1]. The collection of subsets ${t_{m} C \cap Λ : m \geq 1}$ of $F$ is therefore finite, so some subset $S \subseteq F$ occurs for infinitely many $m$ ; fix such an infinite subsequence.

3.1step 2.1

Along that subsequence $t_{m} \to λ_{i}$ and $\dim span S = \dim span (t_{m} C \cap Λ) \geq i$ .

4.1F3step 3.1

Every $v \in S$ lies in $t_{m} C$ for all $m$ of the subsequence, that is $v / t_{m} \in C$ ; since $v / t_{m} \to v / λ_{i}$ and $C$ is closed by [F3], also $v / λ_{i} \in C$ , so $v \in λ_{i} C$ . Hence $S \subseteq λ_{i} C \cap Λ$ and $\dim span (λ_{i} C \cap Λ) \geq \dim span S \geq i$ , which is clause 1.

5.1F1step 4.1algebra

(Adapted basis.) Fix an enumeration $v_{1}, \dots, v_{N}$ of the finite set $F = (t_{0} + 1) C \cap Λ$ of step 2.1. Define $a_{1}$ to be the first $v_{k}$ with $v_{k} \in λ_{1} C \cap Λ$ and $v_{k} \neq 0$ , and for $k \geq 2$ define $a_{k}$ to be the first $v_{j}$ with $v_{j} \in λ_{k} C \cap Λ$ and $v_{j} \notin span {a_{1}, \dots, a_{k - 1}}$ . Each step succeeds: $\dim span (λ_{k} C \cap Λ) \geq k$ by step 4.1, while $span {a_{1}, \dots, a_{k - 1}} \subseteq span (λ_{k} C \cap Λ)$ because $a_{j} \in λ_{j} C \subseteq λ_{k} C$ for $j \leq k$ by [F1]. The recursion is a definition by the least index $j$ in a fixed finite list, so it selects nothing.

6.1F4step 5.1

By construction $a_{k} \in λ_{k} C \cap Λ$ and $a_{k} \notin span {a_{1}, \dots, a_{k - 1}}$ for every $k$ , so $a_{1}, \dots, a_{n}$ are linearly independent vectors of $Λ$ with $a_{k} \in λ_{k} C$ .

7.1F1step 6.1

(Flag equality.) Fix $i$ and put $r : = # {j : λ_{j} \leq λ_{i}}$ , so $r \geq i$ and $λ_{r} \leq λ_{i} < λ_{r + 1}$ when $r < n$ . For $j \leq r$ one has $λ_{j} \leq λ_{i}$ , hence $a_{j} \in λ_{j} C \subseteq λ_{i} C$ ; the $a_{1}, \dots, a_{r}$ are independent by step 6.1, so $\dim span (λ_{i} C \cap Λ) \geq r$ . If the dimension exceeded $r$ , then $\dim span (t C \cap Λ) \geq r + 1$ for $t = λ_{i}$ , whence $λ_{r + 1} \leq λ_{i}$ by definition of the infimum, contradicting $λ_{r + 1} > λ_{i}$ (and for $r = n$ the dimension is at most $n = r$ ). Hence the dimension equals $r$ and, since $span {a_{1}, \dots, a_{r}} \subseteq span (λ_{i} C \cap Λ)$ is a subspace of the same dimension $r$ , the two agree; that is $span (λ_{i} C \cap Λ) = span {a_{j} : λ_{j} \leq λ_{i}}$ , which is clause 2.

7.2F1step 6.1

(Interior flag.) Let $v \in Λ \cap int (λ_{i} C)$ and put $p : = # {j : λ_{j} < λ_{i}}$ , so $λ_{p} < λ_{i} = λ_{p + 1}$ when $p < n$ and $λ_{j} \leq λ_{p}$ for $j \leq p$ . If $p = n$ then $span {a_{j} : λ_{j} < λ_{i}} = span {a_{1}, \dots, a_{n}} = R^{n}$ contains $v$ ; so assume $p < n$ .

7.3F1F3step 1.2step 6.1

Let $ε_{0}$ be as in step 1.2. If $p = 0$ , choose any $0 < ε < ε_{0}$ . If $p > 0$ , choose $0 < ε < \min (ε_{0}, \min_{λ_{j} < λ_{i}} (λ_{i} - λ_{j}))$ ; the inner minimum is then over a nonempty finite set of positive numbers. In either case step 1.2 gives $v \in (λ_{i} - ε) C$ , and for each $j \leq p$ convexity of $C$ with $0 \in C$ gives $a_{j} / (λ_{i} - ε) = \frac{λ_{j}}{λ_{i} - ε} \cdot \frac{a_{j}}{λ_{j}} \in C$ , since $λ_{j} / (λ_{i} - ε) < 1$ .

8.1F1step 7.2step 1.2step 7.3

Suppose $v \notin span {a_{j} : λ_{j} < λ_{i}} = span {a_{1}, \dots, a_{p}}$ . By steps 1.2 and 7.3 the independent family $a_{1}, \dots, a_{p}$ together with $v$ lies in $(λ_{i} - ε) C \cap Λ$ , so $\dim span ((λ_{i} - ε) C \cap Λ) \geq p + 1$ ; by definition of the infimum $λ_{p + 1} \leq λ_{i} - ε < λ_{i} = λ_{p + 1}$ , a contradiction. Therefore $v \in span {a_{j} : λ_{j} < λ_{i}}$ , which is clause 3.

9.1step 4.1step 7.1step 8.1∎

Clauses 1, 2 and 3 are steps 4.1, 7.1 and 8.1 respectively.

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Triangular Borel maps scale Euclidean volume

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , let $a_{1}, \dots, a_{n} > 0$ be real numbers, and for $i = 1, \dots, n$ let $ψ_{i} : R^{n - i} \to R$ be a Borel function, where $R^{0}$ is a one-point space so that $ψ_{n}$ is a constant. Define the triangular map

$T : R^{n} ⟶ R^{n}, T (x)_{i} = a_{i} x_{i} + ψ_{i} (x_{i + 1}, \dots, x_{n}) (i = 1, \dots, n) .$

Then $T$ is a bijection, $T$ and $T^{- 1}$ are Borel maps, $T (E)$ is a Borel set for every Borel $E \subseteq R^{n}$ , and

$vol (T (E)) = (\prod_{i = 1}^{n} a_{i}) vol (E) .$

Equivalently, the inverse triangular map scales volume by $(\prod_{i} a_{i})^{- 1}$ , and both identities hold with $+ \infty$ allowed.

Facts & Assumptions

Given: The Axiom of Choice, an integer $n \geq 1$ , positive reals $a_{1}, \dots, a_{n}$ , Borel functions $ψ_{i}$ as in the statement, and a Borel set $E \subseteq R^{n}$ . Put $φ_{i} : = ψ_{i} / a_{i}$ and $S_{i} (x) : = x + φ_{i} (x_{i + 1}, \dots, x_{n}) e_{i}$ for $i = 1, \dots, n$ , and let $A (x)_{i} : = a_{i} x_{i}$ be the diagonal scaling.

[A1]

The Axiom of Choice implies the Axiom of Countable Choice (AC implies DC implies countable choice), so the countable-choice hypotheses of [F2] and [F4] are discharged for the whole argument; no further choice is used.

[F1]

Tonelli's theorem: for sigma-finite measure spaces and a product-measurable $f \geq 0$ , the integral over the product equals either iterated integral (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F2]

Under the identification $R^{m + n} = R^{m} \times R^{n}$ , the product measure $λ_{m} \times λ_{n}$ agrees with Lebesgue measure $λ_{m + n}$ on every Borel set (On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}).

[F3]

Lebesgue measure is translation invariant: $λ_{n} (E + h) = λ_{n} (E)$ for every Lebesgue measurable $E$ , and $E$ is measurable if and only if $E + h$ is (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F4]

For nonzero real $c$ , $λ_{n} (c E) = ∣ c ∣^{n} λ_{n} (E)$ for every Lebesgue measurable $E$ , and $E$ is measurable if and only if $c E$ is (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

Proof

1.1given

The factors satisfy $T = A \circ S$ for the shear $S$ given by $S (x)_{i} = x_{i} + φ_{i} (x_{i + 1}, \dots, x_{n})$ : indeed $A (S (x))_{i} = a_{i} S (x)_{i} = a_{i} x_{i} + ψ_{i} (x_{i + 1}, \dots, x_{n})$ .

1.2given

For points written as $(u, t, v) \in R^{i - 1} \times R \times R^{n - i}$ , the shear is $S_{i} (u, t, v) = (u, t + φ_{i} (v), v)$ , a bijection whose inverse $(u, t, v) \mapsto (u, t - φ_{i} (v), v)$ is Borel because $φ_{i}$ is Borel; hence $S_{i} (F)$ is Borel for every Borel $F$ , and for a Borel set $F$ the $t$ -section at fixed $(u, v)$ is the translate of the section of $F$ by $φ_{i} (v)$ .

1.3given

The sections of a Borel set $F \subseteq R^{i - 1} \times R \times R^{n - i}$ are Borel sets, because they are the preimages of $F$ under the continuous maps $t \mapsto (u, t, v)$ ; in particular they are Lebesgue measurable and [F3] applies to them.

1.4given

The diagonal scaling factors as $A = D_{1} \circ \dots \circ D_{n}$ with $D_{i}$ multiplying only the $i$ -th coordinate by $a_{i}$ , and each $D_{i}$ is an invertible linear bijection whose inverse is Borel, so $D_{i} (F)$ is Borel for every Borel $F$ .

2.1F1F2F3step 1.2step 1.3

For each $i$ and every Borel $F \subseteq R^{n}$ one has $vol (S_{i} (F)) = vol (F)$ : writing $f = 1_{S_{i} (F)}$ and using [F2] and [F1], the volume is the iterated integral $\int \int \int f (u, t, v) d u d t d v$ , whose $t$ -integrand at fixed $(u, v)$ equals $1_{F} (u, t - φ_{i} (v), v)$ , and its integral over $t$ equals the $t$ -length of the section of $F$ at $(u, v)$ by [F3] and step 1.3; integrating the unchanged section lengths over $(u, v)$ with [F1] returns $vol (F)$ .

2.2A1F4step 1.4

For each $i$ and every Borel $F \subseteq R^{n}$ , with $f = 1_{D_{i} (F)}$ the $t$ -integrand at fixed $(u, v)$ equals $1_{F} (u, t / a_{i}, v)$ , whose integral over $t$ is the length of the section of $F$ scaled by $a_{i}$ by [F4] in dimension one; the countable-choice hypothesis is supplied by [A1].

2.3step 1.2step 1.1

The shear $S = S_{n} \circ \dots \circ S_{1}$ : applying $S_{1}, \dots, S_{n}$ in that order changes the $i$ -th coordinate by $φ_{i} (x_{i + 1}, \dots, x_{n})$ while the higher coordinates are still the original ones, and $S$ is a Borel bijection with Borel inverse.

3.1F1F2step 2.2

Integrating the section identity of step 2.2 over the remaining coordinates with [F1] and [F2] gives $vol (D_{i} (F)) = a_{i} vol (F)$ for every Borel $F$ .

3.2step 2.1step 2.3

For every Borel $F$ one has $vol (S (F)) = vol (F)$ and $S (F)$ Borel, by applying step 2.1 to the factors of the composition in step 2.3.

4.1step 1.1step 1.4step 3.1step 3.2

Consequently $T = A \circ S$ satisfies $vol (T (E)) = vol (A (S (E))) = (\prod_{i} a_{i}) vol (S (E)) = (\prod_{i} a_{i}) vol (E)$ , and $T (E)$ is Borel, by steps 1.4, 3.1, 3.2 and the factorization of step 1.1.

5.1step 4.1algebra∎

The backward recursion $x_{i} = a_{i}^{- 1} (y_{i} - ψ_{i} (x_{i + 1}, \dots, x_{n}))$ , run from $i = n$ down to $i = 1$ , exhibits $T^{- 1}$ as a composition of Borel functions, so $T$ is a bijection with Borel inverse; applying the identity of step 4.1 to $T^{- 1}$ , which is again a triangular map with coefficients $a_{i}^{- 1}$ and Borel data from the same recursion, gives $vol (T^{- 1} (F)) = (\prod_{i} a_{i})^{- 1} vol (F)$ for Borel $F$ , and $T$ and $T^{- 1}$ being Borel in both directions makes each a Borel isomorphism.

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Successive-minima volume deformation and collision avoidance

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $C \subseteq R^{n}$ be compact, convex, centrally symmetric with nonempty interior, let $Λ$ be a full lattice, let $λ_{1} \leq \dots \leq λ_{n}$ be the successive minima of $C$ with respect to $Λ$ (Successive minima of a convex body), and let $a_{1}, \dots, a_{n} \in Λ$ be an adapted basis as in clause 2 of Attained successive minima and adapted flag. Put $U : = int (C)$ and $λ_{0} : = 0$ , and write $x = \sum_{i} x_{i} a_{i}$ for the coordinates of $x$ in the real basis $a_{1}, \dots, a_{n}$ . For $x \in U$ and $1 \leq j \leq n$ let

$F_{j} (x) : = {z \in C : z_{i} = x_{i} for i \geq j}$

be the slice of $C$ through $x$ parallel to $span (a_{1}, \dots, a_{j - 1})$ ; define $φ_{1} (x) : = x$ and, for $j \geq 2$ , let $φ_{j} (x)$ be the centroid of $F_{j} (x)$ , that is the mean vector of $F_{j} (x)$ with respect to $(j - 1)$ -dimensional Lebesgue measure on its affine hull. Define

$Φ (x) : = \sum_{j = 1}^{n} (λ_{j} - λ_{j - 1}) φ_{j} (x), x \in U .$

Then:

each $φ_{j} : U \to C$ is Borel, its $i$ -th coordinate equals $x_{i}$ for $i \geq j$ , and for $i < j$ its $i$ -th coordinate is a Borel function of $(x_{j}, \dots, x_{n})$ alone;
$Φ$ is Borel and odd, and in coordinates $Φ_{i} (x) = λ_{i} x_{i} + ψ_{i} (x_{i + 1}, \dots, x_{n})$ for Borel functions $ψ_{i} : R^{n - i} \to R$ ;
$vol (Φ (U)) = (\prod_{i = 1}^{n} λ_{i}) vol (C)$ ;
no two distinct points of $Φ (U)$ differ by an element of $2 Λ$ , and $Φ (U) \cap Λ = {0}$ .

Convexity of the image $Φ (U)$ is not asserted.

Facts & Assumptions

Given: The Axiom of Choice, a compact convex centrally symmetric body $C \subseteq R^{n}$ with nonempty interior, a full lattice $Λ$ , the successive minima $λ_{1} \leq \dots \leq λ_{n}$ , an adapted basis $a_{1}, \dots, a_{n} \in Λ$ , and $U = int (C)$ .

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), which discharges the Countable Choice hypotheses of the product-measure fact [F4] and of the volume-scaling fact [F6], invoked in steps 5.1 and 2.2 respectively; the triangular map fact [F3] is applied under the Axiom of Choice assumed in the statement, and the only arbitrary pick below is the single point $x_{0} \in U$ fixed in step 2.2, which requires no choice principle.

[F1]

$λ_{i} = \inf {t > 0 : \dim span (t C \cap Λ) \geq i}$ , $0 < λ_{1} \leq \dots \leq λ_{n} < \infty$ , $s C \subseteq t C$ for $0 < s < t$ , and $t C$ is compact and convex for every $t > 0$ (Successive minima of a convex body).

[F2]

$a_{1}, \dots, a_{n}$ are linearly independent vectors of $Λ$ with $a_{i} \in λ_{i} C$ , $span (λ_{i} C \cap Λ) = span {a_{j} : λ_{j} \leq λ_{i}}$ , and $span (int (λ_{i} C) \cap Λ) \subseteq span {a_{j} : λ_{j} < λ_{i}}$ (Attained successive minima and adapted flag).

[F3]

A triangular Borel map $T (x)_{i} = a_{i} x_{i} + ψ_{i} (x_{i + 1}, \dots, x_{n})$ with $a_{i} > 0$ and $ψ_{i}$ Borel is a Borel bijection of $R^{n}$ with Borel inverse, sends Borel sets to Borel sets, and $vol (T (E)) = (\prod_{i} a_{i}) vol (E)$ for every Borel $E$ (Triangular Borel maps scale Euclidean volume).

[F4]

Tonelli: for product-measurable $f \geq 0$ the partial integrals $y \mapsto \int f (x, y) d μ (x)$ are measurable, and iterated integrals agree (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product); the product measure agrees with Lebesgue measure on Borel sets (On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}). For a signed first-moment integrand, apply Tonelli separately to its positive and negative parts; for the compact set $C^{'}$ in step 5.1 both parts have bounded support and finite integrals.

[F5]

If $K$ is nonempty, compact and convex in an affine subspace $H$ of dimension $m \geq 1$ , and has positive $m$ -dimensional relative volume, its centroid with respect to relative Lebesgue measure on $H$ lies in $K$ . Indeed, choose an affine isometry $ψ : R^{m} \to H$ and put $K_{0} : = ψ^{- 1} (K)$ ; relative measure and centroids correspond to ordinary Lebesgue measure and centroids on $K_{0}$ . Its coordinate functions are integrable because $K_{0}$ is compact. If its centroid $c_{0}$ were outside the closed convex set $K_{0}$ , strict separation would give $u \neq 0$ and $b$ with $⟨ u, z ⟩ \leq b < ⟨ u, c_{0} ⟩$ for every $z \in K_{0}$ (A point outside a nonempty closed convex set is strictly separated from it, Integrable real and complex functions, and their integrals), while linearity of the integral gives $⟨ u, c_{0} ⟩ = λ_{m} (K_{0})^{- 1} \int_{K_{0}} ⟨ u, z ⟩ d z \leq b$ , a contradiction (The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F6]

Translations and nonzero dilations of Lebesgue measurable sets are measurable and satisfy $λ_{n} (E + h) = λ_{n} (E)$ , $λ_{n} (t E) = ∣ t ∣^{n} λ_{n} (E)$ ; invertible linear images satisfy $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it, A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F7]

For a measure $μ$ and measurable sets $A_{k} ↑ A$ one has $μ (A) = \lim_{k} μ (A_{k}) = \sup_{k} μ (A_{k})$ : this follows from countable additivity by writing $A = A_{1} ⊔ ⨆_{k} (A_{k + 1} ∖ A_{k})$ (Measures on sigma-algebras).

[F8]

$C$ is convex and closed, $- C = C$ , $0 \in C$ , and the interior $U$ of a convex set is convex; if $u \in U$ , $v \in C$ and $0 \leq s < 1$ , then $(1 - s) u + s v \in U$ , because $B (u, δ) \subseteq C$ for some $δ > 0$ and convexity gives $B ((1 - s) u + s v, (1 - s) δ) \subseteq C$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

Proof

1.1F2

The vectors $a_{1}, \dots, a_{n}$ form a real basis by [F2], so $x = \sum_{i} x_{i} a_{i}$ is a well-defined coordinate representation; $L (t) : = \sum_{i} t_{i} a_{i}$ is an invertible linear map.

1.2F1F8

$U = int (C)$ is open, convex, nonempty, bounded and symmetric ( $U = - U$ ), and $0 \in U$ ; also $int (t C) = t U$ for every $t > 0$ .

1.3given

(Collision avoidance.) Let $x, y \in U$ be distinct with $Φ (x) - Φ (y) \in 2 Λ$ , and put $μ : = (Φ (x) - Φ (y)) / 2 \in Λ$ . Let $k$ be the largest index with $x_{k} \neq y_{k}$ .

1.4given

(No nonzero lattice point in the image.) Let $ν \in Φ (U) \cap Λ$ , say $ν = Φ (x)$ with $x \in U$ , and suppose $x \neq 0$ ; let $k$ be the largest index with $x_{k} \neq 0$ .

2.1F1F8step 1.2

For $x \in U$ and $j \geq 2$ the slice $F_{j} (x)$ of the statement is compact and convex (an intersection of the convex set $C$ with the affine subspace ${z_{i} = x_{i}, i \geq j}$ ), it contains $x$ , and it has positive $(j - 1)$ -dimensional volume: $B (x, δ) \subseteq U$ for some $δ > 0$ , and the relative ball $B (x, δ) \cap (x + span (a_{1}, \dots, a_{j - 1}))$ lies in $F_{j} (x)$ . For $j = 1$ , $F_{1} (x) = {x}$ .

2.2F6F7F8A1step 1.2

(The interior has the same volume as the body.) Fix $x_{0} \in U$ and put $C_{k} : = (1 - 1 / k) C + (1 / k) x_{0}$ for $k \geq 2$ ; every scale $1 - 1 / k$ is positive. Each $C_{k}$ is compact and convex, and $C_{k} \subseteq U$ by [F8]. To see the sequence is increasing, for $c \in C$ set $c^{'} : = (1 - 1 / k^{2}) c + (1 / k^{2}) x_{0} \in C$ ; then $(1 - 1 / k) c + (1 / k) x_{0} = (1 - 1 / (k + 1)) c^{'} + (1 / (k + 1)) x_{0}$ , so $C_{k} \subseteq C_{k + 1}$ . For every $z \in U$ , the points $c_{k} : = z + (z - x_{0}) / (k - 1)$ tend to $z$ , so for all sufficiently large $k$ openness of $U$ gives $c_{k} \in U \subseteq C$ and $z = (1 - 1 / k) c_{k} + (1 / k) x_{0} \in C_{k}$ . Thus $U = ⋃_{k \geq 2} C_{k}$ , and [F7] and [F6] give $vol (U) = \lim_{k \to \infty} vol (C_{k}) = \lim_{k \to \infty} (1 - 1 / k)^{n} vol (C) = vol (C)$ ; the Countable Choice hypothesis of [F6] is supplied by [A1].

3.1F5step 2.1

By [F5] applied to the compact convex slice $F_{j} (x)$ , its centroid $φ_{j} (x)$ lies in $F_{j} (x) \subseteq C$ ; in particular $φ_{1} (x) = x \in U$ .

4.1step 2.1step 3.1algebra

For $z \in F_{j} (x)$ one has $z_{i} = x_{i}$ whenever $i \geq j$ ; hence the $i$ -th coordinate of $φ_{j} (x)$ equals $x_{i}$ for $i \geq j$ , and for $i < j$ the integral defining that coordinate is taken over the fibre of $C$ over $(x_{j}, \dots, x_{n})$ , so it depends only on those coordinates.

4.2F6F8step 3.1algebra

$φ_{j}$ is odd: $x \mapsto - x$ maps $U$ onto $U$ and $F_{j} (x)$ onto $F_{j} (- x) = - F_{j} (x)$ . On the affine hull of each slice this reflection is an affine isometry whose linear part has determinant of absolute value $1$ , so it preserves relative Lebesgue measure by [F6]. Changing variables in the centroid integral therefore gives $φ_{j} (- x) = - φ_{j} (x)$ . This uses the paired-slice identity $F_{j} (- x) = - F_{j} (x)$ and does not require an individual slice $F_{j} (x)$ to be symmetric.

5.1F4A1step 2.1step 4.1

(Borelness of the centroids.) In the coordinate model of step 1.1 write $C^{'} = L^{- 1} (C)$ , which is compact because $L$ is a homeomorphism, and write $U^{'} = L^{- 1} (U)$ . Fix $j \geq 2$ and split $t = (s, τ)$ with $s \in R^{j - 1}$ and $τ \in R^{n - j + 1}$ . The functions $V_{j} (τ) : = \int_{R^{j - 1}} 1_{C^{'}} (s, τ) d s$ and $M_{j, i} (τ) : = \int_{R^{j - 1}} s_{i} 1_{C^{'}} (s, τ) d s$ are Borel by Tonelli [F4], with the signed moment split into positive and negative parts; compactness of $C^{'}$ makes their supports bounded. Let $A_{j}$ be the matrix with columns $a_{1}, \dots, a_{j - 1}$ . The restriction of $L$ to the first $j - 1$ coordinates scales intrinsic fibre measure by the constant $J_{j} = \sqrt{\det (A_{j}^{T} A_{j})} > 0$ , independent of $τ$ . Thus this factor cancels in the centroid ratios, and for $i < j$ the $i$ -th coordinate of $φ_{j}$ in the $a$ -basis is $M_{j, i} (τ) / V_{j} (τ)$ . The projection of the open set $U^{'}$ to the $τ$ -coordinates is open, and $V_{j} (τ) > 0$ there by step 2.1. Hence these ratios are Borel on that open set; extending them by $0$ outside gives globally Borel functions of $τ$ .

5.2step 4.2

$Φ$ is odd, because each $φ_{j}$ is odd by step 4.2; in particular $Φ (0) = 0$ .

5.3step 4.1step 1.3

For $j > k$ the coordinates $(x_{j}, \dots, x_{n})$ and $(y_{j}, \dots, y_{n})$ agree, so $φ_{j} (x) = φ_{j} (y)$ by step 4.1; hence $Φ (x) - Φ (y) = \sum_{j \leq k} (λ_{j} - λ_{j - 1}) (φ_{j} (x) - φ_{j} (y))$ and $μ = \sum_{j \leq k} (λ_{j} - λ_{j - 1}) u_{j}$ with $u_{j} : = (φ_{j} (x) - φ_{j} (y)) / 2$ .

6.1step 4.1step 5.1algebra

Define $Φ (x) : = \sum_{j = 1}^{n} (λ_{j} - λ_{j - 1}) φ_{j} (x)$ for $x \in U$ . In coordinates, for fixed $i$ the coordinates of $φ_{j}$ with index $i < j$ depend only on $(x_{j}, \dots, x_{n})$ by step 4.1, while for $j \leq i$ the $i$ -th coordinate of $φ_{j}$ equals $x_{i}$ ; hence $Φ_{i} (x) = \sum_{j \leq i} (λ_{j} - λ_{j - 1}) x_{i} + \sum_{j > i} (λ_{j} - λ_{j - 1}) (φ_{j} (x))_{i} = λ_{i} x_{i} + ψ_{i} (x_{i + 1}, \dots, x_{n})$ , where $ψ_{i}$ is Borel by step 5.1.

6.2F1F8step 1.2step 5.3

Here $u_{1} = (x - y) / 2 \in U$ by step 1.2, and $u_{j} = (φ_{j} (x) + (- φ_{j} (y))) / 2 \in C$ for $j \geq 2$ by step 3.1 and [F8]. The weights $λ_{j} - λ_{j - 1}$ are nonnegative and sum to $λ_{k}$ , with positive first weight $λ_{1}$ on the interior point $u_{1}$ . Repeated application of [F8] (or induction on the finite number of terms) puts the convex combination $μ / λ_{k}$ in $U$ , that is $μ \in int (λ_{k} C)$ by step 1.2.

7.1F1F3F6step 1.1step 6.1

By steps 6.1 and 5.2 and [F3] applied with $a_{i} = λ_{i} > 0$ to the Borel set $E = U^{'}$ , the image $Φ (U)$ is Borel and the conjugate $Φ^{'} = L^{- 1} \circ Φ \circ L$ satisfies $vol (Φ^{'} (U^{'})) = (\prod_{i} λ_{i}) vol (U^{'})$ ; conjugating by the invertible linear map $L$ and using [F6] gives $vol (Φ (U)) = (\prod_{i} λ_{i}) vol (U)$ .

7.2step 4.1step 6.2

The $a_{k}$ -coordinate of $μ$ is $λ_{k} (x_{k} - y_{k}) / 2 \neq 0$ : for $j \leq k$ the $a_{k}$ -coordinate of $φ_{j} (x) - φ_{j} (y)$ is $x_{k} - y_{k}$ by step 4.1. Every vector in either $span {a_{1}, \dots, a_{k - 1}}$ or $span {a_{j} : λ_{j} < λ_{k}}$ has zero $a_{k}$ -coordinate, so $μ$ lies in neither span.

7.3step 4.1step 4.2step 5.2step 6.2step 1.4

Since $Φ (0) = 0$ by step 5.2 and $Φ (0) = \sum_{j} (λ_{j} - λ_{j - 1}) φ_{j} (0)$ with $φ_{j} (0) = 0$ by step 4.2, the same computation as steps 5.3 and 6.2 with $y = 0$ gives $ν = \sum_{j \leq k} (λ_{j} - λ_{j - 1}) φ_{j} (x) \in int (λ_{k} C)$ , and the $a_{k}$ -coordinate of $ν$ is $λ_{k} x_{k} \neq 0$ .

8.1step 7.1step 2.2

Combining steps 7.1 and 2.2 gives $vol (Φ (U)) = (\prod_{i} λ_{i}) vol (C)$ , which is clause 3.

8.2F2step 6.2step 7.2

But $μ \in Λ \cap int (λ_{k} C)$ , so clause 3 of [F2] forces $μ \in span {a_{j} : λ_{j} < λ_{k}}$ , contradicting step 7.2. Hence no two distinct points of $Φ (U)$ differ by an element of $2 Λ$ .

9.1F2step 7.3

Again clause 3 of [F2] would put $ν$ in $span {a_{j} : λ_{j} < λ_{k}}$ , contradicting the nonzero $a_{k}$ -coordinate. Hence $x = 0$ , and since $Φ (0) = 0$ the only lattice point of $Φ (U)$ is $0$ , which is clause 4 together with step 8.2.

10.1step 4.1step 5.1step 6.1step 5.2step 8.1step 8.2step 9.1∎

Clause 1 is step 4.1 with step 5.1, clause 2 is steps 6.1 and 5.2, clause 3 is step 8.1, and clause 4 is steps 8.2 and 9.1.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Minkowski second theorem on successive minima

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $C \subseteq R^{n}$ be compact, convex, centrally symmetric (A convex subset of $R^{m}$ contains every line segment between two of its points) with nonempty interior, and let $Λ \subseteq R^{n}$ be a full lattice with $covol (Λ) > 0$ (Full Euclidean lattice and covolume). Let $λ_{1} \leq \dots \leq λ_{n}$ be the successive minima of $C$ with respect to $Λ$ (Successive minima of a convex body). Then

$\frac{2^{n}}{n!} covol (Λ) \leq (\prod_{i = 1}^{n} λ_{i}) vol (C) \leq 2^{n} covol (Λ) .$

Facts & Assumptions

Given: A compact convex centrally symmetric $C \subseteq R^{n}$ with nonempty interior, a full lattice $Λ$ with $covol (Λ) > 0$ , and the successive minima $λ_{1} \leq \dots \leq λ_{n}$ of $C$ with respect to $Λ$ .

[A1]

The Axiom of Choice implies the Axiom of Countable Choice (AC implies DC implies countable choice), which supplies the hypotheses of the linear-change-of-variables fact [F6] invoked in steps 1.3 and 4.1, and of the Lebesgue-measure and Tonelli facts [F7] invoked in steps 2.1 and 3.1; no other selection is made.

[F1]

The successive minima are defined by $λ_{i} = \inf {t > 0 : \dim span (t C \cap Λ) \geq i}$ , and for $C$ compact with nonempty interior and $Λ$ full one has $0 < λ_{1} \leq \dots \leq λ_{n} < \infty$ ; also $λ_{0} : = 0$ by convention (Successive minima of a convex body).

[F2]

There exist linearly independent $a_{1}, \dots, a_{n} \in Λ$ with $a_{i} \in λ_{i} C$ for every $i$ (Attained successive minima and adapted flag).

[F3]

With $U : = int (C)$ the centroid map $Φ : U \to R^{n}$ of Successive-minima volume deformation and collision avoidance is Borel measurable, has $vol (Φ (U)) = (\prod_{i} λ_{i}) vol (C)$ , and no two distinct points of $Φ (U)$ differ by an element of $2 Λ$ (Successive-minima volume deformation and collision avoidance).

[F4]

Blichfeldt's principle: for a Lebesgue measurable $S \subseteq R^{n}$ with $vol (S) > covol (Λ)$ there are distinct $x, y \in S$ with $x - y \in Λ$ (Blichfeldt lattice-point principle).

[F5]

If $b_{1}, \dots, b_{n}$ is a $Z$ -basis of a full lattice $Λ$ , then $covol (Λ) = ∣ \det (b_{1}, \dots, b_{n}) ∣$ ; consequently $covol (2 Λ) = ∣ \det (2 b_{1}, \dots, 2 b_{n}) ∣ = 2^{n} covol (Λ)$ (Full Euclidean lattice and covolume, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , The determinant of a triangular matrix is the product of its diagonal entries).

[F6]

Assume the Axiom of Countable Choice. For a linear $T : R^{n} \to R^{n}$ with matrix $A$ , if $\det A \neq 0$ then $T [E]$ is measurable and $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E)$ for every Lebesgue measurable $E$ ; if $\det A = 0$ , then $T [E]$ is measurable and null for every $E \subseteq R^{n}$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F7]

Under the Axiom of Countable Choice, product Lebesgue measure agrees with Euclidean Lebesgue measure on Borel sets, and Tonelli's theorem permits iterated integration; thus the volume of a Borel subset of $R^{n}$ can be computed by its coordinate integrals (for $n = 1$ , use the one-dimensional integral directly) (On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product). Also under Countable Choice, $λ_{n}$ is a complete measure and is therefore additive over finite unions of pairwise disjoint measurable sets (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F8]

A convex set contains every convex combination of finitely many of its points, and central symmetry means $- C = C$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[F9]

For $A \in M_{n} (Z)$ with $\det A \neq 0$ the index $[Z^{n} : A Z^{n}]$ equals $∣ \det A ∣$ and is a positive integer; for real square matrices $\det (A B) = \det (A) \det (B)$ and $\det (2 I_{n}) = 2^{n}$ (The index of a full-rank subgroup of $Z^{n}$ is the absolute determinant of a generating matrix, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , The determinant of a triangular matrix is the product of its diagonal entries).

Proof

1.1F1given

The minima satisfy $0 < λ_{1} \leq \dots \leq λ_{n} < \infty$ , so every $λ_{i}$ is a positive finite real number.

1.2given

Let $Δ : = {y \in R^{n} : \sum_{i = 1}^{n} ∣ y_{i} ∣ \leq 1}$ and $S : = {y \in R^{n} : y_{i} \geq 0, \sum_{i} y_{i} \leq 1}$ .

1.3F6A1algebra

For each sign vector $ε = (ε_{1}, \dots, ε_{n}) \in {\pm 1}^{n}$ let $ε S : = {(ε_{1} y_{1}, \dots, ε_{n} y_{n}) : y \in S}$ . These $2^{n}$ measurable sets cover $Δ$ . If $ε \neq ε^{'}$ , choose $i$ with $ε_{i} \neq ε_{i}^{'}$ ; then $ε S \cap ε^{'} S$ lies in the coordinate hyperplane $H_{i} = {y : y_{i} = 0}$ . The projection onto $H_{i}$ is singular and has image $H_{i}$ , so [F6] and [A1] give $λ_{n} (H_{i}) = 0$ . Thus the sign pieces overlap only on null sets. Each sign map is an invertible diagonal linear map with determinant of absolute value $1$ , so [F6] and [A1] give $λ_{n} (ε S) = λ_{n} (S)$ .

1.4F3F5

For the upper bound, [F3] gives $vol (Φ (U)) = (\prod_{i} λ_{i}) vol (C)$ for the Borel set $Φ (U)$ , and $covol (2 Λ) = 2^{n} covol (Λ)$ by [F5].

2.1F7step 1.2

By Tonelli's theorem [F7] applied to the indicator of the simplex, $λ_{n} (S) = \int_{0}^{1} \int_{0}^{1 - x_{1}} \dots \int_{0}^{1 - x_{1} - \dots - x_{n - 1}} 1 d x_{n} \dots d x_{1} = \frac{1}{n!}$ .

2.2step 1.2algebra

$Δ = conv {\pm e_{1}, \dots, \pm e_{n}}$ : every $y$ with $\sum_{i} ∣ y_{i} ∣ \leq 1$ is $\sum_{i} y_{i} e_{i}$ , a convex combination of the vectors $\pm e_{i}$ after moving negative coefficients and adding the origin, and conversely every convex combination of the $\pm e_{i}$ satisfies the inequality.

2.3F2step 1.1

Choose linearly independent $a_{1}, \dots, a_{n} \in Λ$ with $a_{i} \in λ_{i} C$ by [F2], and let $B$ be the matrix with columns $a_{1} / λ_{1}, \dots, a_{n} / λ_{n}$ ; by step 1.1 the columns are well defined, and they are linearly independent, so $B$ is invertible and $P : = B [Δ] = {B y : \sum_{i} ∣ y_{i} ∣ \leq 1}$ is measurable.

2.4F3F4F5step 1.4

If $vol (Φ (U)) > covol (2 Λ)$ then [F4] applied to the lattice $2 Λ$ produces distinct $x, y \in Φ (U)$ with $x - y \in 2 Λ$ , contradicting the collision-free clause of [F3]; hence $vol (Φ (U)) \leq covol (2 Λ) = 2^{n} covol (Λ)$ .

3.1F7step 1.3step 2.1

Disjointifying the finite cover in step 1.3 changes each piece only by a null set, so finite additivity and steps 1.3 and 2.1 give $vol (Δ) = \sum_{ε} λ_{n} (ε S) = 2^{n} / n!$ .

3.2F8step 2.2step 2.3

Each $a_{i} / λ_{i}$ lies in $C$ , and $- a_{i} / λ_{i}$ lies in $C$ by central symmetry; hence every convex combination of the $2 n$ points $\pm a_{i} / λ_{i}$ lies in $C$ by [F8]. Since $B [Δ] = {\sum_{i} y_{i} (a_{i} / λ_{i}) : \sum_{i} ∣ y_{i} ∣ \leq 1} = conv {\pm a_{i} / λ_{i}}$ by the same convex-combination identity as step 2.2, we have $P \subseteq C$ .

3.3F5step 2.3

Let $b_{1}, \dots, b_{n}$ be a $Z$ -basis of $Λ$ and $M = (b_{1} \dots b_{n})$ , so $covol (Λ) = ∣ \det M ∣$ ; each $a_{i} \in Λ$ has $a_{i} = M c_{i}$ with a unique $c_{i} \in Z^{n}$ , and $A = M D$ for $D = (c_{1} \dots c_{n}) \in M_{n} (Z)$ .

4.1F6F9A1step 3.1step 2.3step 3.2

Therefore $vol (C) \geq vol (P) = ∣ \det B ∣ vol (Δ) = 2^{n} ∣ \det B ∣ / n!$ by [F6], whose Countable Choice hypothesis is supplied by [A1], and steps 3.1 and 3.2; writing $A = (a_{1} \dots a_{n})$ we have $B = A diag (1 / λ_{1}, \dots, 1 / λ_{n})$ , so $\det B = \det A / \prod_{i} λ_{i}$ by [F9], and multiplying the volume inequality by $\prod_{i} λ_{i} > 0$ gives $(\prod_{i} λ_{i}) vol (C) \geq 2^{n} ∣ \det A ∣ / n!$ .

4.2F9step 2.3step 3.3

Since $A$ is invertible and $\det A = \det M \det D$ by [F9], also $\det D \neq 0$ ; hence $∣ \det D ∣ = [Z^{n} : D Z^{n}]$ is a positive integer by [F9], in particular at least $1$ .

5.1F5step 4.1step 3.3step 4.2

It follows that $∣ \det A ∣ = ∣ \det M ∣ ∣ \det D ∣ \geq ∣ \det M ∣ = covol (Λ)$ , and combining with step 4.1 gives $(\prod_{i} λ_{i}) vol (C) \geq 2^{n} ∣ \det A ∣ / n! \geq (2^{n} / n!) covol (Λ)$ .

6.1step 5.1step 2.4∎

Steps 5.1 and 2.4 combine into $(2^{n} / n!) covol (Λ) \leq (\prod_{i} λ_{i}) vol (C) \leq 2^{n} covol (Λ)$ .

Remarks

The two bounds have different shapes. The lower bound is geometric: the adapted vectors $a_{i} \in λ_{i} C$ turn the cross-polytope of side data into a subset of $C$ , and the determinant comparison against a lattice basis produces the index factor $∣ \det D ∣ \geq 1$ . The upper bound is measure theoretic: the centroid deformation of [F3] has volume $(\prod_{i} λ_{i}) vol (C)$ and avoids collisions modulo $2 Λ$ , so Blichfeldt's principle bounds that volume by $covol (2 Λ)$ . For $Λ = Z^{n}$ , the cube $C = [- 1, 1]^{n}$ attains the upper bound: all $λ_{i} = 1$ and $vol (C) = 2^{n}$ . The cross-polytope $C = {x : \sum_{i} ∣ x_{i} ∣ \leq 1}$ attains the lower bound: again all $λ_{i} = 1$ , since $C$ contains the standard basis vectors and no $t C$ with $t < 1$ contains a nonzero lattice point, while $vol (C) = 2^{n} / n!$ by step 3.1. The cube attains both bounds only when $n = 1$ .

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Number-field integer rings and ideals are full lattices

Statement

Let $K$ be a number field (Number field) of degree $n = [K : Q]$ , let $O_{K}$ be its ring of integers (Ring of integers), and let $σ : K \to R^{n}$ be the unscaled Minkowski embedding (Unscaled Minkowski embedding). Then:

$σ (O_{K})$ is a full lattice in $R^{n}$ (Full Euclidean lattice and covolume);
for every nonzero fractional $O_{K}$ -ideal $I$ (Fractional ideals), the image $σ (I)$ is a full lattice in $R^{n}$ .

No choice principle is used: both lattices are exhibited by explicit $Z$ -bases. The proof fixes one basis for the given ring of integers, considers one ideal or subgroup at a time, and at each finite induction stage selects one lift from a single nonempty fiber; it does not select simultaneously from an arbitrary family of nonempty sets.

Facts & Assumptions

Given: A number field $K$ of degree $n = [K : Q]$ , its ring of integers $O_{K}$ , and the unscaled Minkowski embedding $σ$ .

[F1]

For every ordered $Q$ -basis $α_{1}, \dots, α_{n}$ of $K$ the real $n \times n$ matrix $A$ whose $j$ -th column is $σ (α_{j})$ is invertible, and $∣ \det A ∣ = 2^{- r_{2}} \sqrt{∣ disc (α_{1}, \dots, α_{n}) ∣} \neq 0$ (Unscaled Minkowski embedding, Embedding determinant formula).

[F2]

$O_{K}$ is a free $Z$ -module of rank $n$ (The ring of integers has rank the degree).

[F3]

Every additive subgroup of $Z$ is $d Z$ for a unique nonnegative integer $d$ , with $d > 0$ when the subgroup is nonzero (Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$ ).

[F4]

An ideal $I ⊴ R$ is an additive subgroup with $r i \in I$ for all $r \in R$ , $i \in I$ ; in particular an ideal $a$ of $O_{K}$ is a $Z$ -submodule of $O_{K}$ , and $u O_{K} \subseteq a$ for every $u \in a$ (Left, right and two-sided ideals).

[F5]

A fractional ideal of $O_{K}$ is a nonzero $O_{K}$ -submodule $I \subseteq K$ for which some $0 \neq d \in O_{K}$ satisfies $d I \subseteq O_{K}$ (Fractional ideals).

[F6]

A full lattice is by definition the $Z$ -span $Z b_{1} \oplus \dots \oplus Z b_{n}$ of a real basis $b_{1}, \dots, b_{n}$ of $R^{n}$ (Full Euclidean lattice and covolume).

Proof

1.1F1givenalgebra

For every ordered $Q$ -basis $α_{1}, \dots, α_{n}$ of $K$ , let $A$ have columns $σ (α_{j})$ . By [F1], $A$ is invertible; hence these images form a real basis of $R^{n}$ .

1.2F2algebrachoose

Choose a $Z$ -basis $α_{1}, \dots, α_{n}$ of $O_{K}$ , which exists by [F2]. A rational relation among the $α_{i}$ , multiplied by a positive common denominator, would be an integer relation, so $Z$ -independence makes them $Q$ -independent. There are $n = [K : Q]$ of them, so they form a $Q$ -basis of $K$ .

1.3given

We prove by induction on $m \geq 0$ that every additive subgroup of $Z^{m}$ has a finite $Z$ -basis with at most $m$ members. The claim holds for $m = 0$ , since its only subgroup is ${0}$ , with empty basis.

2.1F3step 1.3

Let $m > 0$ , assume the claim for $m - 1$ , and project $H \leq Z^{m}$ onto its first coordinate. By [F3], the image is $d Z$ for a nonnegative integer $d$ . If $d = 0$ , $H$ lies in the last $m - 1$ coordinates, so induction gives a basis with at most $m - 1$ members.

2.2F6step 1.1step 1.2algebra

By steps 1.1 and 1.2, the vectors $σ (α_{1}), \dots, σ (α_{n})$ form a real basis. Additivity of $σ$ gives $σ (O_{K}) = Z σ (α_{1}) \oplus \dots \oplus Z σ (α_{n})$ , so this is a full lattice by [F6]. This proves clause 1.

3.1step 1.3step 2.1choosealgebra

If $d > 0$ , choose $h \in H$ whose first coordinate is $d$ . The kernel $H_{0}$ of the projection, viewed in $Z^{m - 1}$ , has a basis $k_{1}, \dots, k_{s}$ by induction, with $s \leq m - 1$ . Every $g \in H$ has first coordinate $q d$ for a unique $q \in Z$ , so $g - q h \in H_{0}$ ; hence $h, k_{1}, \dots, k_{s}$ span $H$ . If $a h + \sum_{i} b_{i} k_{i} = 0$ with integer coefficients, the first coordinate gives $a d = 0$ , hence $a = 0$ , and independence of the basis of $H_{0}$ gives every $b_{i} = 0$ . Together with the $d = 0$ case, this proves the induction claim.

4.1F2F4step 1.2step 1.3step 3.1choose

Let $a \subseteq O_{K}$ be a nonzero integral ideal. It is an additive subgroup by [F4]. Using the basis of $O_{K}$ from step 1.2 to identify it with $Z^{n}$ , steps 1.3, 2.1, and 3.1 give a $Z$ -basis $β_{1}, \dots, β_{r}$ of $a$ with $r \leq n$ . Choose $0 \neq u \in a$ ; then $u O_{K} \subseteq a$ by [F4].

5.1F4F6step 4.1step 2.2algebra

For any nonzero $v \in K$ , coordinate multiplication by the values of the embeddings at $v$ defines a block-diagonal real map $L_{v}$ . Its real blocks are the nonzero scalars $σ_{i} (v)$ ; a complex block $τ_{j} (v) = a + i b$ is represented by $(\begin{array}{cc} a & - b \\ b & a \end{array})$ , whose determinant is $a^{2} + b^{2} = ∣ τ_{j} (v) ∣^{2} > 0$ because each embedding is injective. Thus $L_{v}$ is invertible. In particular, for the element $u$ chosen in step 4.1, $σ (u x) = L_{u} σ (x)$ and $L_{u} (σ (O_{K})) = σ (u O_{K}) \subseteq σ (a)$ . Applying $L_{u}$ to the basis in step 2.2 gives a real basis, whose integer span is a full lattice by [F6]; hence $σ (a)$ spans $R^{n}$ .

6.1F1F6step 4.1step 5.1algebra

Since $β_{1}, \dots, β_{r}$ are $Z$ -independent, they are $Q$ -independent: a rational relation, after multiplication by a positive common denominator, is an integer relation and therefore has all coefficients zero. Additivity gives $σ (a) = Z σ (β_{1}) + \dots + Z σ (β_{r})$ , so these images span it over $R$ . Step 5.1 forces $r \geq n$ , while step 4.1 gives $r \leq n$ . Thus $r = n$ , the $β_{i}$ form a $Q$ -basis of $K$ , and [F1] makes their images a real basis. Therefore $σ (a) = Z σ (β_{1}) \oplus \dots \oplus Z σ (β_{n})$ is a full lattice.

7.1F5step 6.1choose

Let $I$ be a nonzero fractional $O_{K}$ -ideal. By [F5], choose $0 \neq d \in O_{K}$ with $b : = d I \subseteq O_{K}$ . The set $b$ is an ideal because $I$ is an $O_{K}$ -submodule, and it is nonzero because multiplication by $d \neq 0$ in the field $K$ is injective. Thus step 6.1 shows that $σ (b)$ is a full lattice.

8.1F5F6step 5.1step 7.1algebra

The real-coordinate multiplication $L_{d}$ is invertible by the block calculation of step 5.1. From $b = d I$ and $σ (d x) = L_{d} σ (x)$ we obtain $L_{d} (σ (I)) = σ (b)$ . If $γ_{1}, \dots, γ_{n}$ is a lattice basis of $σ (b)$ from step 7.1, then $L_{d}^{- 1} γ_{1}, \dots, L_{d}^{- 1} γ_{n}$ is a real basis and its integer span is $σ (I)$ ; thus $σ (I)$ is a full lattice by [F6]. This proves clause 2.

9.1step 2.2step 8.1∎

Clause 1 is step 2.2 and clause 2 is step 8.1, so both assertions of the statement hold.

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Covolume of an integral ideal lattice

Statement

Let $K$ be a number field of degree $n$ with $r_{2}$ complex places (Number field, Unscaled Minkowski embedding), ring of integers $O_{K}$ and discriminant $d_{K} \neq 0$ (Number-field discriminant is well-defined and nonzero). Let $σ : K \to R^{n}$ be the unscaled Minkowski embedding and let $a \subseteq O_{K}$ be a nonzero integral ideal, with absolute norm $N a = ∣ O_{K} / a ∣$ (The absolute norm of an integral ideal). Then

$covol (σ (a)) = 2^{- r_{2}} \sqrt{∣ d_{K} ∣} \cdot N a .$

The formula is for integral ideals only. It is not applied below to an ideal that is merely fractional; such an ideal is first multiplied by a positive integer (or by an element of $O_{K}$ ) to become integral.

Facts & Assumptions

Given: A number field $K$ of degree $n$ , its ring of integers $O_{K}$ , discriminant $d_{K} \neq 0$ , the unscaled Minkowski embedding $σ$ , and a nonzero ideal $a \subseteq O_{K}$ .

[F1]

The image $σ (a)$ of a nonzero integral ideal is a full lattice, and $a$ has a $Z$ -basis $α_{1}, \dots, α_{n}$ ; moreover $σ (O_{K})$ is a full lattice with integral basis $β_{1}, \dots, β_{n}$ of $O_{K}$ (Number-field integer rings and ideals are full lattices, Integral and power integral bases).

[F2]

For an ordered $Q$ -basis $α_{1}, \dots, α_{n}$ of $K$ , the real $n \times n$ matrix $A$ with columns $σ (α_{j})$ satisfies $∣ \det A ∣ = 2^{- r_{2}} \sqrt{∣ disc (α_{1}, \dots, α_{n}) ∣}$ , the real determinant being obtained from the full complex embedding matrix by replacing each conjugate pair of rows by its real and imaginary parts (Unscaled Minkowski embedding).

[F3]

$disc (α_{1}, \dots, α_{n}) = \det (ψ_{i} (α_{j}))^{2}$ for the full list of embeddings $ψ_{1}, \dots, ψ_{n}$ , and this determinant is nonzero (Embedding determinant formula).

[F4]

If $β_{j} = \sum_{i} a_{i j} α_{i}$ for two ordered bases, then $disc (β_{1}, \dots, β_{n}) = \det (A)^{2} disc (α_{1}, \dots, α_{n})$ (Change of basis for discriminants).

[F5]

For every integral basis $β_{1}, \dots, β_{n}$ of $O_{K}$ one has $disc (β_{1}, \dots, β_{n}) = d_{K} \neq 0$ (Discriminant of a basis and order, Number-field discriminant is well-defined and nonzero).

[F6]

$N a = ∣ O_{K} / a ∣$ is finite (The absolute norm of an integral ideal, A nonzero number-field ideal has finite quotient).

[F7]

For $C \in M_{n} (Z)$ with $\det C \neq 0$ , the subgroup $C Z^{n} \subseteq Z^{n}$ has finite index $∣ \det C ∣$ (The index of a full-rank subgroup of $Z^{n}$ is the absolute determinant of a generating matrix).

[F8]

$covol (Λ) = ∣ \det B ∣$ for a $Z$ -basis $b_{1}, \dots, b_{n}$ of a full lattice $Λ$ with matrix $B = (b_{1} \dots b_{n})$ (Full Euclidean lattice and covolume).

Proof

1.1F1F8given

By [F1] and [F8], $covol (σ (a)) = ∣ \det A ∣$ , where $A$ is the matrix with columns $σ (α_{j})$ for a $Z$ -basis $α_{1}, \dots, α_{n}$ of $a$ ; this basis is a $Q$ -basis of $K$ because $σ$ is injective and $σ (a)$ is a full lattice.

1.2F1algebra

Let $β_{1}, \dots, β_{n}$ be an integral basis of $O_{K}$ [F1]. Each $α_{j}$ lies in $a \subseteq O_{K}$ , so $α_{j} = \sum_{i} c_{i j} β_{i}$ with uniquely determined integers $c_{i j}$ ; let $C = (c_{i j})$ .

2.1F1step 1.1step 1.2algebra

The matrix $C$ has $\det C \neq 0$ : if $\det C = 0$ there is a nonzero rational vector $y$ with $C y = 0$ , whence $\sum_{j} y_{j} α_{j} = 0$ , contradicting $Q$ -linear independence of the $α_{j}$ from step 1.1.

3.1F6F7step 1.2step 2.1

(Index.) The map $φ : Z^{n} \to O_{K}$ , $(m_{i}) \mapsto \sum_{i} m_{i} β_{i}$ , is a $Z$ -linear bijection with $φ (C Z^{n}) = a$ ; hence $[O_{K} : a] = [Z^{n} : C Z^{n}] = ∣ \det C ∣$ by [F7], and by [F6] this index is $N a$ .

4.1F4F5step 3.1

By [F4] applied to $α_{j} = \sum_{i} c_{i j} β_{i}$ and [F5], $disc (α_{1}, \dots, α_{n}) = \det (C)^{2} d_{K}$ , so $∣ disc (α_{1}, \dots, α_{n}) ∣ = (N a)^{2} ∣ d_{K} ∣$ by step 3.1.

5.1F2F3step 1.1step 4.1

Steps 1.1, [F2] and [F3] give $covol (σ (a)) = ∣ \det A ∣ = 2^{- r_{2}} \sqrt{∣ disc (α_{1}, \dots, α_{n}) ∣}$ , and step 4.1 evaluates the discriminant, so $covol (σ (a)) = 2^{- r_{2}} \sqrt{(N a)^{2} ∣ d_{K} ∣} = 2^{- r_{2}} \sqrt{∣ d_{K} ∣} N a$ .

6.1step 5.1∎

Step 5.1 is the asserted formula.

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Archimedean product region, volume and norm bound

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let integers $r_{1}, r_{2} \geq 0$ with $n = r_{1} + 2 r_{2} \geq 1$ and a real $t > 0$ be given, and in $R^{r_{1}} \times C^{r_{2}} ≅ R^{n}$ put

$X_{t} = {(x, z) : \sum_{i = 1}^{r_{1}} ∣ x_{i} ∣ + 2 \sum_{j = 1}^{r_{2}} ∣ z_{j} ∣ \leq t} .$

Then:

$X_{t}$ is compact, convex and centrally symmetric with nonempty interior;
$vol (X_{t}) = 2^{r_{1}} (\frac{π}{2})^{r_{2}} \frac{t^{n}}{n!}$ ;
every point of $X_{t}$ satisfies $\prod_{i = 1}^{r_{1}} ∣ x_{i} ∣ \prod_{j = 1}^{r_{2}} ∣ z_{j} ∣^{2} \leq (\frac{t}{n})^{n}$ .

Facts & Assumptions

Given: The Axiom of Choice, integers $r_{1}, r_{2} \geq 0$ with $n = r_{1} + 2 r_{2} \geq 1$ and a real $t > 0$ , with $X_{t}$ as in the statement and the identification of Unscaled Minkowski embedding.

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), which discharges the Countable Choice hypotheses of the volume facts [F6], [F1] and [F3], invoked in steps 1.4, 2.1 and 3.1 respectively; no further choice is used.

[F1]

Polar coordinates: for Borel $f \geq 0$ on $R^{m}$ , $\int_{R^{m}} f d λ_{m} = \int_{0}^{\infty} \int_{S^{m - 1}} f (r ω) r^{m - 1} d σ (ω) d r$ , where $σ$ is the polar surface set function on $S^{m - 1}$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F2]

Tonelli's theorem for sigma-finite product spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F3]

Under $R^{p + q} = R^{p} \times R^{q}$ , the product measure agrees with Lebesgue measure on Borel sets (On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}).

[F4]

AM-GM: for $a_{1}, \dots, a_{n} \geq 0$ with $n \geq 1$ , $\prod_{k} a_{k} \leq (\frac{1}{n} \sum_{k} a_{k})^{n}$ (The arithmetic mean, geometric mean inequality).

[F5]

The unit disc has area $λ_{2} (B^{2}) = π$ : the unit-ball volume formula $V_{m} (1) = π^{m / 2} / Γ (m / 2 + 1)$ at $m = 2$ gives $π / Γ (2) = π$ (The closed form for the volume of the unit $n$ -ball).

[F6]

Invertible linear maps scale Lebesgue measure by the absolute value of their determinant (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F7]

The polar surface set function of The polar surface set function on the unit sphere is defined by $σ (E) = n λ_{n} ({r ω : ω \in E, 0 < r \leq 1})$ ; for $E = S^{m - 1}$ the set on the right is the unit ball $B^{m}$ up to the null set ${0}$ , so $σ (S^{m - 1}) = m λ_{m} (B^{m})$ (The polar surface set function on the unit sphere).

Proof

1.1given

The function $N (x, z) = \sum_{i} ∣ x_{i} ∣ + 2 \sum_{j} ∣ z_{j} ∣$ is continuous, convex and even, so $X_{t} = N^{- 1} ([0, t])$ is closed, convex and centrally symmetric; it is bounded because every coordinate of a point of $X_{t}$ has absolute value at most $t$ , hence compact, and the origin is interior because a small ball around the origin satisfies $\sum_{i} ∣ x_{i} ∣ + 2 \sum_{j} ∣ z_{j} ∣ < t$ .

1.2algebra

(Weighted simplex integral.) For integers $p \geq 0$ and nonnegative integer weights $c_{1}, \dots, c_{p}$ , define $J_{p} (t) = \int_{y_{1}, \dots, y_{p} \geq 0, \sum_{k} y_{k} \leq t} \prod_{k} y_{k}^{c_{k}} d y$ ; then $J_{p} (t) = t^{p + C} \prod_{k} c_{k}! / (p + C)!$ with $C = \sum_{k} c_{k}$ .

1.3F5F7given

For each complex coordinate the polar surface value is $σ (S^{1}) = 2 λ_{2} (B^{2}) = 2 π$ , by [F7] with $m = 2$ and [F5].

1.4F6A1given

(Sign splitting.) The region ${(x, z) : \sum_{i} ∣ x_{i} ∣ + 2 \sum_{j} ∣ z_{j} ∣ \leq t}$ is the union over the $2^{r_{1}}$ sign choices of the pieces with prescribed signs of $x_{1}, \dots, x_{r_{1}}$ , and coordinate reflections carry each piece to the piece with all signs positive while preserving Lebesgue measure by [F6], whose Countable Choice hypothesis is supplied by [A1]; intersections lie in coordinate hyperplanes, which have measure zero.

1.5F4givenalgebra

For $(x, z) \in X_{t}$ apply [F4] with $n$ arguments equal to $∣ x_{1} ∣, \dots, ∣ x_{r_{1}} ∣$ and to the two copies each of $∣ z_{1} ∣, \dots, ∣ z_{r_{2}} ∣$ : their sum is at most $t$ , so their product satisfies $\prod_{i} ∣ x_{i} ∣ \prod_{j} ∣ z_{j} ∣^{2} \leq (t / n)^{n}$ .

2.1A1F1step 1.3given

(Radial reduction.) Using step 1.3 and the polar formula [F1] with $m = 2$ , the substitution $u = 2 ρ$ gives $\int_{C} F (2 ∣ z ∣) d z = 2 π \int_{0}^{\infty} F (2 ρ) ρ d ρ = (π / 2) \int_{0}^{\infty} F (u) u d u$ for Borel $F \geq 0$ , the Countable Choice hypothesis of [F1] being supplied by [A1].

2.2F2step 1.2algebra

(Induction for step 1.2.) The identity of step 1.2 is proved by induction on $p$ : for $p = 0$ both sides are $1$ ; for $p \geq 1$ Tonelli slices the last variable, $J_{p} (t) = \int_{0}^{t} y^{c_{p}} J_{p - 1} (t - y) d y$ , and the induction hypothesis reduces the claim to the one-variable identity $\int_{0}^{t} y^{a} (t - y)^{b} d y = a! b! t^{a + b + 1} / (a + b + 1)!$ , which follows by induction on $b$ from $\int_{0}^{t} y^{a} d y = t^{a + 1} / (a + 1)$ and $y^{a} (t - y)^{b + 1} = t y^{a} (t - y)^{b} - y^{a + 1} (t - y)^{b}$ , both elementary antiderivative computations for polynomials on a compact interval.

3.1F2F3A1step 2.1step 1.4

Applying step 2.1 in each complex coordinate and [F2] together with [F3] to the resulting iterated integrals, then applying step 1.4 to the real coordinates, gives $vol (X_{t}) = 2^{r_{1}} (π / 2)^{r_{2}} D (t)$ with $D (t) = J_{r_{1} + r_{2}} (t)$ for the weight vector with $c_{k} = 0$ on the first $r_{1}$ indices and $c_{k} = 1$ on the remaining $r_{2}$ indices, the Countable Choice hypothesis of [F3] being supplied by [A1].

4.1step 1.2step 3.1step 2.2algebra

For the weight vector of step 3.1 one has $p + C = (r_{1} + r_{2}) + r_{2} = n$ , so $D (t) = t^{n} / n!$ and $vol (X_{t}) = 2^{r_{1}} (π / 2)^{r_{2}} t^{n} / n!$ .

5.1step 1.1step 4.1step 1.5∎

Step 1.1 proves the compactness, convexity and symmetry clause, step 4.1 the volume formula and step 1.5 the norm bound, so the three assertions of the statement hold.

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Small nonzero element in a number-field ideal

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of degree $n = [K : Q]$ and signature $(r_{1}, r_{2})$ , and let $a \subseteq O_{K}$ be a nonzero integral ideal with absolute norm $N a$ . Then there is $0 \neq α \in a$ with

$∣ N_{K / Q} (α) ∣ \leq (\frac{4}{π})^{r_{2}} \frac{n!}{n^{n}} \sqrt{∣ d_{K} ∣} \cdot N a .$

Facts & Assumptions

Given: A number field $K$ of degree $n$ and signature $(r_{1}, r_{2})$ , so $n = r_{1} + 2 r_{2}$ , with ring of integers $O_{K}$ and nonzero integral ideal $a \subseteq O_{K}$ of absolute norm $N a$ (Unscaled Minkowski embedding).

[F1]

Minkowski convex-body theorem at equality: under the Axiom of Choice, if $C \subseteq R^{n}$ is compact, convex and centrally symmetric and $vol (C) \geq 2^{n} covol (Λ)$ for a full lattice $Λ$ , then $C$ contains a nonzero point of $Λ$ (Minkowski convex-body theorem at equality, Full Euclidean lattice and covolume).

[F2]

For $t > 0$ the set $X_{t} = {(x, z) \in R^{r_{1}} \times C^{r_{2}} : \sum_{i} ∣ x_{i} ∣ + 2 \sum_{j} ∣ z_{j} ∣ \leq t}$ is compact, convex and centrally symmetric, has $vol (X_{t}) = 2^{r_{1}} (π / 2)^{r_{2}} t^{n} / n!$ , and every point of $X_{t}$ satisfies $\prod_{i} ∣ x_{i} ∣ \prod_{j} ∣ z_{j} ∣^{2} \leq (t / n)^{n}$ (Archimedean product region, volume and norm bound).

[F3]

For a nonzero integral ideal $a$ the image $σ (a)$ under the unscaled Minkowski embedding is a full lattice in $R^{n}$ with $covol (σ (a)) = 2^{- r_{2}} \sqrt{∣ d_{K} ∣} \cdot N a$ (Number-field integer rings and ideals are full lattices, Covolume of an integral ideal lattice).

[F4]

For $0 \neq α \in K$ the norm is $N_{K / Q} (α) = \prod_{i = 1}^{r_{1}} σ_{i} (α) \prod_{j = 1}^{r_{2}} τ_{j} (α) {\overset{ˉ}{τ}}_{j} (α)$ , so $∣ N_{K / Q} (α) ∣ = \prod_{i} ∣ σ_{i} (α) ∣ \prod_{j} ∣ τ_{j} (α) ∣^{2}$ (Norm and trace from embeddings, with the inseparable exponent in the norm formula).

Proof

1.1F2given

Put $t : = (n! (4 / π)^{r_{2}} \sqrt{∣ d_{K} ∣} N a)^{1 / n} > 0$ , a positive real number, and let $X_{t}$ be the region of [F2].

2.1F2step 1.1

By [F2] the set $X_{t}$ is compact, convex and centrally symmetric.

2.2F2F3step 1.1algebra

By [F2] and [F3], $vol (X_{t}) = 2^{r_{1}} (π / 2)^{r_{2}} t^{n} / n! = 2^{r_{1}} (π / 2)^{r_{2}} (4 / π)^{r_{2}} \sqrt{∣ d_{K} ∣} N a = 2^{r_{1}} 2^{r_{2}} \sqrt{∣ d_{K} ∣} N a = 2^{n} 2^{- r_{2}} \sqrt{∣ d_{K} ∣} N a = 2^{n} covol (σ (a))$ , using $n = r_{1} + 2 r_{2}$ .

3.1F1F3step 2.1step 2.2

Applying [F1] to the compact convex centrally symmetric set $X_{t}$ and the full lattice $Λ = σ (a)$ of positive covolume, whose volume equals $2^{n} covol (Λ)$ by step 2.2, gives a nonzero $α \in a$ with $σ (α) \in X_{t}$ .

4.1F2F4step 3.1

Since $σ (α) \in X_{t}$ , the product bound of [F2] reads $\prod_{i} ∣ σ_{i} (α) ∣ \prod_{j} ∣ τ_{j} (α) ∣^{2} \leq (t / n)^{n}$ , and by [F4] the left side is $∣ N_{K / Q} (α) ∣$ .

5.1step 1.1step 4.1algebra∎

Therefore $∣ N_{K / Q} (α) ∣ \leq (t / n)^{n} = \frac{n! (4 / π)^{r_{2}} \sqrt{∣ d_{K} ∣} N a}{n^{n}} = (\frac{4}{π})^{r_{2}} \frac{n!}{n^{n}} \sqrt{∣ d_{K} ∣} N a$ , and $0 \neq α \in a$ .

Remarks

The choice of $t$ makes the volume of $X_{t}$ exactly $2^{n}$ times the covolume, which is why the equality form of Minkowski's theorem is needed and produces the constant $\sqrt{∣ d_{K} ∣}$ rather than a strict inequality. The factor $(4 / π)^{r_{2}}$ is the ratio between the volume of the $ℓ^{1} \oplus ℓ^{2}$ region of [F2] and the covariantly normalized volume, and it carries the unscaled real/imaginary convention. Passing to a fractional ideal requires multiplying by a denominator first, as recorded on the covolume theorem.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Minkowski bound for ideal classes

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of degree $n$ and signature $(r_{1}, r_{2})$ , and put

$M_{K} : = (\frac{4}{π})^{r_{2}} \frac{n!}{n^{n}} \sqrt{∣ d_{K} ∣} .$

Then every class in the ideal class group $Cl (O_{K})$ (The ideal class group) contains an integral ideal $b \subseteq O_{K}$ with $N b \leq M_{K}$ .

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ , its ring of integers $O_{K}$ , and a class $[I] \in Cl (O_{K})$ represented by a nonzero fractional ideal $I$ .

[F1]

Under the Axiom of Choice, $O_{K}$ is a Dedekind domain (Rings of integers are Dedekind domains).

[F2]

A fractional ideal of $O_{K}$ is a nonzero $O_{K}$ -submodule $I \subseteq K$ for which some $0 \neq d \in O_{K}$ satisfies $d I \subseteq O_{K}$ ; its inverse is $I^{- 1} = (O_{K} : I)$ , products and colons of fractional ideals are fractional ideals, and every nonzero fractional ideal of a Dedekind domain is invertible, so that $I I^{- 1} = O_{K}$ and the nonzero fractional ideals form a group under multiplication (Fractional ideals, Products, colons, and inverse candidates for fractional ideals, The basic operations on fractional ideals are well defined, Invertible fractional ideals, Every nonzero fractional ideal of a Dedekind domain is invertible).

[F3]

$Cl (O_{K})$ is the quotient of the group of nonzero fractional ideals by the subgroup of nonzero principal fractional ideals, and multiplication descends to the quotient (The ideal class group, The ideal class group quotient is well defined).

[F4]

Small nonzero element in an integral ideal: for every nonzero integral ideal $c \subseteq O_{K}$ there is $0 \neq β \in c$ with $∣ N_{K / Q} (β) ∣ \leq M_{K} N c$ (Small nonzero element in a number-field ideal).

[F5]

For $0 \neq α \in O_{K}$ the principal ideal $(α)$ satisfies $N ((α)) = ∣ N_{K / Q} (α) ∣$ , and for nonzero integral ideals $a, b$ one has $N (a b) = N a N b$ (The norm of a principal integral ideal, Ideal norm is multiplicative).

[F6]

$(α) \subseteq c$ exactly when $α \in c$ (The ideal generated by a subset and principal ideals).

Proof

1.1F1F2F3given

By [F1] the ring $O_{K}$ is Dedekind, so the fractional ideals and the class group of [F2] and [F3] are available, and the class $[I]$ has a nonzero fractional representative $I$ .

2.1F2F3step 1.1

By the denominator condition in [F2] applied to the fractional ideal $I^{- 1}$ , there is $0 \neq u \in O_{K}$ with $b : = u I^{- 1} \subseteq O_{K}$ ; $b$ is a nonzero integral ideal, and $[b] = [I]^{- 1}$ in $Cl (O_{K})$ because $u$ contributes the principal class.

3.1F4step 2.1

Applying [F4] to the nonzero integral ideal $b$ gives $0 \neq β \in b$ with $∣ N_{K / Q} (β) ∣ \leq M_{K} N b$ .

4.1F2F6step 3.1

By [F6] the membership $β \in b$ says $(β) \subseteq b$ ; multiplying this inclusion by the fractional ideal $b^{- 1}$ and using $b b^{- 1} = O_{K}$ from [F2] gives $a : = (β) b^{- 1} \subseteq O_{K}$ , a nonzero integral ideal because $(β) \neq 0$ and $b^{- 1} \neq 0$ .

5.1F3step 2.1step 4.1

In the class group, $[a] = [(β)] [b]^{- 1} = [b]^{- 1} = [I]$ , since the principal fractional ideal $(β)$ represents the identity class.

5.2F5step 3.1step 4.1algebra

From $a = (β) b^{- 1}$ we get the identity of integral ideals $a b = (β)$ ; both factors are nonzero integral ideals, so [F5] gives $N a N b = N ((β)) = ∣ N_{K / Q} (β) ∣ \leq M_{K} N b$ , and dividing by the positive integer $N b$ yields $N a \leq M_{K}$ .

6.1step 5.1step 5.2∎

Thus the integral ideal $a$ lies in the class $[I]$ and satisfies $N a \leq M_{K}$ ; since the class was arbitrary, every class of $Cl (O_{K})$ contains such an ideal.

Remarks

The preliminary denominator $u$ is what makes the argument work without a norm theory for fractional ideals: it converts $I^{- 1}$ into an integral ideal, the small-element theorem is applied there, and the factor $(β)$ then produces the integral representative $a = (β) b^{- 1}$ in the class $[I]$ . The class direction is $[a] = [b]^{- 1} = [I]$ , not $[I]^{- 1}$ . Because $M_{K}$ depends only on the signature and discriminant, the theorem bounds every class by one numerical constant; this is the input to both the finiteness of the class group and the generation by small prime ideals.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Finitely many ideals of bounded norm

Statement

Let $K$ be a number field and let $B \geq 1$ be real. Then only finitely many nonzero integral ideals $a \subseteq O_{K}$ satisfy $N a \leq B$ .

Facts & Assumptions

Given: A number field $K$ with ring of integers $O_{K}$ , a real number $B \geq 1$ , and an integer $n = [K : Q]$ .

[F1]

For every nonzero integral ideal $a \subseteq O_{K}$ the absolute norm $N a = ∣ O_{K} / a ∣$ is a finite cardinal (The absolute norm of an integral ideal, A nonzero number-field ideal has finite quotient).

[F2]

$O_{K}$ is a free $Z$ -module of rank $n$ (The ring of integers has rank the degree).

[F3]

If $G$ is a finite group and $H \leq G$ then $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ); hence the order of every element of $G$ divides $∣ G ∣$ , because the cyclic subgroup it generates has that order.

Proof

1.1F1F3given

Let $a$ be a nonzero integral ideal with $N a = m$ ; by [F1] the additive group $O_{K} / a$ has exactly $m$ elements, so by [F3] the order of each of its elements divides $m$ , and for $x \in O_{K}$ this gives $m (x + a) = a$ , that is $m x \in a$ , so $m O_{K} \subseteq a$ .

1.2F2algebra

Fix a $Z$ -basis $e_{1}, \dots, e_{n}$ of $O_{K}$ , which exists by [F2]; every class of $O_{K} / m O_{K}$ has exactly one representative $\sum_{i} a_{i} e_{i}$ with $0 \leq a_{i} < m$ , because subtracting suitable multiples of $m$ from the coordinates gives existence, while $\sum_{i} b_{i} e_{i} \in m O_{K}$ with $∣ b_{i} ∣ < m$ for all $i$ forces each $b_{i}$ to be an integer multiple of $m$ , hence $b_{i} = 0$ , giving uniqueness; thus $∣ O_{K} / m O_{K} ∣ = m^{n}$ .

2.1step 1.1algebra

Conversely every nonzero integral ideal $a$ with $m O_{K} \subseteq a$ has image $a / m O_{K}$ an ideal of the quotient ring $O_{K} / m O_{K}$ , and distinct ideals $a, b$ containing $m O_{K}$ have distinct images: if $a / m O_{K} = b / m O_{K}$ and $x \in a$ , then $x + m O_{K} \in b / m O_{K}$ , so $x \in b + m O_{K} \subseteq b$ , and interchanging $a$ and $b$ gives equality.

3.1step 2.1step 1.2algebra

A finite ring has only finitely many ideals, since its underlying set has only finitely many subsets, so for each fixed $m$ there are finitely many nonzero integral ideals with $N a = m$ by steps 2.1 and 1.2.

4.1step 3.1algebra∎

A nonzero integral ideal of norm $\leq B$ has norm equal to one of the positive integers $m \leq B$ , of which there are only the finitely many values $1 \leq m \leq ⌊ B ⌋$ , and a finite union of finite sets is finite, so only finitely many nonzero integral ideals satisfy $N a \leq B$ .

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Finiteness of the number-field class group

Statement

Assume the Axiom of Choice (The Axiom of Choice). For every number field $K$ , the ideal class group $Cl (O_{K})$ (The ideal class group) is finite.

Facts & Assumptions

Given: The Axiom of Choice and a number field $K$ of degree $n$ and signature $(r_{1}, r_{2})$ , with Minkowski constant $M_{K} = (4 / π)^{r_{2}} (n! / n^{n}) \sqrt{∣ d_{K} ∣}$ .

[F1]

Minkowski bound: every class in $Cl (O_{K})$ contains an integral ideal $b \subseteq O_{K}$ with $N b \leq M_{K}$ (Minkowski bound for ideal classes).

[F2]

For every real $B \geq 1$ there are only finitely many nonzero integral $O_{K}$ -ideals $b$ with $N b \leq B$ (Finitely many ideals of bounded norm).

[F3]

$Cl (O_{K})$ is the quotient of the group of nonzero fractional ideals by the subgroup of nonzero principal fractional ideals, and multiplication descends to a well-defined product on it; in particular there is a canonical class map $b \mapsto [b]$ from nonzero integral ideals to $Cl (O_{K})$ (The ideal class group, The ideal class group quotient is well defined).

Proof

1.1F2given

Put $B : = \max {M_{K}, 1} \geq 1$ ; by [F2] the set $S$ of nonzero integral ideals $b \subseteq O_{K}$ with $N b \leq B$ is finite.

2.1F2step 1.1

$S$ contains $O_{K}$ itself, of norm $1 \leq B$ , but only its finiteness is used below.

2.2F3step 1.1

By [F3] let $φ : S \to Cl (O_{K})$ be the class map $φ (b) = [b]$ .

3.1F1step 1.1step 2.2

The map $φ$ is surjective: for any class $[J] \in Cl (O_{K})$ , [F1] supplies an integral ideal $b$ with $[b] = [J]$ and $N b \leq M_{K} \leq B$ , so $b \in S$ and $φ (b) = [J]$ .

4.1step 1.1step 3.1∎

A set admitting a surjection from the finite set $S$ is finite, so $Cl (O_{K})$ is finite.

Remarks

The proof is a pure surjectivity argument: finiteness of the class group follows from the bound on representatives and the finiteness of ideals of bounded norm, with no further structure of the group used. The Choice assumption is inherited from the Minkowski bound route; the finiteness of ideals of bounded norm itself is choice-free. The set $S$ is explicit in terms of $O_{K}$ , the signature, and $∣ d_{K} ∣$ through $M_{K}$ .

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-10-02Open item page →

Class group generated by small prime ideals

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of degree $n$ and signature $(r_{1}, r_{2})$ , and let $M_{K} = (4 / π)^{r_{2}} (n! / n^{n}) \sqrt{∣ d_{K} ∣}$ . Then $Cl (O_{K})$ is generated by the classes of the nonzero prime ideals $p \subseteq O_{K}$ with $N p \leq M_{K}$ .

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ , its ring of integers $O_{K}$ , and a class $[J] \in Cl (O_{K})$ .

[F1]

Under the Axiom of Choice $O_{K}$ is a Dedekind domain (Rings of integers are Dedekind domains).

[F2]

Minkowski bound: every class in $Cl (O_{K})$ contains an integral ideal $b$ with $N b \leq M_{K}$ (Minkowski bound for ideal classes, The ideal class group).

[F3]

Every nonzero fractional ideal of a Dedekind domain factors uniquely as a product of powers of nonzero prime ideals, with only finitely many nonzero exponents and with nonnegative exponents on integral ideals (Unique factorization of nonzero fractional ideals into prime powers).

[F4]

For nonzero integral ideals $a, b$ one has $N (a b) = N a N b$ (Ideal norm is multiplicative), and for a nonzero prime ideal $p$ there is a rational prime $p$ with $N p = p^{f} \geq 2$ (The norm of a prime ideal).

[F5]

Multiplication of classes is well defined on $Cl (O_{K})$ and is compatible with products of fractional ideals (The ideal class group, The ideal class group quotient is well defined).

Proof

1.1F2given

Let $[J] \in Cl (O_{K})$ . By [F2] there is an integral ideal $b$ with $[b] = [J]$ and $N b \leq M_{K}$ ; in particular $b$ is nonzero.

2.1F1F3step 1.1

By [F1] and [F3], the nonzero integral ideal $b$ has a unique factorisation $b = p_{1}^{e_{1}} \dots p_{r}^{e_{r}}$ with the $p_{i}$ nonzero prime ideals of $O_{K}$ and integers $e_{i} \geq 1$ .

3.1F4step 1.1step 2.1algebra

By [F4], $N b = \prod_{i = 1}^{r} (N p_{i})^{e_{i}}$ with each $N p_{i} \geq 2$ ; since every factor of the positive integer $N b$ is at most $N b$ , each $N p_{i} \leq N b \leq M_{K}$ .

3.2F5step 1.1step 2.1

By [F5] the class of the product is the product of the classes: $[J] = [b] = \prod_{i = 1}^{r} [p_{i}]^{e_{i}}$ in $Cl (O_{K})$ .

4.1step 3.1step 3.2∎

Steps 3.1 and 3.2 exhibit every class $[J]$ as a product of classes of nonzero prime ideals $p$ with $N p \leq M_{K}$ ; hence those classes generate $Cl (O_{K})$ .

Remarks

Finite generation alone does not imply finiteness of a group. In the preceding Finiteness of the number-field class group, [F2] supplies an integral ideal of norm at most $M_{K}$ representing every class. The finite set of these bounded-norm ideals therefore surjects onto $Cl (O_{K})$ , which proves finiteness.

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Nontrivial number fields have discriminant of absolute value greater than one

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of degree $n = [K : Q] > 1$ . Then $∣ d_{K} ∣ > 1$ ; in particular $d_{K}$ is neither $1$ nor $- 1$ .

Facts & Assumptions

Given: The Axiom of Choice and a number field $K$ of degree $n > 1$ with signature $(r_{1}, r_{2})$ , so that $0 \leq r_{2} \leq n / 2$ .

[F1]

Minkowski bound: every class of $Cl (O_{K})$ contains an integral ideal $b$ with $N b \leq M_{K} = (4 / π)^{r_{2}} (n! / n^{n}) \sqrt{∣ d_{K} ∣}$ (Minkowski bound for ideal classes, The ideal class group).

[F2]

For a nonzero integral ideal $b$ the absolute norm $N b = ∣ O_{K} / b ∣$ is a finite positive integer, hence at least $1$ (The absolute norm of an integral ideal, A nonzero number-field ideal has finite quotient).

[F3]

Bernoulli's inequality: $(1 + x)^{n} \geq 1 + n x$ for $x \geq - 1$ and natural $n$ (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ ).

[F4]

Gregory-Leibniz: for every natural $N$ , $π / 4 = \sum_{k = 0}^{N} (- 1)^{k} / (2 k + 1) + R_{N}$ with $R_{N} = (- 1)^{N + 1} \int_{0}^{1} x^{2 N + 2} / (1 + x^{2}) d x$ (The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+...).

Proof

1.1F1F2given

The principal class of $Cl (O_{K})$ exists, so by [F1] it contains an integral ideal $b$ with $N b \leq M_{K}$ ; by [F2] the norm $N b$ is a positive integer, so $N b \geq 1$ and therefore $1 \leq (4 / π)^{r_{2}} (n! / n^{n}) \sqrt{∣ d_{K} ∣}$ .

1.2F4algebra

Take $N = 1$ and $N = 2$ in [F4]: $π / 4 = 1 - 1 / 3 + R_{1}$ with $R_{1} = \int_{0}^{1} x^{4} / (1 + x^{2}) d x > 0$ gives $π > 8 / 3 > 2$ , and $π / 4 = 1 - 1 / 3 + 1 / 5 + R_{2}$ with $R_{2} = - \int_{0}^{1} x^{6} / (1 + x^{2}) d x < 0$ gives $π < 52 / 15 < 4$ . Hence $2 / π < 1$ and $4 / π > 1$ .

2.1step 1.2algebra

Put $U_{m} : = (4 / π)^{m / 2} m! / m^{m}$ for $m \geq 2$ . Then $U_{2} = (4 / π) \cdot 2 / 4 = 2 / π < 1$ by step 1.2.

2.2F3step 1.2algebra

For $m \geq 2$ , Bernoulli's inequality [F3] with $x = 1 / m$ gives $(1 + 1 / m)^{m} \geq 2$ , so $(m / (m + 1))^{m} \leq 1 / 2$ and $\frac{U_{m + 1}}{U_{m}} = \frac{2}{\sqrt{π}} {(\frac{m}{m + 1})}^{m} \leq \frac{1}{\sqrt{π}} < 1$ .

3.1step 2.1step 2.2algebra

Consequently $U_{m} \leq U_{2} (\sqrt{π})^{- (m - 2)} < 1$ for every $m \geq 2$ ; in particular $U_{n} < 1$ .

4.1step 3.1givenalgebra

Since $r_{2} \leq n / 2$ and $4 / π > 1$ by step 1.2, $(4 / π)^{r_{2}} \leq (4 / π)^{n / 2}$ , so $c : = (4 / π)^{r_{2}} n! / n^{n} \leq U_{n} < 1$ with $c > 0$ .

5.1step 1.1step 4.1algebra

Step 1.1 gives $1 \leq c \sqrt{∣ d_{K} ∣}$ with $0 < c < 1$ , so $\sqrt{∣ d_{K} ∣} \geq 1 / c > 1$ and hence $∣ d_{K} ∣ > 1$ .

6.1step 5.1given∎

Thus every number field of degree $n > 1$ has $∣ d_{K} ∣ > 1$ , so its discriminant is neither $1$ nor $- 1$ ; the degree-one case is excluded by the hypothesis.

Remarks

The estimate compares the Minkowski constant against the smallest possible norm of an integral ideal, namely $1$ . Two elementary inequalities drive it: the two-sided bound $2 < π < 4$ , extracted here from the Gregory-Leibniz series with two and three terms respectively (so that $2 / π < 1$ and $4 / π > 1$ ), and Bernoulli's inequality $(1 + 1 / m)^{m} \geq 2$ , which makes the auxiliary sequence $U_{m}$ strictly decreasing. The hypothesis $n > 1$ is essential: $d_{Q} = 1$ , so the conclusion fails for the degree-one field.

CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Every nontrivial number field has a ramified finite prime

Statement

Assume the Axiom of Choice (The Axiom of Choice). Every finite number-field extension $K / Q$ with $[K : Q] > 1$ has a rational prime that ramifies in $K$ .

Facts & Assumptions

Given: The Axiom of Choice and a number field $K$ of degree $n = [K : Q] > 1$ .

[F1]

The preceding corollary gives $∣ d_{K} ∣ > 1$ (Nontrivial number fields have discriminant of absolute value greater than one).

[F2]

$O_{K}$ is a free $Z$ -module of rank $n$ , so it has an integral basis $α_{1}, \dots, α_{n}$ (The ring of integers has rank the degree).

[F3]

For an integral basis, $d_{K} = \det ({Tr}_{K / Q} (α_{i} α_{j}))_{i, j}$ is a nonzero signed integer, independent of the basis (Discriminant of a basis and order, Number-field discriminant is well-defined and nonzero).

[F4]

For $x \in K$ , the trace ${Tr}_{K / Q} (x)$ is the trace of the $Q$ -linear operator of multiplication by $x$ on $K$ (The norm $N_{K / F}$ and trace ${Tr}_{K / F}$ of a finite field extension).

[F5]

Ramification data: $p O_{K} = \prod_{P ∣ p} P^{e (P / p)}$ is a finite product of powers of distinct nonzero primes, and $p$ is ramified in $K$ exactly when some $e (P / p) > 1$ ; each residue field $O_{K} / P$ is finite (Integral ideal factorisation in a number field, in ZF, Ramification index, Primes above and residue degree, Splitting and ramification terminology).

[F6]

Chinese remainder theorem: for pairwise comaximal ideals $I_{1}, \dots, I_{r}$ of a commutative ring $R$ , the canonical map $R \to \prod_{i} R / I_{i}$ induces $R / \prod_{i} I_{i} ≅ \prod_{i} R / I_{i}$ (Chinese remainder theorem for pairwise comaximal ideals).

[F7]

The trace pairing of a finite separable field extension $L / F$ , $(x, y) \mapsto {Tr}_{L / F} (x y)$ , is nondegenerate (The trace pairing in a finite separable extension is nondegenerate).

[F8]

For a bilinear form on a finite-dimensional vector space, nondegeneracy is equivalent to invertibility of its matrix in a basis (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[F9]

Every finite field is perfect, and every algebraic extension of a perfect field is separable (Fields of characteristic zero, finite fields, and algebraically closed fields are perfect, Every algebraic extension of a perfect field is separable).

[F10]

A nilpotent endomorphism of a finite-dimensional vector space has trace $0$ : over an algebraic closure the characteristic polynomial splits, every eigenvalue of a nilpotent operator vanishes, and the trace is the sum of the eigenvalues with multiplicity (If $χ_{T} (x) = \prod_{i < n} (x - λ_{i})$ in $F [x]$ , then $tr (T) = \sum_{i < n} λ_{i}$ : trace is the sum of the eigenvalues counted with algebraic multiplicity).

[F11]

Under the Axiom of Choice, $O_{K}$ is a Dedekind domain, and every nonzero ideal of a Dedekind domain is invertible (Rings of integers are Dedekind domains, Every nonzero fractional ideal of a Dedekind domain is invertible).

[F12]

Every integer greater than $1$ has a prime divisor (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime).

[F13]

Ramification is detected by the discriminant: a rational prime $p$ ramifies in $K / Q$ if and only if $p ∣ d_{K}$ (Ramification is detected by the number-field discriminant).

Proof

1.1F1F2F3F12choose

Fix an integral basis $α_{1}, \dots, α_{n}$ of $O_{K}$ , which exists by [F2]. By [F1] and [F3] the integer $∣ d_{K} ∣$ is greater than $1$ , so [F12] gives a rational prime $p$ dividing $d_{K}$ .

1.2F2algebra

Put $A : = O_{K} / p O_{K}$ . Since $O_{K}$ is free with $Z$ -basis $α_{1}, \dots, α_{n}$ by [F2], the classes ${\overset{ˉ}{α}}_{1}, \dots, {\overset{ˉ}{α}}_{n}$ form an $F_{p}$ -basis of $A$ ; in particular $\dim_{F_{p}} A = n$ .

1.3F5F6algebra

Factorisation: by [F5], write $p O_{K} = P_{1}^{e_{1}} \dots P_{r}^{e_{r}}$ with distinct nonzero primes $P_{i}$ and $e_{i} \geq 1$ . Distinct maximal ideals satisfy $P_{i} + P_{j} = O_{K}$ ; choosing $u + v = 1$ with $u \in P_{i}$ , $v \in P_{j}$ and expanding $(u + v)^{e_{i} + e_{j} - 1}$ exhibits every term as an element of $P_{i}^{e_{i}} + P_{j}^{e_{j}}$ , so $1$ lies in that sum and the powers are pairwise comaximal. Applying [F6] to the ideals $P_{i}^{e_{i}}$ gives an isomorphism $A ≅ \prod_{i = 1}^{r} A_{i}$ with $A_{i} : = O_{K} / P_{i}^{e_{i}}$ .

1.4F5F11algebra

Reducedness of the factors: ideals of $R / I$ correspond to ideals of $R$ containing $I$ , so the maximal ideals of $A_{i}$ are the images of maximal ideals of $O_{K}$ containing $P_{i}^{e_{i}}$ ; a maximal ideal containing $P_{i}^{e_{i}}$ contains the prime $P_{i}$ , hence equals it, and $m_{i} : = P_{i} / P_{i}^{e_{i}}$ is the unique maximal ideal of $A_{i}$ , with $A_{i} / m_{i} = O_{K} / P_{i}$ a finite field by [F5]. If $e_{i} = 1$ then $A_{i} = O_{K} / P_{i}$ is a field and reduced. If $e_{i} \geq 2$ then $P_{i}^{e_{i}} ⊊ P_{i}$ : otherwise $P_{i}^{e_{i}} = P_{i}$ , and multiplying by the inverse ideal $P_{i}^{- 1}$ , which exists by [F11], gives $P_{i}^{e_{i} - 1} = O_{K} \subseteq P_{i}$ , a contradiction; so some $x \in P_{i} ∖ P_{i}^{e_{i}}$ has nonzero image in $A_{i}$ with $x^{e_{i}} \in P_{i}^{e_{i}}$ , a nonzero nilpotent. Therefore $A_{i}$ is reduced exactly when $e_{i} = 1$ , and since a finite product of nonzero rings is reduced exactly when each factor is, $A$ is reduced exactly when all $e_{i} = 1$ ; by [F5] this is exactly the case that $p$ is unramified.

2.1F3F4F8step 1.2algebra

Trace form and discriminant: for $x \in O_{K}$ , multiplication by $x$ on $O_{K}$ has matrix with integer entries in the basis $α_{i}$ and trace ${Tr}_{K / Q} (x)$ by [F4]. Reducing modulo $p$ shows that multiplication by $\overset{ˉ}{x}$ on $A$ has $F_{p}$ -trace ${Tr}_{K / Q} (x) m o d p$ . Hence $T (\overset{ˉ}{x}, \overset{ˉ}{y}) : = {Tr}_{K / Q} (x y) m o d p$ defines an $F_{p}$ -bilinear form on $A$ whose matrix in the basis ${\overset{ˉ}{α}}_{i}$ is $({Tr}_{K / Q} (α_{i} α_{j}) m o d p)$ , with determinant $d_{K} m o d p$ by [F3]. By [F8] this form is degenerate exactly when that determinant vanishes, that is, exactly when $p ∣ d_{K}$ .

2.2F7F8F9F10step 1.3step 1.4algebra

Trace form versus reducedness over the perfect field $F_{p}$ : (a) if every $e_{i} = 1$ , then $A ≅ \prod_{i} F_{i}$ with $F_{i} = O_{K} / P_{i}$ a finite field; each $F_{i} / F_{p}$ is finite, hence separable by [F9], so each factor trace pairing is nondegenerate by [F7]. Multiplication by an element of the product acts blockwise on the direct sum $⨁_{i} F_{i}$ , so the trace form of $A$ is the orthogonal direct sum of the factor pairings; a vector orthogonal to everything has every component orthogonal to its own factor, hence is zero, and by [F8] the form is nondegenerate. (b) if some $e_{i} > 1$ , choose $0 \neq \overset{ˉ}{x}$ in the nilpotent maximal ideal of the factor $A_{i}$ as in step 1.4; for every $\overset{ˉ}{y} \in A$ the product $\overset{ˉ}{x} \overset{ˉ}{y}$ is nilpotent, so multiplication by it is a nilpotent endomorphism and has trace $0$ by [F10]. Thus $\overset{ˉ}{x} \neq 0$ lies in the radical of $T$ and $T$ is degenerate. Consequently $T$ is nondegenerate exactly when $A$ is reduced.

3.1F5F13step 2.1step 1.4step 2.2

Combining steps 2.1, 1.4 and 2.2, for the rational prime $p$ the following are equivalent: $p ∣ d_{K}$ ; the trace form $T$ on $A = O_{K} / p O_{K}$ is degenerate; $A$ is not reduced; some ramification index exceeds $1$ ; and $p$ ramifies in $K$ . This verifies the published ramification-discriminant criterion [F13] for this field and prime in full.

4.1F13step 1.1step 3.1∎

By step 1.1 the prime $p$ divides $d_{K}$ , so step 3.1, equivalently the criterion [F13], shows that $p$ ramifies in $K$ . Therefore every number field of degree $n > 1$ has a rational prime that ramifies in it.

Remarks

The corollary is the contrapositive of the statement that a number field unramified at every finite prime has $∣ d_{K} ∣ = 1$ . The proof spells out the ramification-discriminant criterion rather than citing it silently: over the finite field $F_{p}$ the discriminant is the determinant of the reduced trace pairing, the residue algebra is the product of the prime-power factors $O_{K} / P_{i}^{e_{i}}$ , and over the perfect residue field that algebra is reduced exactly when all ramification indices are $1$ . Only finite primes are involved; no archimedean place enters the discriminant. The published criterion (Ramification is detected by the number-field discriminant) is used as stated and re-verified by steps 1.3, 1.4, 2.1, 2.2 and 3.1.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Bounded roots give finitely many monic integer polynomials

Statement

For every integer $n \geq 1$ and every real $R \geq 1$ , the set of monic polynomials in $Z [X]$ of degree at most $n$ whose complex roots, counted with multiplicity, all have modulus at most $R$ is finite.

Facts & Assumptions

Given: An integer $n \geq 1$ and a real $R \geq 1$ .

[F1]

If $α_{1}, \dots, α_{m} \in C$ and $f (X) = \prod_{j = 1}^{m} (X - α_{j}) = X^{m} + c_{m - 1} X^{m - 1} + \dots + c_{0}$ , then expanding the product and comparing coefficients gives $c_{m - k} = (- 1)^{k} \sum_{∣ S ∣ = k} \prod_{j \in S} α_{j}$ for $1 \leq k \leq m$ , the sum being over the $k$ -element subsets $S \subseteq {1, \dots, m}$ . In particular $∣ c_{m - k} ∣ \leq (\binom{m}{k}) R^{k}$ whenever every $∣ α_{j} ∣ \leq R$ .

Proof

1.1F1given

Fix $m$ with $1 \leq m \leq n$ and let $f (X) = X^{m} + c_{m - 1} X^{m - 1} + \dots + c_{0} \in Z [X]$ be monic of degree $m$ with roots $α_{1}, \dots, α_{m} \in C$ , counted with multiplicity; by [F1] each coefficient is an integer satisfying $∣ c_{m - k} ∣ \leq (\binom{m}{k}) R^{k}$ when $∣ α_{j} ∣ \leq R$ for all $j$ .

1.2given

The degree-zero case contributes only the constant polynomial $1$ , which has no roots, so the root condition holds for it vacuously.

2.1step 1.1algebra

Thus every $c_{m - k}$ lies in the intersection $Z \cap [- (\binom{m}{k}) R^{k}, (\binom{m}{k}) R^{k}]$ , an integer interval whose endpoints depend only on $m$ , $k$ and $R$ , and such an interval contains at most $2 (\binom{m}{k}) R^{k} + 1$ integers, a finite number because $R \geq 1$ .

3.1step 2.1algebra

For each fixed $m$ the coefficient vector $(c_{0}, \dots, c_{m - 1})$ therefore ranges over a product of $m$ finite sets, which is finite, and the monic degree- $m$ polynomials inject into that product by their coefficient vector, so there are finitely many of them.

4.1step 3.1step 1.2algebra∎

The set in the statement is the union over the finitely many degrees $0 \leq m \leq n$ of the corresponding pieces, and a finite union of finite sets is finite, so the statement holds.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Bounded primitive integral element for Hermite-Minkowski

Statement

Assume the Axiom of Choice (The Axiom of Choice). Fix an integer $n \geq 2$ and a real number $B \geq 1$ . Let $K$ be a number field of degree $n = [K : Q]$ whose discriminant satisfies $∣ d_{K} ∣ \leq B$ . Then there is an integral element $α \in O_{K}$ with $K = Q (α)$ such that every conjugate of $α$ has modulus at most $\sqrt{B + 2}$ .

Facts & Assumptions

Given: The Axiom of Choice, an integer $n \geq 2$ , a real number $B \geq 1$ , and a number field $K$ of degree $n$ with $∣ d_{K} ∣ \leq B$ . Write $(r_{1}, r_{2})$ for the signature of $K$ , so $n = r_{1} + 2 r_{2}$ , and $D : = ∣ d_{K} ∣$ , so $1 \leq D \leq B$ .

[A1]

The Axiom of Choice implies the Axiom of Countable Choice (AC implies DC implies countable choice), which is the choice hypothesis of the volume fact [F5], invoked in steps 2.2 and 2.3; the strict Minkowski theorem [F2] is applied under the Axiom of Choice assumed in the statement, and no other selection is made in this proof.

[F1]

$σ (O_{K})$ is a full lattice in $R^{n}$ with $covol (σ (O_{K})) = 2^{- r_{2}} \sqrt{∣ d_{K} ∣}$ (Number-field integer rings and ideals are full lattices, Covolume of an integral ideal lattice, Full Euclidean lattice and covolume).

[F2]

Minkowski convex-body theorem, strict form: under the Axiom of Choice, a Lebesgue measurable convex centrally symmetric $C \subseteq R^{n}$ with $λ_{n} (C) > 2^{n} covol (Λ)$ contains a nonzero point of the full lattice $Λ$ (Minkowski convex-body theorem, strict form, Full Euclidean lattice and covolume).

[F3]

The unscaled Minkowski embedding is $σ (x) = (σ_{1} (x), \dots, σ_{r_{1}} (x), τ_{1} (x), \dots, τ_{r_{2}} (x))$ with each complex coordinate $τ_{j} (x)$ split into its real and imaginary parts, and it is injective (Unscaled Minkowski embedding).

[F4]

The embeddings of $K$ into $C$ over $Q$ are the $r_{1}$ real embeddings $σ_{i}$ , and the two members $τ_{j}, {\overset{ˉ}{τ}}_{j}$ of each complex conjugate pair. For every $0 \neq α \in O_{K}$ the norm is $N_{K / Q} (α) = \prod_{i = 1}^{r_{1}} σ_{i} (α) \prod_{j = 1}^{r_{2}} τ_{j} (α) {\overset{ˉ}{τ}}_{j} (α)$ , this product is nonzero because embeddings are injective field homomorphisms, and $N_{K / Q} (α) \in Z$ ; hence $∣ N_{K / Q} (α) ∣ \geq 1$ and $∣ N_{K / Q} (α) ∣ = \prod_{i} ∣ σ_{i} (α) ∣ \prod_{j} ∣ τ_{j} (α) ∣^{2}$ (Norm and trace from embeddings, with the inseparable exponent in the norm formula, Trace and norm of an algebraic integer).

[F5]

Euclidean volume is multiplicative over Borel product sets (On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product), and the open disc of radius $ρ$ in $R^{2}$ has area $π ρ^{2}$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F6]

A product of convex sets is convex, and a product of sets each symmetric about the origin is centrally symmetric; bounded open intervals and open discs are convex and symmetric about the origin (A convex subset of $R^{m}$ contains every line segment between two of its points).

[F7]

For the finite tower $Q \subseteq Q (α) \subseteq K$ , restriction ${Hom}_{Q} (K, C) \to {Hom}_{Q} (Q (α), C)$ is surjective and every fibre has cardinality $[K : Q (α)]_{s}$ , the separable degree (Restriction partitions embeddings in a finite tower into extension fibres).

[F8]

Fields of characteristic zero are perfect and algebraic extensions of perfect fields are separable; hence $K / Q (α)$ is separable and $[K : Q (α)]_{s} = [K : Q (α)]$ (Fields of characteristic zero, finite fields, and algebraically closed fields are perfect, Every algebraic extension of a perfect field is separable).

[F9]

For $N = 1$ , the Gregory--Leibniz finite-remainder formula has partial sum $1 - 1 / 3 = 2 / 3$ and remainder $\int_{0}^{1} x^{4} / (1 + x^{2}) d x > 0$ . For example, the integrand is at least $1 / 32$ on $[1 / 2, 1]$ , so the remainder is at least $1 / 64$ . Thus $π / 4 > 2 / 3$ and $π > 8 / 3 > 2$ (The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+...).

[F10]

A finite tower of field extensions satisfies $[K : Q] = [K : Q (α)] [Q (α) : Q]$ (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ).

[F11]

A rational algebraic integer is an integer (The rational algebraic integers are exactly the integers).

Proof

1.1F3given

Exactly one of the cases $r_{1} \geq 1$ and $r_{1} = 0$ holds; in the second case $n = r_{1} + 2 r_{2} = 2 r_{2} \geq 2$ gives $r_{2} \geq 1$ . We treat the two cases in turn and produce the same conclusion in each.

1.2F3given

Case $r_{1} \geq 1$ . Let $X$ be the set of $(x_{1}, \dots, x_{r_{1}}, z_{1}, \dots, z_{r_{2}}) \in R^{r_{1}} \times C^{r_{2}}$ with $∣ x_{1} ∣ < \sqrt{D + 1}$ , $∣ x_{i} ∣ < 1$ for $2 \leq i \leq r_{1}$ , and $∣ z_{j} ∣ < 1$ for all $j$ ; in the real coordinates of [F3] this is $∣ x_{1} ∣ < \sqrt{D + 1}$ , $∣ x_{i} ∣ < 1$ , $(Re z_{j})^{2} + (Im z_{j})^{2} < 1$ .

1.3F3given

Case $r_{1} = 0$ . Here $n = 2 r_{2}$ with $r_{2} \geq 1$ . Let $Y$ be the set of $(z_{1}, \dots, z_{r_{2}}) \in C^{r_{2}}$ with $∣ Re z_{1} ∣ < 1$ , $∣ Im z_{1} ∣ < \sqrt{D + 1}$ , and $∣ z_{j} ∣ < 1$ for $j \geq 2$ ; written in real coordinates, $Y$ is the product of the rectangle $(- 1, 1) \times (- \sqrt{D + 1}, \sqrt{D + 1})$ in the first complex coordinate with $r_{2} - 1$ open unit discs.

2.1F6step 1.2

$X$ is a product of bounded open intervals and open discs, hence open and therefore Lebesgue measurable, and by [F6] it is convex and centrally symmetric.

2.2A1F5step 1.2

By [F5] and step 1.2, $λ_{n} (X) = (2 \sqrt{D + 1}) \cdot 2^{r_{1} - 1} \cdot π^{r_{2}} = 2^{r_{1}} π^{r_{2}} \sqrt{D + 1}$ .

2.3A1F5F6step 1.3

$Y$ is open and hence measurable, convex and centrally symmetric by [F6], and by [F5] it has $λ_{n} (Y) = (2 \cdot 2 \sqrt{D + 1}) \cdot π^{r_{2} - 1} = 4 π^{r_{2} - 1} \sqrt{D + 1}$ .

2.4step 1.3givenalgebra

Every $(z_{1}, \dots, z_{r_{2}}) \in Y$ has $∣ z_{1} ∣^{2} = (Re z_{1})^{2} + (Im z_{1})^{2} < 1 + (D + 1) = D + 2 \leq B + 2$ by the defining coordinate bounds and $D \leq B$ . Thus $∣ z_{1} ∣ < \sqrt{B + 2}$ .

3.1F1F9step 2.2algebra

By [F1] and step 2.2, $λ_{n} (X) / (2^{n} covol (σ (O_{K}))) = (π / 2)^{r_{2}} \sqrt{1 + 1 / D}$ . If $r_{2} = 0$ , the square-root factor is greater than $1$ ; if $r_{2} > 0$ , [F9] gives $π / 2 > 1$ and again the ratio is greater than $1$ . Thus $λ_{n} (X) > 2^{n} covol (σ (O_{K}))$ .

3.2F1F2F9step 2.3algebra

By [F1] and step 2.3, $λ_{n} (Y) / (2^{n} covol (σ (O_{K}))) = 2 (π / 2)^{r_{2} - 1} \sqrt{1 + 1 / D} > 1$ , since $r_{2} \geq 1$ , [F9] gives $π / 2 > 1$ , and $D \geq 1$ . Applying [F2] gives in this case an element $0 \neq α \in O_{K}$ with $σ (α) \in Y$ .

4.1F1F2step 2.1step 3.1

Applying [F2] with $Λ = σ (O_{K})$ and $C = X$ , whose hypotheses are verified in steps 2.1 and 3.1, gives in this case an element $0 \neq α \in O_{K}$ with $σ (α) \in X$ .

4.2F4step 3.2

In the totally complex case with $r_{2} \geq 2$ , every $j \geq 2$ has $0 < ∣ τ_{j} (α) ∣ < 1$ . There is at least one such factor, so $Q : = \prod_{j \geq 2} ∣ τ_{j} (α) ∣^{2} < 1$ . By [F4], $∣ N_{K / Q} (α) ∣ = ∣ τ_{1} (α) ∣^{2} Q \geq 1$ , hence $∣ τ_{1} (α) ∣ > 1$ .

4.3F4F10F11step 3.2step 2.4

If $r_{1} = 0$ and $r_{2} = 1$ , then $[K : Q] = 2$ . The element from step 3.2 is nonzero and satisfies $∣ Re τ_{1} (α) ∣ < 1$ . If $α \in Q$ , [F11] makes $α \in Z$ ; since an embedding fixes $Q$ , this would give $∣ α ∣ < 1$ and hence $α = 0$ , a contradiction. Therefore $α \notin Q$ , so $[Q (α) : Q] > 1$ . By the tower law [F10] this degree divides $[K : Q] = 2$ , and thus $K = Q (α)$ . Its two complex embeddings give the two conjugates, which are distinct because $α$ generates $K$ ; both have modulus less than $\sqrt{B + 2}$ by step 2.4 and conjugation.

5.1F4step 4.1

For this $α$ in the real-embedding case and every $i \geq 2$ one has $∣ σ_{i} (α) ∣ < 1$ , and for every $j$ one has $∣ τ_{j} (α) ∣ < 1$ . There are $r_{1} - 1 + 2 r_{2} = n - 1 \geq 1$ factors in $P : = \prod_{i \geq 2} ∣ σ_{i} (α) ∣ \prod_{j} ∣ τ_{j} (α) ∣^{2}$ , each positive and less than $1$ , so $P < 1$ and $∣ N_{K / Q} (α) ∣ = ∣ σ_{1} (α) ∣ P$ . Since the norm has absolute value at least $1$ by [F4], necessarily $∣ σ_{1} (α) ∣ > 1$ .

5.2F7F8step 1.3step 4.2

If $r_{1} = 0$ and $r_{2} \geq 2$ , every embedding other than $τ_{1}$ and ${\overset{ˉ}{τ}}_{1}$ sends $α$ to a value of modulus $< 1$ . The values $τ_{1} (α)$ and ${\overset{ˉ}{τ}}_{1} (α)$ have modulus $> 1$ by step 4.2 and are distinct: equality would make $τ_{1} (α)$ real, contrary to $∣ Re τ_{1} (α) ∣ < 1$ . Thus the fibre of [F7] over $τ_{1} ∣_{Q (α)}$ is the singleton ${τ_{1}}$ , so $[K : Q (α)]_{s} = 1$ , and [F8] gives $[K : Q (α)] = 1$ , that is, $K = Q (α)$ .

6.1F4step 1.2step 5.1

So in this case every embedding $φ$ of $K$ other than $σ_{1}$ sends $α$ to a complex number of modulus $< 1$ , while $∣ σ_{1} (α) ∣ > 1$ ; in particular $φ (α) = σ_{1} (α)$ holds only for $φ = σ_{1}$ . Also $∣ σ_{1} (α) ∣ < \sqrt{D + 1} \leq \sqrt{B + 2}$ by step 1.2, and $D + 1 \leq B + 2$ since $D \leq B$ .

7.1F7F8step 6.1

The fibre of the restriction map [F7] over $σ_{1} ∣_{Q (α)}$ is the set ${φ : φ (α) = σ_{1} (α)}$ , and the fibre is nonempty because it contains $σ_{1}$ ; by step 6.1 it is the singleton ${σ_{1}}$ . Hence $[K : Q (α)]_{s} = 1$ by [F7], and [F8] upgrades this to $[K : Q (α)] = 1$ , that is, $K = Q (α)$ .

8.1step 5.1step 2.4step 6.1step 5.2step 4.3step 7.1∎

Steps 7.1, 5.2, and 4.3 cover respectively the real-embedding case, the totally complex case with $r_{2} \geq 2$ , and the totally complex quadratic case; each gives $K = Q (α)$ for the constructed integral $α$ . In the real-embedding case step 6.1 bounds the distinguished real conjugate and all others have modulus $< 1$ . In the totally complex cases step 2.4 bounds $τ_{1} (α)$ and its conjugate, while all other conjugates have modulus $< 1$ by the chosen window. Since $B \geq 1$ , these bounds are all at most $\sqrt{B + 2}$ .

Remarks

The two windows are the ones used by Milne: the real case enlarges the first real coordinate, and the totally complex case enlarges the imaginary part of the first complex coordinate while keeping its real part in $(- 1, 1)$ . For $r_{2} \geq 2$ , the norm makes the first conjugate pair the unique values outside the unit circle, and the asymmetry separates the pair. For $r_{2} = 1$ , the strict real-coordinate bound rules out a rational integral element, and degree two then makes the nonzero element primitive. The uniform bound $\sqrt{B + 2}$ absorbs both coordinate bounds. This lemma is the analytic input to the Hermite-Minkowski finiteness theorem proved later on this page; the finiteness of the possible minimal polynomials there is a separate, purely algebraic step.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Hermite-Minkowski finiteness

Statement

Assume the Axiom of Choice (The Axiom of Choice). For every pair of positive integers $n$ and $B$ , only finitely many $Q$ -isomorphism classes of number fields $K$ of degree $n = [K : Q]$ (Number field) satisfy $∣ d_{K} ∣ \leq B$ .

Facts & Assumptions

Given: Positive integers $n$ and $B$ .

[F1]

Bounded primitive integral element: for $n \geq 2$ and real $B \geq 1$ , every number field $K$ of degree $n$ with $∣ d_{K} ∣ \leq B$ has $α \in O_{K}$ with $K = Q (α)$ such that every conjugate of $α$ has modulus at most $\sqrt{B + 2}$ (Bounded primitive integral element for Hermite-Minkowski).

[F2]

For every integer $m \geq 1$ and real $R \geq 1$ the set of monic polynomials in $Z [X]$ of degree at most $m$ whose complex roots, counted with multiplicity, all have modulus at most $R$ is finite (Bounded roots give finitely many monic integer polynomials).

[F3]

For $α \in K$ , one has $α \in O_{K}$ if and only if the monic minimal polynomial $m_{α}$ of $α$ over $Q$ lies in $Z [X]$ ; the degree of $m_{α}$ is $[Q (α) : Q]$ (Minimal-polynomial criterion for algebraic integers).

[F4]

If $f \in Q [X]$ is monic and irreducible and $β$ is a complex root of $f$ , then there is a field homomorphism $Q [X] / (f) \to C$ fixing $Q$ and sending $x + (f)$ to $β$ ; applied to the minimal polynomial $m_{α}$ of $K = Q (α)$ , whose quotient is $Q$ -isomorphic to $Q (α)$ by $x + (m_{α}) \mapsto α$ , it embeds $K$ into $C$ sending $α$ to $β$ (Universal property of adjoining a root of an irreducible polynomial).

[F5]

For a finite field extension $K / Q$ , one has $[K : Q] = 1$ if and only if $K = Q$ (A finite extension has degree one if and only if the two fields are equal).

Proof

1.1F5given

If $n = 1$ then every degree-one number field $K$ satisfies $[K : Q] = 1$ , hence $K = Q$ by [F5]; all such fields form the single $Q$ -isomorphism class of $Q$ .

1.2F2given

Now assume $n \geq 2$ . Since $B$ is a positive integer, $B \geq 1$ , and $R : = \sqrt{B + 2} \geq 1$ . Let $P$ be the set of monic $f \in Z [X]$ with $\deg f \leq n$ all of whose complex roots have modulus at most $R$ .

1.3F1given

Let $K$ be any number field of degree $n$ with $∣ d_{K} ∣ \leq B$ . By [F1] applied under the Axiom of Choice assumed in the statement, there is $α \in O_{K}$ with $K = Q (α)$ and every conjugate of $α$ of modulus at most $R$ . Let $f : = m_{α}$ be its minimal polynomial over $Q$ .

2.1F2F4step 1.2

By [F2] the set $P$ is finite. Let $Q \subseteq P$ be the subset of those $f$ that are irreducible in $Q [X]$ and have degree exactly $n$ , and define $g (f)$ to be the $Q$ -isomorphism class of the field $Q [X] / (f)$ ; this is well defined because for irreducible $f$ of degree $n$ the quotient is a field extension of $Q$ of degree $n$ .

2.2F3F4step 1.3

By [F3] the polynomial $f$ is monic of degree $[Q (α) : Q] = [K : Q] = n$ with integer coefficients; it is irreducible in $Q [X]$ , and $K ≅ Q [X] / (f)$ as extensions of $Q$ .

3.1F4step 1.3step 2.2

Every complex root $β$ of $f$ is a conjugate of $α$ : by [F4] there is an embedding $K \to C$ fixing $Q$ and sending $α$ to $β$ , so $β$ is one of the conjugates of step 1.3 and $∣ β ∣ \leq R$ . Hence $f \in Q$ and the class of $K$ equals $g (f)$ , which lies in the image $g (Q)$ .

4.1step 2.1step 3.1

Every $Q$ -isomorphism class of a degree- $n$ number field with $∣ d_{K} ∣ \leq B$ therefore belongs to the image of the finite set $Q$ under $g$ , and an image of a finite set is finite; so only finitely many such classes exist for $n \geq 2$ .

5.1step 1.1step 4.1∎

Combining the case $n = 1$ of step 1.1 with the case $n \geq 2$ of step 4.1 gives the result for all positive integers $n$ and $B$ .

Remarks

The proof uses no choice beyond the Axiom of Choice already assumed in the statement and in [F1]: the finite set $Q$ of candidate minimal polynomials is constructed explicitly, and a class is counted only when some integral primitive element realizes it. Two distinct polynomials in $Q$ may define the same field; this only shrinks the image. The bounded-root lemma is what makes the candidate set finite, and the primitive-element lemma is what bounds the minimal polynomial of every eligible field by $\sqrt{B + 2}$ .

5 · Examples, counterexamples and false statements

None yet.