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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 $\sqrt{B + 2}$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

J. S. Milne, Algebraic Number Theory v3.08 (standard reference, not scraped)
Brian Conrad and Aaron Landesman, Math 154 Algebraic Number Theory (standard reference, not scraped)