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Signature constant rules out discriminant ±1

Example

Assume the Axiom of Choice. Write the Minkowski numerical constant of a signature $(r_{1}, r_{2})$ with $n = r_{1} + 2 r_{2} > 1$ as $C_{n, r_{2}} = (4 / π)^{r_{2}} n! / n^{n}$ . Then $C_{n, r_{2}} < 1$ , and consequently the inequality $1 \leq C_{n, r_{2}} \sqrt{∣ d_{K} ∣}$ that the class bound produces for a field of that signature forces $∣ d_{K} ∣ > 1$ . The two cases of degree $2$ are $C_{2, 0} = 1 / 2$ and $C_{2, 1} = 2 / π$ , and for $n \geq 3$ the bound $r_{2} \leq n / 2$ reduces the constant to the auxiliary sequence $U_{n} = (4 / π)^{n / 2} n! / n^{n} < 1$ .

Facts & Assumptions

Given: The Axiom of Choice and a signature $(r_{1}, r_{2})$ with $n = r_{1} + 2 r_{2} > 1$ , together with a number field $K$ of that signature.

[F1]

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

[F2]

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

[F3]

Gregory-Leibniz with $N = 1$ and $N = 2$ : $π / 4 = 1 - 1 / 3 + R_{1}$ with $R_{1} = \int_{0}^{1} x^{4} / (1 + x^{2}) d x > 0$ and $π / 4 = 1 - 1 / 3 + 1 / 5 + R_{2}$ with $R_{2} = - \int_{0}^{1} x^{6} / (1 + x^{2}) d x < 0$ , so $8 / 3 < π < 52 / 15 < 4$ ; in particular $2 / π < 1$ and $4 / π > 1$ (The Gregory-Leibniz series: pi over four equals 1-1/3+1/5-1/7+...).

[F4]

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

[F5]

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

Proof

1.1F3algebra

Degree two: $C_{2, 0} = (4 / π)^{0} \cdot 2! / 2^{2} = 1 / 2 < 1$ , while $C_{2, 1} = (4 / π) \cdot 2! / 2^{2} = 2 / π < 1$ by [F3].

1.2F3F4algebra

Auxiliary sequence: put $U_{m} : = (4 / π)^{m / 2} m! / m^{m}$ for $m \geq 2$ . Then $U_{2} = 2 / π < 1$ by [F3]; for $m \geq 2$ Bernoulli's inequality [F4] with $x = 1 / m$ gives $(1 + 1 / m)^{m} \geq 2$ , so $U_{m + 1} / U_{m} = \frac{2}{\sqrt{π}} {(\frac{m}{m + 1})}^{m} \leq \frac{1}{\sqrt{π}} < 1$ by [F3], and therefore $U_{m} \leq U_{2} (\sqrt{π})^{- (m - 2)} < 1$ for every $m \geq 2$ .

2.1F3step 1.2givenalgebra

General signature: $4 / π > 1$ by [F3], so $C_{n, r_{2}} = (4 / π)^{r_{2}} n! / n^{n} \leq (4 / π)^{n / 2} n! / n^{n} = U_{n} < 1$ by step 1.2 and the hypothesis $r_{2} \leq n / 2$ ; thus $C_{n, r_{2}} < 1$ for every $n > 1$ .

3.1F1F2step 1.1step 2.1algebra

Class bound and conclusion: by [F1] the principal class contains an integral ideal $b$ with $N b \leq C_{n, r_{2}} \sqrt{∣ d_{K} ∣}$ ; by [F2] $N b$ is a positive integer, so $1 \leq N b \leq C_{n, r_{2}} \sqrt{∣ d_{K} ∣}$ . Since $0 < C_{n, r_{2}} < 1$ by steps 1.1 and 2.1, dividing gives $\sqrt{∣ d_{K} ∣} \geq 1 / C_{n, r_{2}} > 1$ , hence $∣ d_{K} ∣ > 1$ .

4.1F5step 1.1step 2.1step 3.1∎

Summary: for every signature with $n > 1$ the numerical constant $C_{n, r_{2}}$ is less than $1$ , so the Minkowski inequality $1 \leq C_{n, r_{2}} \sqrt{∣ d_{K} ∣}$ forces $∣ d_{K} ∣ > 1$ ; the degree-two constants are $C_{2, 0} = 1 / 2$ and $C_{2, 1} = 2 / π$ . This records exactly where the signature factor $(4 / π)^{r_{2}}$ enters and recovers the conclusion of [F5] from the class bound alone.

Remarks

The example isolates the arithmetic of the constant: the factor $4 / π$ is larger than $1$ , so the worst case for a given degree is the maximal number $r_{2} \leq n / 2$ of conjugate pairs, and at $n = 2$ the two constants $1 / 2$ and $2 / π$ are already smaller than the smallest possible ideal norm. Only the elementary bounds $2 < π < 4$ and Bernoulli's inequality are used; the value of the constant is never needed beyond strict comparison with $1$ . Signatures with $n > 1$ that are not realized by any number field cause no difficulty, since the statement is conditional on a field of the given signature existing.

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Sources

J. S. Milne, Algebraic Number Theory v3.08 (standard reference, not scraped)
William A. Stein, Algebraic Number Theory: A Computational Approach (standard reference, not scraped)