Lemma 4 — a normalized CM ideal lattice

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Statement

Retain the notation of [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_3|Lemma 3]]. Fix a fractional ideal $I$ of $K$ and a nonzero $\alpha\in N_{K/F}(I)$. View $I$ through its archimedean embedding as a lattice

$$\Lambda=I\subset \prod_{v\in\Sigma_{F,\infty}}K_v \cong\mathbb C^d\cong\mathbb R^{2d}.$$

For $x=(x_v)_v$, define

$$\|x\|=\max_{v\in\Sigma_{F,\infty}} \frac{|x_v|}{\sqrt{|\alpha|_v}}. \tag{1}$$

Choose one infinite place $v_0$. Let $\pi:I\to\mathbb C\cong\mathbb R^2$ be its field embedding followed by division by $\sqrt{|\alpha|_{v_0}}$. Then $\pi$ is injective, the least nonzero lattice norm $\rho$ satisfies

$$\rho\geq \#(N_{K/F}(I)/(\alpha))^{-1/(2d)}, \tag{2}$$

and every $\beta\in I$ satisfying $\beta c(\beta)=\alpha$ obeys

$$\|\beta\|=1, \qquad |\pi(\beta)|=1. \tag{3}$$

Proof

If $\beta c(\beta)=\alpha$, then at every infinite place $v$,

$$|\beta|_v=\sqrt{|\beta c(\beta)|_v}=\sqrt{|\alpha|_v}.$$

This proves (3). For arbitrary nonzero $\beta\in I$, take square roots in Lemma 3:

$$\prod_{v\in\Sigma_{F,\infty}} \frac{|\beta|_v}{\sqrt{|\alpha|_v}} =\sqrt{ \frac{\#(I/(\beta))}{\#(N_{K/F}(I)/(\alpha))}} \geq\#(N_{K/F}(I)/(\alpha))^{-1/2}. \tag{4}$$

The maximum of $d$ nonnegative numbers is at least their geometric mean. Applying that observation to (4) gives (2). The selected field embedding is injective, so the same is true of $\pi$.

Source scope

This is Lemma 4 on physical p. 5 of the arXiv v1 manuscript. The explicit nonzero qualification is forced by the normalizing denominators. The printed statement does not assert that $\pi$ is injective; that clause is added here, with its one-line proof, because Lemma 5 needs it to apply Lemma 2.

Used by. [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_5|Lemma 5]].

Bears on. Problem 90.