Lemma 3 — ideal indices and archimedean sizes

---

Statement

Let $F$ be a totally real number field, let $K/F$ be a totally imaginary quadratic extension, and put $d=[F:\mathbb Q]$. Let $c$ be the nontrivial member of $\operatorname{Gal}(K/F)$. For a fractional ideal $I$ of $K$, let $N_{K/F}(I)$ denote the fractional ideal of $F$ generated by the elements $x c(x)$ with $x\in I$. Write $\Sigma_{F,\infty}$ for the infinite places of $F$.

If $0\neq\beta\in I$ and $0\neq\alpha\in N_{K/F}(I)$, then

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

Proof

At finite places use product-formula-normalized absolute values; at real and complex places use the usual absolute value, so a complex factor is squared in the product formula. For a nonarchimedean place $w$ of $K$, let $|I|_w$ be the absolute value of a generator of $I\otimes_{\mathcal O_K}\mathcal O_{K_w}$. Localizing the finite quotient, then applying the product formula, gives

$$\begin{aligned} \#(I/(\beta)) &=\prod_{w\nmid\infty}\frac{|I|_w}{|\beta|_w}\\ &=\left(\prod_{w\nmid\infty}|I|_w\right) \left(\prod_{v\in\Sigma_{F,\infty}}|\beta|_v^2\right). \end{aligned} \tag{2}$$

The analogous calculation over $F$ is

$$\#(N_{K/F}(I)/(\alpha)) =\left(\prod_{v\nmid\infty}|N_{K/F}(I)|_v\right) \left(\prod_{v\in\Sigma_{F,\infty}}|\alpha|_v\right). \tag{3}$$

At every finite place of $F$, the contribution of the norm ideal equals the product of the contributions of the places of $K$ above it. Therefore

$$\prod_{v\nmid\infty}|N_{K/F}(I)|_v =\prod_{w\nmid\infty}|I|_w. \tag{4}$$

Dividing (2) by (3) and using (4) proves (1).

Source scope

This is Lemma 3 on physical pp. 4--5 of the arXiv v1 manuscript. The explicit nonzero qualifications are forced by the finite quotients in the printed formula.

Used by. [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_4|Lemma 4]].

Bears on. Problem 90.