Lemma 7 — split-prime ideals give a large norm fibre

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Statement

Let $K/F$ be CM, with conjugation $c$ and $d=[F:\mathbb Q]$. Use the norm ideal $N_{K/F}(I)$ defined in [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_3|Lemma 3]], and write $h^-(K)=h(K)/h(F)$. Let $S_F$ be a finite set of prime ideals of $\mathcal O_F$, all split in $K/F$, and let $k:S_F\to\mathbb Z_{>0}$. There are a fractional ideal $I$ of $K$ and a nonzero $\alpha\in N_{K/F}(I)$ such that

$$\#\{\beta\in I:\beta c(\beta)=\alpha\} \geq \frac{\prod_{\mathfrak p\in S_F}(k(\mathfrak p)+1)} {2^d h^-(K)} \tag{1}$$

and

$$\#(N_{K/F}(I)/(\alpha)) =\prod_{\mathfrak p\in S_F} \#(\mathcal O_F/\mathfrak p)^{k(\mathfrak p)}. \tag{2}$$

Proof

Let $\mathcal L$ be the set of integral ideals $J$ of $K$ with

$$N_{K/F}(J)=\prod_{\mathfrak p\in S_F} \mathfrak p^{k(\mathfrak p)}. \tag{3}$$

For each $\mathfrak p$, write $\mathfrak p\mathcal O_K=\mathfrak p_1\mathfrak p_2$. Its contribution to $J$ can be

$$\mathfrak p_1^j\mathfrak p_2^{k(\mathfrak p)-j}, \qquad 0\leq j\leq k(\mathfrak p).$$

Unique factorization of ideals gives

$$|\mathcal L|=\prod_{\mathfrak p\in S_F}(k(\mathfrak p)+1). \tag{4}$$

Fix $J_0\in\mathcal L$. Since $N_{K/F}(JJ_0^{-1})=(1)$, the assignment

$$J\longmapsto[(JJ_0^{-1},1)]\in G_K \tag{5}$$

is defined. By [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_6|Lemma 6]], one fiber contains at least $|\mathcal L|/|G_K|$ ideals. Choose a representative $(J_m,u)$ of its common class. For every $J$ in that fiber, the defining equivalence in $G_K$ supplies a $\beta\in K^\times$ with

$$\beta J_m=JJ_0^{-1}, \qquad u\beta c(\beta)=1. \tag{6}$$

Set

$$I=J_0^{-1}J_m^{-1}, \qquad \alpha=u^{-1}. \tag{7}$$

Because $J$ is integral, (6) puts $\beta$ in $I$, and its second equation gives $\beta c(\beta)=\alpha$. Also $(u)=N_{K/F}(J_m)$, so $(\alpha)=N_{K/F}(J_m)^{-1}$. The ideal $N_{K/F}(J_0)$ is integral, which shows that $\alpha\in N_{K/F}(I)$, and multiplication of fractional-ideal quotients gives

$$\begin{aligned} \#(N_{K/F}(I)/(\alpha)) &=\#\left( \frac{N_{K/F}(J_0)^{-1}N_{K/F}(J_m)^{-1}} {N_{K/F}(J_m)^{-1}} \right)\\ &=\#(\mathcal O_F/N_{K/F}(J_0))\\ &=\prod_{\mathfrak p\in S_F} \#(\mathcal O_F/\mathfrak p)^{k(\mathfrak p)}, \end{aligned}$$

proving (2).

Equation (6) recovers $J$ from $\beta$, so ideals in the fiber give distinct $\beta$ after choosing one for each. They give both $\beta$ and $-\beta$, which are distinct. Lemma 6 and (4) therefore yield

$$2\frac{|\mathcal L|}{|G_K|} \geq \frac{\prod_{\mathfrak p\in S_F}(k(\mathfrak p)+1)} {2^d h^-(K)},$$

which is (1).

Source scope

This is Lemma 7 on physical pp. 6--7 of the arXiv v1 manuscript.

Used by. [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_8|Lemma 8]].

Bears on. Problem 90.