Lemma 8 — the Galois split-prime specialization

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Statement

Assume the CM field $K$ is Galois over $\mathbb Q$, so its totally real subfield $F$ is Galois as well. Put $d=[F:\mathbb Q]$. Let $S_{\mathbb Q}$ be a finite set of rational primes and let $k:S_{\mathbb Q}\to\mathbb Z_{>0}$. Suppose every prime of $F$ over each $p\in S_{\mathbb Q}$ splits in $K/F$. Let $e_p$ and $f_p$ be the common ramification index and inertia degree of $p$ in $F/\mathbb Q$.

There are a fractional ideal $I$ of $K$ and a nonzero $\alpha\in N_{K/F}(I)$ for which

$$\#\{\beta\in I:\beta c(\beta)=\alpha\} \geq \frac{ \prod_{p\in S_{\mathbb Q}} (k(p)+1)^{d/(e_p f_p)}} {2^d h^-(K)} \tag{1}$$

and

$$\#(N_{K/F}(I)/(\alpha)) =\prod_{p\in S_{\mathbb Q}}p^{k(p)d/e_p}. \tag{2}$$

Proof

Apply [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_7|Lemma 7]] to the set of all primes $\mathfrak p$ of $F$ over the selected rational primes, assigning $k(\mathfrak p)=k(p)$. Galoisness gives exactly $d/(e_p f_p)$ primes over $p$, each with residue-field size $p^{f_p}$. Equation (1) of Lemma 7 immediately gives (1), while its quotient formula gives

$$\prod_{\mathfrak p\mid p} \#(\mathcal O_F/\mathfrak p)^{k(p)} =(p^{f_p})^{k(p)d/(e_pf_p)} =p^{k(p)d/e_p}$$

for each $p$, proving (2).

Source scope

This is Lemma 8 on physical pp. 7--8 of the arXiv v1 manuscript.

Used by. [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/proposition_10|Proposition 10]].

Bears on. Problem 90.