Lemma 11 — the constrained unramified pro-2 group

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Statement

Let $T$ be a finite set of odd rational primes and let $S_{\mathbb Q}$ be another finite set of rational primes. Assume that an odd number of members of $T$ are congruent to $3$ modulo $4$. Put

$$P_T=\prod_{q\in T}q, \qquad Q_0=\mathbb Q(\sqrt{P_T}), \qquad M_T=\mathbb Q(\sqrt q:q\in T). \tag{1}$$

Let $G$ be the Galois group over $Q_0$ of the compositum of all Galois extensions of $Q_0$ which

• have degree a power of $2$;
• are unramified at every finite place and totally real;
• give inertia degree at most $2$ to every prime over $p\in S_{\mathbb Q}$; and
• give inertia degree at most $1$ when $p$ is inert in $Q_0/\mathbb Q$.

Write $d(G)$ for the minimal number of pro-$2$ generators and $r(G)$ for the minimal number of relations in a presentation on $d(G)$ generators. Then:

1. $M_T/Q_0$ is everywhere unramified and totally real, with $\operatorname{Gal}(M_T/Q_0)\cong(\mathbb Z/2\mathbb Z)^{|T|-1}$. Its inertia degrees at the selected primes are at most $2$, and at most $1$ when the rational prime is inert in $Q_0$.
2. $d(G)\geq|T|-1$.
3. One has

$$r(G)\leq d(G)+|S_{\mathbb Q}| +\#\{p\in S_{\mathbb Q}:p\text{ splits in }Q_0\}+2. \tag{2}$$

4. In particular, $G$ is infinite if

$$|T|+|S_{\mathbb Q}| +\#\{p\in S_{\mathbb Q}:p\text{ splits in }Q_0\}+1 \leq\frac{(|T|-1)^2}{4}. \tag{3}$$

Proof of (1)

The full multiquadratic extension $M_T/\mathbb Q$ has group $(\mathbb Z/2\mathbb Z)^{|T|}$. Since $Q_0$ is its quadratic subfield cut out by the product of all the square classes, $M_T/Q_0$ has the asserted rank $|T|-1$.

If $|T|=1$, then $M_T=Q_0$ and unramifiedness is immediate. For $|T|\ge2$ and each $q\in T$, the quadratic extension $Q_0(\sqrt q)/Q_0$ lies in the biquadratic field generated over $\mathbb Q$ by $\sqrt q$ and $\sqrt{P_T/q}$. A place that ramifies in this relative quadratic extension must lie over a rational place ramified in both $\mathbb Q(\sqrt q)$ and $\mathbb Q(\sqrt{P_T/q})$. No odd prime ramifies in both because $q$ and $P_T/q$ are coprime. At $2$, exactly one of $q$ and $P_T/q$ fails to be $1$ modulo $4$: the total number of prime factors that are $3$ modulo $4$ is odd. Thus $2$ does not ramify in both quadratic fields. Each $Q_0(\sqrt q)/Q_0$ is therefore unramified, and so is their compositum $M_T/Q_0$. It is totally real because all $q$ are positive.

Every element of $\operatorname{Gal}(M_T/\mathbb Q)$ has order at most two, so every rational-prime inertia degree in $M_T/\mathbb Q$ is at most two. If $p$ is inert in $Q_0/\mathbb Q$, its inertia degree there is already two; multiplicativity in the tower makes its relative inertia degree in $M_T/Q_0$ equal to one. This verifies all local constraints in the definition of $G$.

Proof of (2)

Part (1) makes $(\mathbb Z/2\mathbb Z)^{|T|-1}$ a quotient of $G$. Generator rank cannot increase on passing to a quotient, so $d(G)\geq|T|-1$.

Proof of (3)

Let $G'$ be the Galois group of the maximal totally real, everywhere finite-unramified pro-$2$ extension of $Q_0$, before the inertia constraints at $S_{\mathbb Q}$ are imposed. The number of primes of $Q_0$ over the rational primes in $S_{\mathbb Q}$ is

$$m=|S_{\mathbb Q}| +\#\{p\in S_{\mathbb Q}:p\text{ splits in }Q_0\}. \tag{4}$$

For a prime over a rational $p$ inert in $Q_0$, impose that its Frobenius be trivial. At each other prime in (4), impose that the square of Frobenius be trivial. These are exactly the residue-degree bounds defining $G$. Thus $G$ is obtained from $G'$ by adjoining at most $m$ pro-$2$ relations. At each step, a new relation either removes one minimal generator or increases the minimal relation count by at most one. Consequently

$$r(G)-d(G)\leq r(G')-d(G')+m. \tag{5}$$

The external input is Neukirch--Schmidt--Wingberg, Cohomology of Number Fields, second edition, Springer Grundlehren 323 (2008), Theorem 10.7.12. The same numbered statement was checked in the authors' corrected electronic version 2.3 (May 2020), printed p. 675 (physical p. 689). In the specialization used by the source, take its $S=T=\varnothing$ and $p=2$, treat $\mathbb C/\mathbb R$ as ramified, and use that the real quadratic field $Q_0$ has $r=r_1+r_2=2$, $\delta=1$, and $\theta=1$. The sums and remaining terms vanish, so the final inequality of that theorem gives $\chi_2(G')\leq3$. Since here

$$\chi(G')=1+r(G')-d(G'),$$

it follows that $r(G')-d(G')\leq2$. Insert this and (4) into (5) to obtain (2).

Proof of (4)

The second external input is the refined Golod--Shafarevich criterion in the form Sawin attributes to Gaschütz and Vinberg: a nontrivial finitely generated pro-$2$ group satisfying

$$r(G)\leq\frac{d(G)^2}{4} \tag{6}$$

is infinite. The manuscript cites Golod--Shafarevich (1964), Vinberg (1965), and Helmut Koch, Zum Satz von Golod-Schafarewitsch, Mathematische Nachrichten 42 (1969), 321--333, DOI 10.1002/mana.19690420413.

By (2), condition (6) follows if

$$|S_{\mathbb Q}|+\#\{p\in S_{\mathbb Q}:p\text{ splits in }Q_0\}+2 \leq\frac{d(G)^2}{4}-d(G). \tag{7}$$

The right side is increasing in the relevant range $d(G)\geq4$. Replacing $d(G)$ by the lower bound $|T|-1$ from part (2) turns (7) exactly into (3). Condition (3) cannot hold in the excluded smaller range, so the replacement loses no case. This proves infinitude.

Source and dependency scope

This is Lemma 11 and equations (9)--(10) on physical pp. 9--11 of the arXiv v1 manuscript. The multiquadratic and quotient arguments are reconstructed. The stated specialization of Neukirch--Schmidt--Wingberg was checked in the identified author-hosted electronic edition. It and the refined Golod--Shafarevich criterion are external theorems and are not proved here.

Used by. [[discrete_geometry/sawin_2026_explicit_lower_bound_unit_distance_problem/lemma_12|Lemma 12]].

Bears on. Problem 90.