A Density of Ramified Primes

Let K be a cyclic number field of odd degree over \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {Q}}$$\end{document}Q with odd narrow class number, such that 2 is inert in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K/{\mathbb {Q}}$$\end{document}K/Q. We define a family of number fields \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{K(p)\}_p$$\end{document}{K(p)}p, depending on K and indexed by the rational primes p that split completely in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K/{\mathbb {Q}}$$\end{document}K/Q, in which p is always ramified of degree 2. Conditional on a standard conjecture on short character sums, the density of such rational primes p that exhibit one of two possible ramified factorizations in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K(p)/{\mathbb {Q}}$$\end{document}K(p)/Q is strictly between 0 and 1 and is given explicitly as a formula in terms of the degree of the extension \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K/{\mathbb {Q}}$$\end{document}K/Q. Our results are unconditional in the cubic case. Our proof relies on a detailed study of the joint distribution of spins of prime ideals.

If K is totally real, then Cl + = Cl if and only if every totally positive unit in O is a square; see Lemma 1. Hence, property (P1) can be restated as (P1) K /Q is Galois, K is totally real, and O × + := {u ∈ O × : u 0} = O × 2 .
Number fields satisfying property (C1) and (P1) were studied by Friedlander, Iwaniec, Mazur, and Rubin [5]. More precisely, Friedlander et al. proved that if σ is a (fixed) generator of Gal(K /Q), then the density of principal prime ideals πO that split in the quadratic extension K ( σ (π))/K is equal to 1/2. Koymans and Milovic [6] extended the results of Friedlander et al. in two different aspects. First, the number field K now needs to satisfy only property (P1), i.e., K /Q need not be cyclic; second, density theorems about the splitting behavior of principal prime ideals are proved for multi-quadratic extensions of the form K ({ σ (π) : σ ∈ })/K , where is a fixed subset of Gal(K /Q) with the property that σ / ∈ whenever σ −1 ∈ .
Our main goal is to further extend these results to a certain setting where = Gal(K /Q) \ {1}; in this setting, we in fact have σ ∈ whenever σ −1 ∈ , and so our work features a new interplay of the Chebotarev Density Theorem and the method of sums of type I and type II. In particular, the densities appearing in our main theorems are of greater complexity than those appearing in [5] or [6]. Another innovation in our work is that by assuming property (C3), we are now also able to study the splitting behavior of all prime ideals, and not only those that are principal. While our generalization of "spin" to non-principal ideals may appear innocuous (see Definition 3), it is of note that it still encodes the relevant splitting information as well as that the study of its oscillations requires new ideas, carried out in Sect. 6.
Let K be a number field satisfying properties (P1) and (C3), and let p be a rational prime that splits completely in K /Q. We will now define an extension K (p)/Q where p ramifies; this extension was first studied by McMeekin [8]. Let p be an unramified prime ideal of degree one in O. Let R + p denote the maximal abelian extension of K unramified at all finite primes other than p; in other words, R + p is the ray class field of K of conductor p∞, where ∞ denotes the product of all real places of K . There is a unique subfield K (p) ⊂ R + p of degree 2 over K when p is prime to 2 (see Lemma 2). Finally, we define K (p) to be the compositum of K (p) over all primes p lying above p, i.e., K (p) = p|p K (p).
As K (p)/Q is Galois, the residue field degree f K (p)/Q (p) of p in K (p)/Q is well-defined. Our goal is to study the distribution of f K (p)/Q (p) as p varies. Note that because p splits completely in K /Q, f K (p)/Q (p) is equal to the residue field degree f K (p)/K (p) of p in K (p)/Q for any prime p of K lying above p. Furthermore, f K (p)/Q (p) = f K (p)/K (p) must divide 2 since [K (p) : K ] is a power of 2 and there are no cyclic subgroups of Gal(K (p)/K ) of order greater than 2.
To state our main results, we now introduce the relevant notions of density. For sets of primes A ⊆ B, we define the density of A restricted to B to be Let P 2 Q denote the set of rational primes co-prime to 2. For a fixed sign, μ ∈ {±}, we define the following sets of rational primes.
Our main results are conditional on the following conjecture, a slight variant of which appears in both [5] and [6]. In the following conjecture, the real number η ∈ (0, 1] plays the role of 1/n from [5, Conjecture C n , p. 738-739]. Conjecture 1 (C η [5]) Let η be a real number satisfying 0 < η ≤ 1. Then there exists a real number δ = δ(η) > 0 such that for all > 0 there exists a real number C = C(η, ) > 0 such that for all integers Q ≥ 3, all real non-principal characters χ of conductor q ≤ Q, all integers N ≤ Q η , and all integers M, we have We note that Conjecture C η is known for η > 1/4, as a consequence of the classical Burgess's inequality [2], and remains open for η ≤ 1/4. Moreover, for sums as above starting at M = 0, Conjecture C η (for any η) is a consequence of the Grand Riemann Hypothesis for the L-function L(s, χ). We are now ready to state our main results. (2 d k /2 + 1) where φ denotes the Euler's totient function. In particular, when n is prime, writing d = d n , The density d(F |S) is determined by the product of densities d(F |R) and d(R|S) where R is the set of primes satisfying a certain Hilbert symbol condition. Toward computing the density d(R|S), the terms s μ arise from counting the number of solutions to this Hilbert symbol condition over (O/4) × /((O/4) × ) 2 . In the cubic case, we have the following unconditional theorem.
Theorem 2 Let K /Q be a cubic cyclic number field and odd class number in which 2 is inert. Then For our main results, we have assumed that K satisfies properties (C1)-(C4). To start, we need properties (P1) and (C3) to define the extensions K (p)/K for primes p that split completely in K /Q. Coincidentally, as mentioned above, property (C3) also allows us to study the splitting behavior of all (not necessarily principal) prime ideals. Property (C2) ensures that Gal(K /Q) contains no involutions. While methods to deal with involutions do exist (see [5,Section 12,p. 745]), incorporating them into our arguments is non-trivial and may pose interesting new challenges in our analytic arguments. Properties (C1) and (C4) simplify our combinatorial arguments and allow us to give explicit density formulas. Removing the assumptions of properties (C1) and (C4) would pose new combinatorial challenges.
To end this section, we give some examples of number fields satisfying (C1)-(C4) so as to convince the reader that our theorems are not vacuous. First, many such fields can be found within the parametric families given by Friedlander et al. in [5, p. 712] and originally due to Shanks [14] and Lehmer [7], namely where α m and β m are roots of the polynomials respectively. Such fields always satisfy properties (P1), (C1), and (C2). We also note that one can use the law of cubic reciprocity to show that the fields Q(α m ) always satisfy property (C4). For small m one can check the remaining properties using Sage or another similar mathematical software package. For instance, if β 7 is any root of then Q(β 7 ) is a totally real cyclic degree-5 number field of class number 1451 where 2 stays inert.
More generally, we can look for special subfields of cyclotomic fields. Let m be a prime number and ζ m a primitive m-th root of unity, so that Q(ζ m )/Q is a cyclic extension of degree ϕ(m), and suppose that n is an odd integer such that ϕ(m) ≡ 0 mod 2n. For instance, we can take n to be a Sophie Germain prime and then take m = 2n + 1 to also be a prime. Suppose also that 2 is inert in Q(ζ m ), i.e., that 2 is a primitive root modulo m. We then define K to be the unique subfield of Q(ζ m + ζ −1 m ) of degree n over Q; K readily satisfies properties (C1), (C2), (C4), while for small n the property that Cl + = Cl and property (C3) can be checked using Sage. For instance, the unique degree-5 subfield of Q(ζ 191 ) has class number 11; it is isomorphic to Q(β 2 ) with β 2 a root of the polynomial g 2 as above.

Two families of number fields
We say a modulus m is narrow whenever it is divisible by all real infinite places. We say a modulus is wide whenever it is not divisible by any infinite place. We say a ray class group or ray class field is narrow or wide whenever its conductor is narrow or wide respectively.
For m an ideal of O, let Cl + m denote the narrow ray class group of conductor m. That is, Cl + m is the ray class group with conductor divisible by all real infinite places with finite part m.
The following lemma leads to several equivalent formulations of property (P1).
Lemma 1 K is any number field. Here, if K is not necessarily totally real, an element is said to be totally positive when it is positive in all real embeddings, and the signature of an element is determined by the signs of the element in each real embedding.
Proof Let K be an arbitrary number field with r 1 real embeddings and r 2 pairs of complex embeddings. That (a) and (b) are equivalent follows from the definitions of the narrow and wide Hilbert class fields. By the exact sequence and canonical isomorphism in [ Lemma 2 Let K be a totally real number field with odd narrow class number h + . Let p be an odd prime of K . Then the narrow ray class field over K of conductor p has a unique subextension that is quadratic over K .
Proof Let O × p,1 denote the totally positive units of K that are congruent to 1 modulo p. The exact sequence from class field theory as in [10, V.1.7] induces the following short exact sequence on the 2−torsion subgroups, where surjectivity of the final map is due to the assumption that h + is odd.
By Lemma 1, all signatures are represented by units.
and u m shares the same signature as u. Therefore the first map is surjective onto the projection to (Z/2) n . Since Cl The first map in this short exact sequence is not surjective because any element of (Z/2) n × (O/p) × [2 ∞ ] of the form (0, x) must come from a square because O × + = (O × ) 2 by Lemma 1. Then the size of Cl + p [2 ∞ ] is nontrivial, so since these are all 2-groups #Cl + p [2 ∞ ] is even.
We may now define the multi-quadratic extension K (p)/K as in Sect. 1. In addition, we define another family of number fields parameterized by prime numbers p for which our results also hold. For both families of number fields, we consider a totally real number field K with odd narrow class number h + . Furthermore, we now impose the condition that K /Q is a Galois extension. Equivalently, we are assuming conditions (P1) and (C3).
In Definition 1, we apply Lemma 2 to ensure the existence of a unique quadratic subextension of the narrow ray class field over K of conductor p. In Definition 2 we will use the fact that for such K , a principal ideal always has a totally positive generator; see Lemma 1.

Definition 1
Given an odd rational prime p that splits completely in K /Q and a prime ideal p ⊂ O lying above p, define K (p) to be the unique quadratic subextension of the narrow ray class field over K of conductor p.
Define K (p) to be the compositum of the fields K (p σ ) as σ ranges over Gal(K /Q).

Definition 2
Given an odd rational prime p that splits completely in K /Q, a prime ideal p ⊂ O lying above p, and a totally positive generator α of the principal ideal p h , we define Define K + (p) to be the compositum of the number fields K + (p σ ) as σ ranges over Gal(K /Q).
Since K is totally real and h + is odd, Lemma 1 implies that O × + = (O × ) 2 , so K + (p) does not depend on the choice of totally positive generator α.
We note that while each of the fields K (p σ ) need not be Galois over Q, their compositum K (p) certainly is. Similarly, K + (p)/Q is Galois, and each of the extensions K + (p σ )/Q need not be.
For an abelian extension of number fields L/E and a prime p of E, let f L/E (p) denote the residue field degree of p in L/E and e L/E (p) the ramification index of p in L/E. In particular, the ramification indices and residue field degrees e K (p)/Q (p), f K (p)/Q (p), e K + (p)/Q (p), and f K + (p)/Q (p) are well-defined.
Since p is assumed to split completely in K /Q, there are n distinct primes in K lying above p, and they are of the form p σ , where p is one such prime and σ ranges over Gal(K /Q).
By Lemma 2, K (p σ )/K is a quadratic extension, and since α generates a prime ideal, K + (p σ )/K is also a quadratic extension. Since K (p σ ) is a subfield of the narrow ray class field over K of conductor p σ , the extension K (p σ )/K is unramified at p τ for all τ = σ in Gal(K /Q). Since h + is odd, K (p σ )/K is ramified at p σ . Therefore e K (p)/Q (p) = 2 and [K (p) : Q] = n2 n where n = [K : Q].
Since p σ is an odd prime, p σ divides the discriminant of K + (p σ )/K and so this extension is ramified at p σ . Since p τ does not divide the discriminant for any τ = σ in Gal(K /Q), The residue field Z/p is cyclic and injects into O K (p) /P where P is a prime of K (p) above p. Therefore f K (p)/Q (p) | 2 because there are no cyclic subextensions of K (p)/K of degree greater than 2, and p is assumed to split completely in K /Q. Similarly, f K + (p)/Q (p) | 2. We summarise in the following Lemma.

Lemma 3
Let K be a totally real number field of degree n that is Galois over Q with odd narrow class number and let p be a prime that splits completely in K /Q. For L = K (p) and for L = K + (p), We will see in Corollary 1 that for a fixed odd rational prime p splitting completely in K /Q, the residue field degrees of p in K (p)/Q and in K + (p)/Q are equal. Hence, to prove Theorem 1, we will prove the analogous results for the family of extensions K + (p)/Q.

The spin of prime ideals
Throughout this section, we will assume K satisfies (P1) and (C3). By Lemma 1, this is equivalent to assuming that K is a totally real number field that is Galois over Q with odd narrow class number, and these conditions imply that O × + = O × 2 . We give the following definition of spin, which extends the definition of spin from [5, (1.1)] in a natural way so that it applies to all odd ideals (not necessary principal).

Definition 3
Let σ ∈ Gal(K /Q) be non-trivial. Given an odd ideal a, we define the spin of a (with respect to σ ) to be where α is any totally positive generator of the principal ideal a h , and where · · denotes the quadratic residue symbol in K .
The assumption O × + = O × 2 is important for two reasons. First, Lemma 1 ensures that the principal ideal a h has a generator α that is totally positive. Second, any two totally positive generators of a h differ by a square, so the value of the quadratic residue symbol defining the spin does not depend on the choice of totally positive generator α.
If a is an odd principal ideal and α 0 is a totally positive generator of a, then α h 0 is a totally positive generator for a h . As h is odd, we have so our definition coincides with that of Friedlander et al. in [5] for odd principal ideals a.

Known results
The main result in [5] can be stated as follows.

Let σ be a generator of the Galois group Gal(K /Q). Then for all real numbers x > 3, we have
Here the implied constant depends only on and K .
Friedlander et al. also proved an analogous result for the case when the summation is restricted to principal prime ideals p with totally positive generators satisfying a suitable congruence condition. By Burgess's inequality, Conjecture C η holds for η = 1/3 with δ = 1 48 , so Theorem 3 holds unconditionally for [K : Q] = 3 where θ = 1 10656 . In [5, Section 11], Friedlander et al. pose some questions about the joint distribution of spin(p, σ ) and spin(p, τ ) as p varies over prime ideals, where σ and τ are two distinct generators of the cyclic group Gal(K /Q). In [6], Koymans and Milovic prove that such spins are distributed independently if n ≥ 5, i.e., that the product spin(p, σ ) spin(p, τ ) oscillates similarly as in Theorem 3. In fact, they prove that the product of spins σ ∈S spin(p, σ ) oscillates as long as the fixed non-empty subset S of Gal(K /Q) satisfies the property that σ / ∈ S whenever σ −1 ∈ S. Moreover, their result holds for number fields K satisfying property (P1) and having arbitrary Galois groups, i.e., not necessarily satisfying property (C1).
The assumption in [6] that σ / ∈ S whenever σ −1 ∈ S is made because spin(p, σ ) and spin(p, σ −1 ) are not independent in the following sense. For a place v of K , let K v denote the completion of K at v. For a, b ∈ K coprime to v, the Hilbert Symbol (a, b) v is defined to be 1 if the equation ax 2 + by 2 = z 2 has a solution x, y, z ∈ K v with at least one of x, y, or z non-zero and −1 otherwise.

Proposition 1 ([5, Lemma 11.1]) Suppose K is a number field satisfying properties (P1) and (C3). Suppose p ⊂ O is a prime ideal and σ ∈ Gal(K /Q)
is an automorphism such that p and p σ are relatively prime. Then where α is a totally positive generator of p h and the product is taken over places v dividing 2.
Proof This is essentially Lemma 11.1 in [5]. The proof uses the fact that Consider v, a finite place not equal to p or p σ , and not dividing 2.
Since v = p, p σ , we have α and α σ are non-zero modulo v. Consider the equation The right hand side and the left hand side each take on (N(v) + 1)/2 values, so there is a solution by the pigeon hole principle. It can not be the case that both x and y are 0. Suppose x ≡ 0 mod v. Since v is prime to 2 and x ≡ 0, Hensel's Lemma implies there exists a solution in the completion at v. Therefore (α, α σ ) v = 1. If y is non-zero, a similar argument works.
In this paper, we study the joint distribution of multiple spins spin(p, σ ), σ ∈ S, in a setting where there are in fact many σ ∈ S such that σ −1 ∈ S as well. From the discussion above, we see that this might involve combining the work of Koymans and Milovic with the study of the products spin(p, σ ) spin(p, σ −1 ) for various σ .

Factorization and spin
The spin of prime ideals is related to the splitting behavior of p in both K + (p) and K (p) as we will see in Proposition 3 and Corollary 1.
Let R + m denote the narrow ray class field over K of conductor m. Let p be an odd prime of K . Recall from Definition 1 that Lemma 2 gives the existence of a unique quadratic subextension of R + p /K , denoted by K (p). We have the following proposition for K satisfying properties (P1) and (C3).
We denote the unit class of u by Proof The assumptions that K satisfies (P1) and (C3) ensure that we can define K (p).
where O L denotes the ring of integers of K ( √ β).
Raising to the power of the class number, where (·/·) denotes the quadratic residue symbol in K .
Proof Since h is odd and b and a are coprime, we have Similarly, By the law of quadratic reciprocity for K [12, Theorem VI.8.3, p. 415], we have where (·, ·) v is the Hilbert symbol on K and the product above is over all places v lying above 2 and infinity. For each infinite place v, we have (α, β) v = 1 since α is totally positive (and thus also positive in the embedding of K into R corresponding to v). For any place v lying above 2, we have (α, β) v = 1 since α is coprime to 2 and any even prime is unramified in . We thus deduce that which in combination with (3) and (2) yields the desired result.
Given a rational prime p, fix a prime p above p and a totally positive generator α of p h . Recall from Definition 2 that K + (p) is the composite of K + (p σ ) as σ varies over all . As before, denote by K (p σ ) the unique quadratic subextension of the narrow ray class field over K of conductor p σ . The factorization of p in K + (p) or K (p) is determined by the factorizations of p in each K + (p σ ) or K (p σ ) respectively, which is in turn determined by the spin of p with respect to σ or σ −1 , respectively. Proposition 3 Assume K satisfies properties (P1) and (C3). For a fixed odd prime p of K that splits completely in K /Q and σ non-trivial in Gal(K /Q), the following are equivalent.
. We will prove that (β/p) = 1 if and only if f L|K (p) = 1. The result will then be established by choosing suitable β and q.
If m > 0 is the minimal positive integer such that q m is principal and if q r is principal for any r > 0, then we can write r = a + ml for l > 0 and 0 ≤ a < m. Then q r = q a q ml and since q r and q m are principal, so is q a . Since m is the minimal such positive integer, a = 0 so m|r. That is, any power of q that is principal must be a power of q m . In particular since h is odd, m is odd and if q 2l is principal, then q l is principal.
Write q h = q m q 2l where m is the minimal positive integer such that q m is principal. By Lemma 1, we can write (γ ) = q m for γ ∈ O with the same signature as β. Then q h = q m q 2l becomes (β) = (γ )(δ) 2 for δ a generator of p l . Since the signatures of β and γ match, β = γ δ 2 . Therefore Here O denotes the ring of integers of K and O L denotes the ring of integers of L. The polynomial discriminant d(1, such that q t is principal. Then m|t but since q is odd, The quadratic residue (γ /p) is equal to 1 exactly when the polynomial Setting β = (uα) σ for u in the unit class u K (p) and q = p σ , by Proposition 2, L = K ( √ β) = K ( (uα) σ ) = K (p σ ). Since K ( (uα) σ ) is contained in R + p σ , no prime above 2 is ramified in the extension K ( (uα) σ )/K , so applying Lemma 4, Alternatively, if we set β = α σ −1 and q = p σ −1 , then L = K ( √ β) = K (p σ −1 ) and since

A consequence of the Chebotarev Density theorem
In this section, we use the Chebotarev Density Theorem to prove that the primes of K are equidistributed in M 4 as defined below, where the mapping takes primes to a totally positive generator considered in M 4 . This contributes toward the density d(R|S) of rational primes p that satisfy the spin relation, spin(p, σ ) spin(p, σ −1 ) = 1 for all non-trivial σ ∈ Gal(K /Q), where p is a prime of K above p, restricted to the rational primes splitting completely in K /Q. We will also give this density restricted modulo 4Z. Theorem 4 and Proposition 5 together give the density of such primes satisfying the spin relation.

Definition 4
For q a power of 2, define Note that M q is a group with a natural action from Gal(K /Q). Observe that {x + 2y : x ∈ R × , y ∈ R} is a set of representatives for U 4 and #U 4 = 2 n (2 n − 1). Therefore elements of U 2 4 are of the form (x + 2y) 2 ≡ x 2 mod 4O for x ∈ R × and y ∈ R. Since #(O/2) × = 2 n − 1 is odd, the squaring map on U 2 = (O/2) × is surjective and so #U 2 4 = 2 n − 1. Therefore #M 4 = #U 4 /#U 2 4 = 2 n . Since M 4 is formed by taking the quotient of U 4 modulo squares, M 4 is a direct product of cyclic groups of order 2. For any α ∈ O coprime to 2, write [α] as the projection of αO in M 4 . Since every x ∈ R × is a square in U 2 , we can write down the isomorphism explicitly as

Lemma 6 Let K be a number field such that 2 is inert in K
Proof Fix some α ∈ O coprime to 2. We claim that (α + 4B, 2) 2 = 1 for some B ∈ O. Since (O/2) × contains all its squareroots, there exist some γ , z ∈ O such that α ≡ γ 2 − 2z 2 mod 4. Write x = γ + 2x for some x ∈ O, set B = x γ + x 2 and y = 1. Then x 2 − (α + 4B)y 2 ≡ 2z 2 mod 8. This proves our claim. Now suppose (α, β) 2 = 1 for all β ∈ O coprime to 2. Then taking B from the above claim, (α + 4B, β) 2 = 1 holds for all β ∈ O coprime to 2 by Lemma 5, and for all β ∈ O divisible by 2, by the above claim. Since the Hilbert symbol is non-degenerate on K × 2 /(K × 2 ) 2 [ For m an ideal of K , let P m K denote the set of prime ideals of O co-prime to m. For K a totally real number field satisfying (C3), we can define the following map.

Definition 5
For q a power of 2, define the map where α ∈ O is a totally positive generator of the principal ideal p h . By Lemma 1 since K is totally real with odd narrow class number, all principal ideals have a totally positive generator and O × + = O × 2 . Since squares are trivial in M 4 by definition the map r q is well-defined. We also note that r q commutes with the Galois action, i.e. r q (p σ ) = r q (p) σ for all σ ∈ Gal(K /Q).
For m an ideal of O, let J m K denote the group of fractional ideals of K prime to m.

Lemma 7 [ [9, Lemma 3.5]] For K totally real with h + odd, the homomorphism J 2 K → M q induced by r q induces a surjective homomorphism,
Proof This result is proven in [9, 3.5] and while stated with more assumptions there, the same proof holds with only the assumptions stated here. For clarity, we expand upon the proof of surjectivity. Let Since i is an isomorphism, there exists an element in (K m /K m,1 )/(K m /K m,1 ) 2 represented by β ∈ K × such that i(β) = (0, X). Since i(β) maps to 0 in the projection to (Z/2) n , β is totally positive. Since β 0, we can choose Let S denote the set of odd primes p of K with inertia degree f K /Q (p) = 1.

Lemma 8 [[9, Lemma 4.3]] Assume K satisfies (P1), (C3), and (C4).
1. For any α ∈ M 4 , the density of primes p of K such that ϕ 4 (p) = α is 1 2 n . That is, 2. Furthermore, the density does not change when we restrict to primes of K that split completely in K /Q. That is, Proof See [9,Lemma 4.3]. There the result is stated with more assumptions, but the same proof holds more generally.
Proof By Lemma 7, the map r 4 : P 2 K → M 4 from Definition 5 induces a surjective group homomorphism Define H:= Art(ker(ϕ 4 )) where Art denotes the Artin map from Cl + 4 to Gal(R + 4 /K ) with R + 4 denoting the narrow ray class field over K of conductor 4. Define L to be the fixed field of H in Gal(R + 4 /K ). Then ϕ 4 induces a canonical isomorphism Then for any α ∈ M 4 , by applying the Chebotarev Density Theorem to the element of Gal(L/K ) corresponding to α via this isomorphism, there exists a prime p ∈ P 2 K with ϕ 4 (p) = α.
Let p and q be odd primes of K such that r 4 (p) = r 4 (q). Let α be a totally positive generator of p h and let β be a totally positive generator of q h , where h is the odd class number of K , which is odd by assumption. Since r 4 (p) = r 4 (q), α ≡ β in M 4 . Then α ≡ βγ 2 mod 4O for some γ ∈ O. Since α σ ≡ β σ (γ σ ) 2 mod 4O for all σ ∈ Gal(K /Q), taking norms N(α) ≡ N(β)N(γ ) 2 mod 4O. Since the norms are in Z, N(α) ≡ N(β) mod 4Z.
We now state an extended version of Lemma 8 that handles the densities restricted to primes of a fixed congruence class modulo 4Z.
For a fixed sign μ ∈ {±}, let S μ denote the set of primes p ∈ S with N(p) ≡ μ1 mod 4Z. In other words S μ is the set of primes of K laying above rational primes in S μ . 4 and for a fixed sign μ ∈ {±}, the density of p ∈ S μ such that ϕ 4 (p) = α is given by

Lemma 9 Assume K satisfies conditions (C1)-(C4). For any α ∈ M
Therefore by [10, V.1.7], the order of Gal(R + 4 /K ) divides h2 n (2 n − 1). As in the proof following Definition 6 and with L as defined there, recall that r 4  For T /E a Galois extension of conductor dividing m, let p be a prime of E, and let τ ∈ Gal(T /E). Let (p, T /E) denote the conjugacy class of Gal(T /E) containing the Frobenius of p where p is a prime of T above p. Let Fix μ ∈ {±}. Let τ − denote the nontrivial element of Gal(F /K ) and let τ + denote the trivial element of Gal(F /K ) so that N(p) = μ1 mod 4 exactly when p ∈ A K F |K (τ μ ).
. Fix α ∈ M 4 and let σ ∈ Gal(L/K ) corresponding to α via the isomorphism induced by r 4 given in the proof following Definition 6. Then r −1 4 (α) = A K L|K (σ ). Letting σ denote the image of σ in the natural surjection to Gal(F /K ), By the Chebotarev Density Theorem, d(A K L|K (σ )) = 1 2 n . Since S has density 1, restricting densities of primes in K to those that split completely in K /Q does not change the density. Therefore ) which is equal to 1 2 by the Chebotarev Density Theorem. Therefore For p ∈ r −1 4 (α), the condition that N(α) = μ1 mod 4 means that N(p) = μ1 mod 4. This is equivalent to the condition that p ∈ A K F |K (τ μ ). Since p ∈ r −1 4 (α) = A K L|K (σ ), the condition that p ∈ A K F |K (τ μ ) is true exactly when σ = τ μ . This completes the proof.
Recall that Proposition 1 states that for p a prime of K with totally positive generator α ∈ O, and for σ ∈ Gal(K /Q) such that p and p σ are relatively prime, This motivates the following definition. Observe that is a well-defined map by Lemma 5. If (6) holds for some α ∈ O, then it holds for α σ for any σ ∈ Gal(K /Q). Therefore (α) = (α σ ) for all σ ∈ Gal(K /Q).
For N ∈ Z + , let R μ,N :={p ∈ R μ : p < N } and S μ,N :={p ∈ S μ : p < N }. We will prove that Then since K is cyclic of odd degree and 2 is inert in K /Q, we can apply Proposition 4 to get that #M 4 = 2 n . Then since half the elements of M 4 are in M + 4 and half in M − 4 , #M + 4 = #M − 4 = 2 n−1 . Let μ denote a fixed sign. Let S μ,N denote the set of primes of K laying above primes in S μ,N and let R μ,N denote the set of primes of K laying above primes in R μ,N . Since primes in S split completely, Let r 4,N denote the restriction of r 4 to S μ,N . Then R μ,N is the disjoint union taken over elements α ∈ ker( μ ), i.e. elements of α ∈ M 4 such that N(α) = μ1 mod 4 and (α) = 1. Therefore # S μ,N ∩ r −1 4,N (α) #S μ,N By Lemma 9, for all α ∈ ker( μ ),

Counting solutions to a Hilbert symbol condition
In this section, we assume that K /Q satisfies (C1)-(C4) and prove formulae for # ker( μ ). Throughout, the degree n := [K : Q] to taken as an odd integer. Fix τ to be a generator of Gal(K /Q). For any α ∈ K , write α (k) :=α τ k for k ∈ Z. Our aim is to count the number of elements in M 4 with a representative α ∈ O K satisfying the spin relation (α, α σ ) 2 = 1 for all non-trivial σ ∈ Gal(K /Q).
By Lemma 6, the property (6) only depends on the class of [α] ∈ M 4 .

The counting problem
Our aim is to obtain the size of the preimage of 0 and h(x) under B. For any polynomial f ,

Lemma 12
For any positive factor k = 1 of n, let d k be the order of 2 in (Z/k) × . Also set d 1 = 1. Consider the following factorisation in F 2 [x], Proof Take f to be an irreducible factor of x n − 1 in F 2 [x]. Let γ be a root of f in an extension of F 2 . Then γ is a primitive k-th root of unity, where k is some integer dividing n. Galois theory on finite fields shows that Gal(F 2 (γ )/F 2 ) is generated by the Frobenius ϕ : x → x 2 . Since ϕ i : x → x 2 i for any i ∈ Z, we see that the order of ϕ must be d k , the order which is closed under inversion precisely when d k is even. Therefore f is self-reciprocal if and only if d k is even. Let A k be the set of distinct irreducible factors of x n − 1 in F 2 [x] which has a primitive k-th root of unity in an extension of F 2 , and M k be a subset of A k containing elements which are self-reciprocal, so 2r k − m k = #A k and m k = #M k . If d k is even, then all f ∈ A k are self-reciprocals, so A k = M k and r k = m k . If d k is odd, then M k = ∅ and m k = 0.
There are φ(k) roots of x n − 1 which are primitive k-th root of unity, so (2r k − m k )d k = φ(k). Now the statement of the Lemma follows from r k = m k when d k is even, and m k = 0 when d k is odd.
We are now ready to prove the formulae for # ker( + ) and # ker( − ).

Proposition 5
For each k = 1 dividing n, let d k be the order of 2 in (Z/k) × . Then If n is a prime, then writing d = d n , In particular, when n = 3, # ker( + ) = 1 and # ker( − ) = 3.

Joint spins
Fix a sign μ ∈ {±}. Recall that S μ is the set of rational primes p ≡ μ1 mod 4 that split completely in K /Q, i.e., unramified and of residue degree 1 in K /Q, and that F μ is the set of p ∈ S μ of residue degree 1 in K (p)/Q. By Corollary 1, a prime p ∈ S μ belongs to F μ if and only if spin(p, σ ) = 1 for all non-trivial σ ∈ Gal(K /Q) and any prime ideal p of K lying above p. Recall that R μ is the set of primes p ∈ S μ such that spin(p, σ ) spin(p, σ −1 ) = 1 for all non-trivial σ ∈ Gal(K /Q) and all prime ideals p of K lying above p, so that F μ ⊂ R μ . In this section, we will prove the following formula for the relative density of F μ in R μ , denoted by d(F μ |R μ ).
Theorem 5 Assume Conjecture C η for η = 2 n(n−1) . Then We cannot apply the results of [6] directly for two reasons. First, the class number h of K need not be 1 -this forces us to relate spin(a, σ ) to quadratic residue symbols involving elements "smaller" than the totally positive generators of a h . Second, the sums (X; H, A) feature the additional restriction that r 4 (p) ∈ A. Since A is a collection of congruence classes modulo 4O, the restriction that r 4 (p) ∈ A is reminiscent of the restriction to a congruence class as in [5,Theorem 1.2,p. 699]. Despite the similarity, there is a technical difference that we will explain.
Fix once and for all a set C consisting of h unramified degree-one prime ideals in O that is a complete set of representatives of ideal classes in the class group of K ; its existence is guaranteed by an application of the Chebotarev Density Theorem to the Hilbert class field of K . Now suppose that a is a non-zero ideal in O coprime to p∈C N(p), and let α denote a totally positive generator of a h . As h is odd, the set {p 2 : p ∈ C} is also a complete set of representatives. Hence there exists p ∈ C such that ap 2 is a principal ideal. Let π denote a totally positive generator of the ideal p h . Let α 0 denote a totally positive generator of ap 2 . Then α h 0 and απ 2 are both totally positive generators of the ideal (ap 2 ) h , so we have since h is odd. Note that for each p ∈ C there is a bijection given by a → α 0 as above, and where D is a set of totally positive elements in O defined in [5, (4 We will now prove the following adaptation of [6, Theorem 1, p. 2]. Theorem 6 With notation as above, let H be a non-empty subset of {τ , . . . , τ n−1 2 }. Assume Conjecture C η holds true for η = 1/(|H|n) with δ = δ(η) > 0 (see [6, p. 7]). Let > 0 be a real number. Then for all X ≥ 2, we have where the implied constant depends only on and K .
Note that the set H above is of size at most n−1 2 . Since Conjecture C η 1 implies Conjecture C η 2 whenever η 1 ≤ η 2 , we see that, conditional on Conjecture C η for η = 2 n(n−1) , Theorem 6 implies (17) for each Gal(K /Q)-orbit A ∈ A 0 and each non-empty subset H ⊂ {τ , . . . , τ n−1 2 }, and hence also Theorem 5. It thus remains to prove Theorem 6. Let F be the integer defined in [6, (2.2), p. 5]; it depends only on K . Moreover, we can choose the sets C a and C b in [6, p. 5] so that their elements are coprime to p∈C N(p). Note that F is divisible by 32.
To deduce Theorem 6, it suffices to prove that because F has only finitely many prime ideal divisors.
The proof of Theorem 6 proceeds via Vinogradov's method, with suitable estimates necessary for the sums of type I where the implied constant depends only on K and .
where the implied constant depends only on K and .

Proof of proposition 6
The proof is very similar to the proof of [6, (2.5), p. 7], so we will outline the additional arguments necessary to prove Proposition 6. For each non-zero ideal a, there exists a prime ideal p ∈ C such that ap 2 is principal. We can thus write where 1(P) is the indicator function of a property P, and II (A) is the set of (ρ 1 , ρ 2 ) ∈ (O/(F )) × × (O/(F )) × such that α 0 ≡ ρ 1 mod F and β 0 ≡ ρ 2 mod F =⇒ (α 0 β 0 ) h ∈ A.
This finishes the proof of Proposition 7 and hence also of Theorem 6.

Proof of main results
We now prove Theorem 1.
Let E denote the set of rational primes p such that for p a prime of K above p, spin(p, σ ) = 1 for all non-trivial σ ∈ Gal(K /Q). For a fixed sign μ ∈ {±}, let E μ denote the set of primes of E congruent to μ1 mod 4. Conjecture 1.1 in [9] made two assertions, one regarding the density d(E|S) of such primes restricted to those splitting completely in K /Q and one regarding the overall density d(E) of such primes. The assertion regarding the restricted density is correct and the assertion regarding the overall density is slightly off due to a very simple error in the case in which p is not assumed to split completely in K /Q. Theorem 7 proves conditionally a slight modification of Conjecture 1.1 in [9] Theorem 7 [9] Let K be a number field with prime degree satisfying properties (C1)-(C4). Assume Conjecture C η holds for η = When n is prime, s = m K n + 1.
Proof If p is a prime of K that does not split completely in K /Q, then for some non-trivial σ ∈ Gal(K /Q), p σ = p so spin(p, σ ) = 0. Therefore E ⊆ S so this E is exactly the F studied in Theorem 1 and E μ = F μ .
In Theorem 2, K satisfies (C1), (C2), and (C4) directly from the assumptions. When K is a cyclic cubic number field with odd class number, by [1, Theorem V] all signatures are represented by units so condition (C3) is satisfied by Lemma 1 because h is odd. It is a consequence of the classical Burgess's inequality [2] that Conjecture C η is true for η = 2 3(3−1) = 1 3 , as is shown in Section 9 of [5]. Therefore Theorem 2 follows from Theorem 1 and is unconditional.
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