Proof. A ring generated over \(\mathbb Z\) by finitely many roots of unity is finitely generated as an abelian group: products of bounded powers of the roots span it. It is torsion-free because it is a subring of \(\mathbb C\), and hence is free of finite rank. Every ideal, as a subgroup of this free abelian group, is finitely generated. Such a ring is therefore Noetherian, as are its localizations. This applies to \(A\) and to \(A[\xi]\), which is a localization of \(\mathbb Z[\eta,\xi]\).

We first determine the maximal ideal of \(A\). Since \(p\nmid L\), the polynomial \(X^L-1\) has no repeated factor over \(\mathbb F_p\). Thus \(\mathbb F_p[X]/(X^L-1)\) is a product of fields. Its quotient \(A_0/pA_0\) is also a product of fields. Localizing at the chosen maximal ideal selects one field factor, so \(A/pA\) is a field and the maximal ideal of \(A\) is \(pA\).

In a Noetherian local domain whose nonzero maximal ideal is generated by \(t\), a nonzero nonunit can be divided by \(t\). This division cannot continue indefinitely: the successive quotients would generate a strictly increasing chain of principal ideals, since equality at one step would, by cancellation, make \(t\) a unit. Thus every nonzero element is a unit times a power of \(t\). Applied to \(A\), this gives its integral valuation, with value \(1\) at \(p\).

We now adjoin \(\xi\). The polynomial \(\Phi_{p^k}(1+Y)\) is monic of degree \(D\), has constant term \(p\), and reduces to \(Y^D\) modulo \(p\). The last assertion follows by reducing \[\Phi_{p^k}(X)(X^{p^{k-1}}-1)=X^{p^k}-1\] in characteristic \(p\). It is irreducible over \(\mathop{\mathrm{Frac}}(A)\) by Eisenstein’s argument. Here is the needed form of that argument. For a nonzero polynomial over \(\mathop{\mathrm{Frac}}(A)\), the minimum valuation of its coefficients is additive under multiplication: after scaling both polynomials to minimum zero, their nonzero reductions have nonzero product over the field \(A/pA\). If a monic integral polynomial factors into monic polynomials, the two minima are at most zero and sum to zero; hence both factors have coefficients in \(A\). A nontrivial factorization of \(\Phi_{p^k}(1+Y)\) would therefore reduce to two monic positive-degree factors of \(Y^D\). Both constant terms would be divisible by \(p\), contradicting the constant term \(p\) of their product.

It follows that \(A[\xi]\simeq A[X]/(\Phi_{p^k}(X))\), giving the asserted basis. Every maximal ideal of this finite \(A\)-algebra contracts to a maximal ideal of \(A\). Indeed, the quotient by such a maximal ideal is a field finite as a module over the corresponding quotient domain of \(A\). The adjugate-matrix argument for multiplication on a finite generating set makes every element of that field integral over the subdomain. Applying a monic relation to the inverse of a nonzero element of the subdomain puts that inverse in the subdomain. The subdomain is therefore a field.

Write \(\pi=\xi-1\) and \(k_0=A/pA\). The reduction \[A[\xi]/pA[\xi]\simeq k_0[Y]/(Y^D)\] has one maximal ideal. Thus \(A[\xi]\) is local with maximal ideal \((p,\pi)\). The equation \(\Phi_{p^k}(1+\pi)=0\) has the form \[\pi^D=-p\,u_\pi,\qquad u_\pi=1+\sum_{j=1}^{D-1}c_j\pi^j,\qquad c_j\in A,\] with the sum empty when \(D=1\). The element \(u_\pi\) has residue one and is a unit. Hence the maximal ideal is \((\pi)\), and \(p\) is a unit times \(\pi^D\). The principal-ideal argument above now shows that every nonzero element of \(A[\xi]\) is a unit times a power of \(\pi\), giving the stated valuation. Since \(p^\ell\) is a unit times \(\pi^{\ell D}\), this description proves (3). The coefficient assertion follows from uniqueness in the free \(A\)-basis. ◻

Proof. For the matrix under consideration, the row norm (1) has already given \(n=2^{2s}u^2\) as in (2). It remains to show \(s=1\).

Project \(C_n\) onto \(C_{2^{2s}}\). The image \(T\) of \(h\) has odd integer coefficients, each being a sum of \(u^2\) signs, and \[TT^*=2^{2s}u^2.\] Suppose \(s\geq2\). We describe an operation starting with \[ T\in\mathbb Z[C_{2^k}],\qquad TT^*=2^{2t}u^2, \qquad k\geq1,\quad t\geq2, \tag{4}\] where every coefficient of \(T\) is odd. Use Lemma 3 with \(A=\mathbb Z_{(2)}\) and a primitive \(2^k\)-th root \(\xi\). Complex conjugation preserves \(A[\xi]\) and sends \(\xi-1\) to the associate \(-\xi^{-1}(\xi-1)\). It therefore preserves the normalized valuation. Evaluating (4) gives \[v(T(\xi))=t,\qquad T(\xi)\in 2^tA[\xi].\]

Write \(T=\sum_{j=0}^{2^k-1}t_jX^j\). Its remainder modulo \(\Phi_{2^k}(X)=1+X^{2^{k-1}}\) is \[\sum_{j=0}^{2^{k-1}-1}(t_j-t_{j+2^{k-1}})X^j.\] The coefficient statement in Lemma 3, with \(\ell=t\), shows that every displayed difference belongs to \(2^t\mathbb Z_{(2)}\). Each difference is an integer, so it is divisible by \(2^t\) in \(\mathbb Z\).

Let \(q:\mathbb Z[C_{2^k}]\to\mathbb Z[C_{2^{k-1}}]\) be the group projection. The coefficients of \(q(T)/2\) are \[\frac{t_j+t_{j+2^{k-1}}}{2}.\] They are odd integers: the two odd summands are congruent modulo \(4\), since \(t\geq2\). Thus \(T'=q(T)/2\) belongs to the smaller integer group ring. Projection commutes with the involution, and division of the projected norm by \(4\) gives \[T'T'^*=2^{2(t-1)}u^2.\]

Starting with \(k=2s\) and \(t=s\), perform this operation \(s-1\) times. After \(j\) steps the indices are \(k=2s-j\) and \(t=s-j\); every step starts with \(t\geq2\). The final element has \(2^{s+1}\) odd integer coefficients and product with its star equal to \(4u^2\). The identity coefficient of that product is the sum of the squares of those coefficients. Each odd square is \(1\) modulo \(8\), and \(2^{s+1}\) is divisible by \(8\). The sum is therefore \(0\) modulo \(8\), whereas \(4u^2\) is \(4\) modulo \(8\). This contradiction proves \(s=1\). ◻

Proof. All the values in the denominators are nonzero, since the product of the two corresponding values is the nonzero scalar \(p^e b\). The ratios therefore begin as elements of \(\mathop{\mathrm{Frac}}(A[\rho])\).

For \(e=1\), the nonnegative integral valuations of \(F(1)\) and \(K(1)\) sum to \(1\), so one is a unit of \(A\). Evaluation at \(\rho\) is congruent to evaluation at \(1\) modulo \(\rho-1\). For that factor, the ratio is therefore a unit of residue \(1\). The two ratios multiply to \(1\), proving the same assertion for the other factor.

Suppose \(e\geq2\). Let \(q:A[C_{p^e}]\to A[C_{p^{e-1}}]\) be the projection sending the chosen generator to the chosen generator. We will show that the projection of one factor is divisible by \(p\) in the smaller group ring. Evaluate at a primitive \(p^e\)-th root \(\xi\), and put \(D=(p-1)p^{e-1}\) as in Lemma 3. The two nonnegative valuations sum to \(e\), so at least one is at least \(1\). Interchange the factors if necessary so that \(F(\xi)\in pA[\xi]\).

Choose a polynomial representative of \(F\) and divide by the monic cyclotomic polynomial: \[F=Q\Phi_{p^e}+R,\qquad Q,R\in A[X],\quad \deg R<D.\] Since \(R(\xi)=F(\xi)\), the coefficient statement in Lemma 3 makes every coefficient of \(R\) divisible by \(p\). Moreover \(q(\Phi_{p^e})=p\). Thus \(q(F)\in pA[C_{p^{e-1}}]\), as required. Define \[F'=q(F)/p,\qquad K'=q(K).\] Then \(F'K'=p^{e-1}b\). Cancellation of \(p\) is valid here because the target group ring is a free \(A\)-module. The evaluations at \(1\) and \(\rho\) factor through \(q\), and scalar division cancels from their ratio. Thus both original comparison ratios are unchanged. Induction proves the assertion, whether the same factor or the other factor is divided at the next step. ◻

Proof. For (i), let \(d\) be the finite rank of the free additive group of \(B\), as in the proof of Lemma 3. Multiplication by an element of \(B\) is thus an integer matrix. On complexifying this representation, the matrices for the generating roots of unity commute and have finite order. Each is diagonalizable, since its minimal polynomial divides some \(X^m-1\), which has distinct complex roots. They are simultaneously diagonalizable: the eigenspaces of one are invariant under the others, and one repeats the argument on those eigenspaces.

On a common eigenvector, multiplication by any \(b\in B\) acts as a scalar \(\sigma(b)\). This is a well-defined unital ring homomorphism because it comes from the multiplication representation, so all relations between the root generators are respected. The multiplication matrix of \(z\) is diagonal in the same basis, with eigenvalues \(\lambda_1,\ldots,\lambda_d\) of modulus one. For each positive integer \(a\), its \(a\)-th power has an integer characteristic polynomial. The coefficient of degree \(d-j\) has absolute value at most \(\binom dj\). Only finitely many such polynomials can occur, so the union of their sets of roots is finite. Each sequence \(\lambda_j^a\) therefore takes only finitely many values. Since \(\lambda_j\ne0\), two equal powers imply that \(\lambda_j\) has finite order. A common multiple of these orders makes the multiplication matrix the identity; applying it to \(1\in B\) proves that a positive power of \(z\) is \(1\).

For (ii), write the order as \(p^a m\) with \(p\nmid m\). If \(m>1\), raising the root to the \(p^a\)-th power gives an element \(\zeta\) of order \(m\) and residue \(1\). In a domain, \((\zeta-1)(1+\zeta+\cdots+\zeta^{m-1})=0\) then implies \[1+\zeta+\cdots+\zeta^{m-1}=0.\] Its reduction says \(m=0\) in the residue field, a contradiction. Thus \(m=1\). ◻

Proof. Fix \(p\in I\) and a maximal ideal \(\mathfrak m\) of \(B\) containing \(p\). Let \[A_0=\mathbb Z[i,\{\rho_q:q\in I\setminus\{p\}\}],\qquad \mathfrak m_0=\mathfrak m\cap A_0,\qquad A=(A_0)_{\mathfrak m_0}.\] Put \[\eta=i\prod_{q\in I\setminus\{p\}}\rho_q,\qquad L=4\prod_{q\in I\setminus\{p\}}q,\] with empty products equal to \(1\). The root orders \(4\) and \(q\in I\setminus\{p\}\) are pairwise coprime, so \(\eta\) is a primitive \(L\)-th root, \(p\nmid L\), and each original root is a power of \(\eta\). Thus \(A_0=\mathbb Z[\eta]\) has the form required by Lemma 3. The ideal \(\mathfrak m_0\) is maximal: it is prime and contains \(p\), and \(A_0/pA_0\) is a product of fields as in that lemma.

Let \(p^e\) be the order of the \(p\)-component, so \(e=v_p(u^2)\). For \(p\notin S\), evaluate every other component in \(xx^*=u^2\) according to \(\chi_S\), leaving the \(p\)-generator as a variable. This ring homomorphism produces two factors \(F,K\in A_0[C_{p^e}]\) with \[FK=p^e b,\qquad b=u^2/p^e\in A^\times.\] Both factors already have coefficients in \(A_0\): in the image of \(x^*\), Gaussian coefficients are conjugated and the evaluated roots are inverted, and these operations preserve \(A_0\). After localization, Lemma 5 therefore applies to these two factors. It gives \[ \frac{x_{S\cup\{p\}}}{x_S}\in A[\rho_p]^\times, \qquad \frac{x_{S\cup\{p\}}}{x_S}\equiv1\pmod{\rho_p-1}. \tag{8}\]

Every denominator inverted in \(A\) lies outside \(\mathfrak m\), so the natural inclusion of \(A[\rho_p]\) in \(B_{\mathfrak m}\) is defined. In the residue field of \(B_{\mathfrak m}\), characteristic \(p\) gives \((\rho_p-1)^p=\rho_p^p-1=0\). Hence \(\rho_p-1\) has residue zero, and the units in (8) retain residue \(1\) in \(B_{\mathfrak m}\). Pairing \(S\) with \(S\cup\{p\}\) yields the identity \[ \Delta(x)=\prod_{\substack{S\subseteq I\\p\notin S}} \left(\frac{x_S}{x_{S\cup\{p\}}}\right)^{\varepsilon_S}. \tag{9}\] Thus \(\Delta(x)\) is a unit of residue \(1\) in \(B_{\mathfrak m}\). The left side of (9) is the same element when the pairing direction changes. Applying the argument for every \(p\mid u\) and every maximal ideal over \(p\) gives this local conclusion at all of those ideals.

At a maximal ideal containing none of the prime divisors of \(u\), the scalar \(u^2\) is a unit. Equation (6) then makes every \(x_S\) a unit there. Consequently \(\Delta(x)\) belongs to \(B_{\mathfrak m}\) for every maximal ideal \(\mathfrak m\) of \(B\). The intersection of these localizations inside \(\mathop{\mathrm{Frac}}(B)\) is \(B\). Indeed, if \(z\notin B\), the ideal \[D_z=\{b\in B:bz\in B\}\] is proper. A maximal ideal \(\mathfrak m\) containing it cannot have \(z\in B_{\mathfrak m}\): such membership would give \(z=a/s\) with \(s\notin\mathfrak m\), whereas \(sz=a\in B\) puts \(s\) in \(D_z\). This proves \(\Delta(x)\in B\).

Every unital homomorphism \(\sigma:B\to\mathbb C\) sends the root generators to roots of unity and commutes with complex conjugation on their integer polynomials. Applying it to (6) gives \(|\sigma(x_S)|=u\). In particular these images are nonzero. Let \(P_0\) and \(Q_0\) be the products of the \(x_S\) with \(\varepsilon_S=1\) and \(\varepsilon_S=-1\), respectively. Then \(\Delta(x)=P_0/Q_0\), and the preceding nonvanishing gives \(\sigma(Q_0)\ne0\). Applying \(\sigma\) to \(Q_0\Delta(x)=P_0\) in \(B\) therefore evaluates the fraction in (7). Since \(I\ne\varnothing\), \[\sum_{S\subseteq I}\varepsilon_S=(1-1)^{|I|}=0.\] It follows that \(|\sigma(\Delta(x))|=1\). Lemma 7(i) makes \(\Delta(x)\) a root of unity.

For each \(p\in I\), there is a maximal ideal of \(B\) containing \(p\): the additive group of \(B\) is free of positive finite rank, so \(B/pB\ne0\). At that ideal \(\Delta(x)\) has residue \(1\), so Lemma 7(ii) makes its order a power of \(p\). The assertions about odd order and two distinct prime divisors follow. ◻