Alternating character products

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

By the lesson Local rings of cyclotomic integers, a circulant Hadamard matrix of order greater than \(4\) would have order \(4u^2\) with \(u>1\) odd. Splitting off the factor \(C_4\) produces elements \(x\) of the group ring \(\mathbb Z[i][C_{u^2}]\) with \(xx^*=u^2\). This lesson studies all such \(x\) [OpenAI-H, Section 4]. For each set \(S\) of primes dividing \(u\), evaluate \(x\) at a character \(\chi_S\) that is nontrivial exactly on the components of those primes, and form the alternating product \(\Delta(x)=\prod_Sx_S^{\pm1}\). Comparing two characters that differ in one prime direction \(p\) gives a unit congruent to \(1\) near \(p\) (Lemma 1.1); since every factor of \(\Delta(x)\) can be paired in every direction, \(\Delta(x)\) is an algebraic integer, of absolute value \(1\) under every embedding, hence a root of unity by Kronecker's criterion (Lemma 2.1), and its order is odd (Proposition 3.1).

We use Circulant Hadamard matrices and group rings and Lemmas 1.1, 2.1 and 3.1 of Local rings of cyclotomic integers; and that commuting complex matrices of finite order are simultaneously diagonalizable (core course Linear Algebra; the argument is recalled in the proof of Lemma 2.1).

1. Comparing two character values

Lemma 1.1 (character comparison; OpenAI). Let \(A\) be as in Lemma 2.1 of Local rings of cyclotomic integers, for a prime \(p\), let \(e\ge1\), and let \(F,K\in A[C_{p^e}]\) satisfy \(FK=p^eb\) with \(b\in A^\times\). For a primitive \(p\)-th root of unity \(\rho\), the ratios \(F(\rho)/F(1)\) and \(K(\rho)/K(1)\) are units of the local ring \(A[\rho]\) with residue \(1\).

Here \(F(\rho)\) is the evaluation at the character sending the chosen generator to \(\rho\), and \(F(1)\) the augmentation.

Proof. The values \(F(1),K(1),F(\rho),K(\rho)\) are nonzero, since their products in pairs are \(p^eb\neq0\).

The case \(e=1\). The valuations of \(F(1),K(1)\in A\) are nonnegative integers with sum \(1\), so one of them, say \(F(1)\), is a unit. Since \(\rho\equiv1\) modulo the maximal ideal \((\rho-1)\) of \(A[\rho]\), \(F(\rho)\equiv F(1)\), so \(F(\rho)/F(1)\) is a unit with residue \(1\). The two ratios multiply to \(\frac{F(\rho)K(\rho)}{F(1)K(1)}=\frac{pb}{pb}=1\), so the other one is its inverse.

The case \(e\ge2\). Let \(\xi\) be a primitive \(p^e\)-th root of unity. In \(A[\xi]\), \(v(F(\xi))+v(K(\xi))=v(p^eb)=e\ge1\), so after renaming, \(v(F(\xi))\ge1\), that is \(F(\xi)\in pA[\xi]\). Write a polynomial representative as \(F=Q\Phi_{p^e}+R\) with \(\deg R<(p-1)p^{e-1}\); then \(R(\xi)=F(\xi)\), and by Lemma 3.1(c) there, all coefficients of \(R\) lie in \(pA\). Let \(q:A[C_{p^e}]\to A[C_{p^{e-1}}]\) be the projection sending generator to generator; it maps \(X^{p^{e-1}}\) to \(1\), so \(q(\Phi_{p^e})=p\), and \(q(F)\in pA[C_{p^{e-1}}]\). Put \(F'=q(F)/p\) and \(K'=q(K)\); then \(F'K'=p^{e-1}b\), dividing by \(p\) in the free \(A\)-module \(A[C_{p^{e-1}}]\). Evaluation at \(1\) and at \(\rho\) factors through \(q\), because \(\rho^{p^{e-1}}=1\); so \(F'(\rho)/F'(1)=F(\rho)/F(1)\) and \(K'(\rho)/K'(1)=K(\rho)/K(1)\). By induction on \(e\), the claim follows. \(\square\)

The exponent of the group must match the valuation of the scalar. For odd \(p\), \(J=\sum_{j<p}X^j\in\mathbb Z[C_p]\) and \(F=p-2J\) satisfy \(J^2=pJ\), \(J^*=J\) and \(FF^*=p^2\), but \(F(\rho)/F(1)=p/(-p)=-1\), whose residue is not \(1\); here the group has order \(p\) and the scalar has valuation \(2\).

2. Kronecker's criterion

Lemma 2.1. (i) (Kronecker) Let \(B\subseteq\mathbb C\) be generated by finitely many roots of unity. If \(z\in B\) and \(|\sigma(z)|=1\) for every ring homomorphism \(\sigma:B\to\mathbb C\), then \(z\) is a root of unity.

(ii) In a local domain whose residue field has characteristic \(p\), a root of unity with residue \(1\) has order a power of \(p\).

Proof. (i) Let \(b_1,\dots,b_d\) be a basis of the free abelian group \(B\) (Lemma 1.1 of Local rings of cyclotomic integers), and let \(M_b\in M_d(\mathbb Z)\) be the matrix of multiplication by \(b\). The matrices of the generating roots of unity commute and have finite order, so each is diagonalizable over \(\mathbb C\) (its minimal polynomial divides some \(X^m-1\), which has distinct roots), and they are simultaneously diagonalizable: the eigenspaces of one are invariant under the others, and one proceeds on each eigenspace. On a common eigenvector, multiplication by each \(b\in B\) acts by a scalar \(\sigma(b)\), and \(b\mapsto\sigma(b)\) is a ring homomorphism \(B\to\mathbb C\). So \(M_z\) is diagonalizable with eigenvalues \(\sigma(z)\) of absolute value \(1\). For each \(a\ge1\), \(M_z^a\) has integer characteristic polynomial with roots of absolute value \(1\), so its coefficients are integers bounded by binomial coefficients; there are finitely many such polynomials, hence finitely many possible eigenvalues of all powers. So each eigenvalue \(\lambda\) has two equal powers, and being nonzero, finite order. For a common multiple \(N\) of these orders \(M_z^N\) is the identity, and \(z^N=M_z^N(1)=1\).

(ii) Write the order as \(p^am\) with \(p\nmid m\). Then \(\zeta=\omega^{p^a}\) has order \(m\) and residue \(1\). If \(m>1\), then \((\zeta-1)(1+\zeta+\dots+\zeta^{m-1})\) equals \(\zeta^m-1=0\), and since \(\zeta\neq1\) in the domain this gives \(1+\zeta+\dots+\zeta^{m-1}=0\), whose residue is \(m\neq0\) in characteristic \(p\). So \(m=1\). \(\square\)

3. The alternating product

Fix an odd \(u>1\), let \(P=C_{u^2}\), and let \(I\) be the set of primes dividing \(u\), which is nonempty. For \(p\in I\), let \(X_p\) generate the \(p\)-component of \(P\), of order \(p^e\) with \(e=v_p(u^2)\), and fix a primitive \(p\)-th root of unity \(\rho_p\). For \(S\subseteq I\), let \(\chi_S\) be the character of \(P\) with \(\chi_S(X_p)=\rho_p\) for \(p\in S\) and \(\chi_S(X_p)=1\) for \(p\notin S\), and for \(x\in\mathbb Z[i][P]\) put \(x_S=\chi_S(x)\). These values lie in \[ B=\mathbb Z[i,\rho_p:p\in I]. \] Put \(\varepsilon_S=(-1)^{|S|}\). If \(xx^*=u^2\), then \(x_S\overline{x_S}=u^2\), so every \(x_S\neq0\), and we define \[ \Delta(x)=\prod_{S\subseteq I}x_S^{\varepsilon_S}\in\operatorname{Frac}(B). \] For \(p\in I\), grouping \(S\) with \(S\cup\{p\}\) for \(p\notin S\) gives the pairing in the direction \(p\), the identity (3.1): \[ \Delta(x)=\prod_{S\subseteq I,\ p\notin S}\Bigl(\frac{x_S}{x_{S\cup\{p\}}}\Bigr)^{\varepsilon_S}. \]

Proposition 3.1 (OpenAI). If \(x\in\mathbb Z[i][P]\) and \(xx^*=u^2\), then \(\Delta(x)\in B\) is a root of unity whose order is a power of \(p\) for every \(p\in I\). In particular its order is odd.

Proof. Near a prime \(p\in I\). Let \(\mathfrak m\) be a maximal ideal of \(B\) containing \(p\). Let \(A_0=\mathbb Z[i,\rho_q:q\in I\setminus\{p\}]\); it is \(\mathbb Z[\eta]\) for \(\eta=i\prod_{q\ne p}\rho_q\), a primitive \(L\)-th root of unity with \(L=4\prod_{q\neq p}q\), and \(p\nmid L\). The ideal \(\mathfrak m_0=\mathfrak m\cap A_0\) contains \(p\) and is maximal, \(A_0/pA_0\) being a product of fields; let \(A=(A_0)_{\mathfrak m_0}\), as in Lemma 2.1 of Local rings of cyclotomic integers. Let \(S\subseteq I\) with \(p\notin S\). Evaluating the components of \(P\) other than the \(p\)-component according to \(\chi_S\), and keeping \(X_p\) as a variable, is a ring homomorphism \(\mathbb Z[i][P]\to A_0[C_{p^e}]\); let \(F\) and \(K\) be the images of \(x\) and \(x^*\). Then \(FK=u^2=p^eb\) with \(b=u^2/p^e\), an integer prime to \(p\), hence a unit of \(A\); and \(F(1)=x_S\), \(F(\rho_p)=x_{S\cup\{p\}}\). By Lemma 1.1, \(x_{S\cup\{p\}}/x_S\) is a unit of \(A[\rho_p]\) congruent to \(1\) modulo \(\rho_p-1\). The ring \(A[\rho_p]\) lies in \(B_{\mathfrak m}\), and \(\rho_p-1\in\mathfrak mB_{\mathfrak m}\) because \((\rho_p-1)^p\equiv\rho_p^p-1=0\) modulo \(p\). By (3.1), \(\Delta(x)\) is a unit of \(B_{\mathfrak m}\) with residue \(1\).

Away from \(u\). If \(\mathfrak m\) contains no prime divisor of \(u\), then \(u^2\notin\mathfrak m\), so each \(x_S\) divides the unit \(u^2\) in \(B_{\mathfrak m}\) and is a unit; hence \(\Delta(x)\in B_{\mathfrak m}\).

Integrality. So \(\Delta(x)\in B_{\mathfrak m}\) for every maximal ideal \(\mathfrak m\) of \(B\). This forces \(\Delta(x)\in B\): for \(z\in\operatorname{Frac}(B)\setminus B\), the ideal \(\{c\in B:cz\in B\}\) is proper, so it lies in some maximal ideal \(\mathfrak m\), and \(z=a/s\) with \(s\notin\mathfrak m\) would put \(s\) in that ideal.

Absolute values. A ring homomorphism \(\sigma:B\to\mathbb C\) maps each root of unity \(\zeta\) to a root of unity, so \(\sigma(\overline\zeta)=\sigma(\zeta^{-1})=\overline{\sigma(\zeta)}\), and \(\sigma\) commutes with complex conjugation on \(B\). Hence \(|\sigma(x_S)|^2=u^2\). Writing \(\Delta(x)=P_0/Q_0\) with \(P_0,Q_0\) the products of the \(x_S\) with \(\varepsilon_S=1\) and \(\varepsilon_S=-1\), we have \(\sigma(Q_0)\neq0\) and \(\sigma(\Delta(x))=\sigma(P_0)/\sigma(Q_0)\), of absolute value \(u^{\sum_S\varepsilon_S}=u^0=1\), since \(\sum_S\varepsilon_S=(1-1)^{|I|}=0\). By Lemma 2.1(i), \(\Delta(x)\) is a root of unity.

Its order. For \(p\in I\), \(B/pB\neq0\), so some maximal ideal contains \(p\), and there \(\Delta(x)\) has residue \(1\); by Lemma 2.1(ii) its order is a power of \(p\). As \(u\) is odd, the order is odd. \(\square\)

If \(u\) has two distinct prime divisors, the order is a power of both, so \(\Delta(x)=1\). If \(u=p^a\), \(\Delta(x)\) can be nontrivial: \(x=p^aX\), with \(X\) a generator of \(P\), has \(xx^*=u^2\) and \(\Delta(x)=x_\varnothing/x_{\{p\}}=\rho_p^{-1}\). The next lesson uses only that the order is odd.

4. Exercises

Exercise 4.1 (easy). For \(I=\{p,q\}\), write \(\Delta(x)\) as a product of two ratios in the direction \(p\) and in the direction \(q\), as in (3.1).

Exercise 4.2 (easy). Verify \(J^2=pJ\) and \(FF^*=p^2\) for \(F=p-2J\) in \(\mathbb Z[C_p]\).

Exercise 4.3 (medium). (a) Show that integrality cannot be dropped in Lemma 2.1(i): \(z=\frac{3+4i}5\) has absolute value \(1\) under both ring homomorphisms \(\mathbb Q(i)\to\mathbb C\) but is not a root of unity. (b) Show that for \(z\in B\), \(|z|=1\) already implies \(|\sigma(z)|=1\) for every ring homomorphism \(\sigma:B\to\mathbb C\).

Exercise 4.4 (medium). Check the example \(\Delta(p^aX)=\rho_p^{-1}\).

5. Solutions

4.1. \(\Delta=\frac{x_\varnothing x_{\{p,q\}}}{x_{\{p\}}x_{\{q\}}}=\frac{x_\varnothing/x_{\{p\}}}{x_{\{q\}}/x_{\{p,q\}}}=\frac{x_\varnothing/x_{\{q\}}}{x_{\{p\}}/x_{\{p,q\}}}\).

4.2. \(XJ=J\), so \(J^2=\sum_jX^jJ=pJ\); \(J^*=\sum_jX^{-j}=J\). Then \((p-2J)^2=p^2-4pJ+4pJ=p^2\).

4.3. (a) \(z\) is a root of \(X^2-\frac65X+1\), its minimal polynomial over \(\mathbb Q\), which does not have integer coefficients; so \(z\) is not an algebraic integer, while roots of unity are. (b) \(z\overline z=1\) in \(B\), and \(\sigma\) commutes with complex conjugation on \(B\) (proof of Proposition 3.1), so \(\sigma(z)\overline{\sigma(z)}=1\).

4.4. Here \(P=C_{p^{2a}}\) is its own \(p\)-component, \(I=\{p\}\), \(x_\varnothing=p^a\) and \(x_{\{p\}}=p^a\rho_p\), so \(\Delta(x)=x_\varnothing x_{\{p\}}^{-1}=\rho_p^{-1}\).

References