Local rings of cyclotomic integers

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

The character values of the group ring elements in this course lie in rings generated by roots of unity. Divisibility by a prime \(p\) in such a ring is controlled by local rings that are as simple as possible: every nonzero element is a unit times a power of a single element. This lesson proves the facts needed [OpenAI-H, Section 2], following the classical description of cyclotomic integers near \(p\) (compare [Turyn, pp. 320–321]): adjoining roots of unity of order prime to \(p\) gives a local ring whose maximal ideal is generated by \(p\) (Lemma 2.1), and adjoining a primitive \(p^k\)-th root of unity \(\xi\) then gives a local ring whose maximal ideal is generated by \(\xi-1\), with \(p\) a unit times \((\xi-1)^D\) (Lemma 3.1). Its last statement turns divisibility of a value \(R(\xi)\) by \(p^\ell\) into divisibility of the coefficients of \(R\). As a first application, Proposition 4.1 proves Turyn's restriction: the order of a circulant Hadamard matrix greater than \(1\) is \(4u^2\) with \(u\) odd.

We use the notation of Circulant Hadamard matrices and group rings, and basic ring theory: ideals, quotients, localization at a prime ideal, and the Chinese remainder theorem for polynomials over a field, as in the core course Abstract Algebra II.

1. Rings generated by roots of unity

Lemma 1.1. Let \(B\subseteq\mathbb C\) be the subring generated by finitely many roots of unity. Then the additive group of \(B\) is free abelian of finite positive rank, \(B\) is Noetherian, every localization of \(B\) is Noetherian, and \(B/pB\neq0\) for every prime \(p\).

Proof. Products of powers of the generators, with exponents below their orders, span \(B\) additively; so \(B\) is finitely generated, and torsion-free as a subgroup of \(\mathbb C\), hence free of finite rank, which is positive since \(1\in B\). An ideal is a subgroup of a finitely generated free abelian group, hence finitely generated; so \(B\) is Noetherian. For a multiplicative set \(S\), every ideal \(\mathfrak a\) of \(S^{-1}B\) is generated by \(\mathfrak a\cap B\) (if \(b/s\in\mathfrak a\), then \(b\in\mathfrak a\cap B\)), which is finitely generated. Finally \(B/pB\) is \((\mathbb Z/p)^{\operatorname{rank}B}\neq0\). \(\square\)

Lemma 1.2. Let \(O\) be a Noetherian local domain whose maximal ideal is generated by an element \(t\neq0\). Every nonzero \(a\in O\) is \(a=ct^j\) for a unit \(c\) and a unique \(j\ge0\).

Proof. If \(a\) is not a unit, \(a=ta_1\); if \(a_1\) is not a unit, \(a_1=ta_2\); and so on. The ideals \((a)\subseteq(a_1)\subseteq(a_2)\subseteq\dots\) increase strictly: \((a_i)=(a_{i+1})\) would give \(a_{i+1}=ea_i=eta_{i+1}\), so \(et=1\), and \(t\) would be a unit. By the Noetherian property the process stops at a unit \(a_j\), and \(a=a_jt^j\). If \(ct^j=c't^{j'}\) with \(j<j'\), cancelling \(t^j\) makes \(c\) a multiple of \(t\), impossible for a unit. \(\square\)

2. Roots of unity of order prime to p

Lemma 2.1. Let \(p\) be a prime, \(\eta\) a primitive \(L\)-th root of unity with \(p\nmid L\) (\(L=1\) allowed), \(A_0=\mathbb Z[\eta]\), \(\mathfrak m_0\) a maximal ideal of \(A_0\) containing \(p\), and \(A=(A_0)_{\mathfrak m_0}\). Then \(A\) is a Noetherian local domain with maximal ideal \(pA\), and every nonzero element of \(A\) is a unit times a power of \(p\).

Proof. \(A_0\) is the image of \(\mathbb Z[X]\) under \(X\mapsto\eta\), and \(X^L-1\) maps to \(0\); so \(A_0/pA_0\) is a quotient of \(\mathbb F_p[X]/(X^L-1)\). Since \(L\) is invertible in \(\mathbb F_p\), \(X^L-1\) is coprime to its derivative \(LX^{L-1}\), so it is a product of distinct irreducible factors, and by the Chinese remainder theorem \(\mathbb F_p[X]/(X^L-1)\) is a finite product of fields. A quotient of a finite product of fields is the product of some of the factors, so \(A_0/pA_0\) is a finite product of fields, and its localization at the image of \(\mathfrak m_0\), which is \(A/pA\), is one of these fields. So \(pA\) is a maximal ideal; it is contained in the unique maximal ideal \(\mathfrak m_0A\) of \(A\), hence equal to it. \(A\) is Noetherian by Lemma 1.1, and Lemma 1.2 with \(t=p\) gives the last statement. \(\square\)

3. A root of unity of p-power order

For \(k\ge1\) let \(\Phi_{p^k}(X)=\sum_{j=0}^{p-1}X^{jp^{k-1}}\), so that \(\Phi_{p^k}(X)(X^{p^{k-1}}-1)=X^{p^k}-1\), and \(D=(p-1)p^{k-1}=\deg\Phi_{p^k}\).

Lemma 3.1. Let \(A\) be as in Lemma 2.1, \(\xi\) a primitive \(p^k\)-th root of unity and \(\pi=\xi-1\).

(a) \(A[\xi]\cong A[X]/(\Phi_{p^k})\); it is a free \(A\)-module with basis \(1,\xi,\dots,\xi^{D-1}\).

(b) \(A[\xi]\) is a Noetherian local domain with maximal ideal \(\pi A[\xi]\), and \(p\) is a unit times \(\pi^D\).

(c) Every nonzero \(z\in A[\xi]\) is a unit times a power \(\pi^j\); put \(v(z)=j/D\), so that \(v(p)=1\), \(v(zz')=v(z)+v(z')\), and for every integer \(\ell\ge0\), \(v(z)\ge\ell\iff z\in p^\ell A[\xi]\). For \(R=\sum_{j<D}r_jX^j\in A[X]\): \(R(\xi)\in p^\ell A[\xi]\iff r_j\in p^\ell A\) for all \(j\).

Proof. Eisenstein. \(\Phi_{p^k}(1+Y)\) is monic of degree \(D\), has constant term \(\Phi_{p^k}(1)=p\), and is congruent to \(Y^D\) modulo \(p\), because in characteristic \(p\), \(\Phi_{p^k}(X)(X^{p^{k-1}}-1)=X^{p^k}-1\) reads \(\Phi_{p^k}(X)(X-1)^{p^{k-1}}=(X-1)^{p^k}\). For a nonzero polynomial over the fraction field of \(A\), let \(\mu\) be the least valuation of its coefficients (Lemma 2.1); \(\mu\) is additive under multiplication, because after scaling both factors to \(\mu=0\) their reductions modulo \(p\) are nonzero polynomials over the field \(A/pA\), whose product is nonzero. So a factorization of the monic \(\Phi_{p^k}(1+Y)\) into monic factors over the fraction field has both factors in \(A[Y]\), and reducing modulo \(p\) gives two monic factors of \(Y^D\), of positive degrees, with constant terms divisible by \(p\); then the constant term \(p\) of the product would be divisible by \(p^2\), which is false. So \(\Phi_{p^k}\) is irreducible over the fraction field of \(A\), and it is the minimal polynomial of \(\xi\).

(a) If \(f\in A[X]\) and \(f(\xi)=0\), divide by the monic \(\Phi_{p^k}\) in \(A[X]\): \(f=Q\Phi_{p^k}+R\), \(\deg R<D\), \(R(\xi)=0\), so \(R=0\) by minimality.

(b) \(A[\xi]\) is a localization of the ring \(\mathbb Z[\eta,\xi]\) generated by roots of unity, hence Noetherian by Lemma 1.1. Let \(\mathfrak M\) be a maximal ideal of \(A[\xi]\). The field \(A[\xi]/\mathfrak M\) is a finitely generated module over the domain \(A/(\mathfrak M\cap A)\), so each of its elements satisfies a monic equation over that domain (determinant trick); for nonzero \(a\) in the domain, the equation \(a^{-m}+c_1a^{-m+1}+\dots+c_m=0\) gives \(a^{-1}=-(c_1+c_2a+\dots+c_ma^{m-1})\), so the domain is a field and \(\mathfrak M\cap A=pA\). Hence \(\mathfrak M\supseteq pA[\xi]\), and \(A[\xi]/pA[\xi]\cong(A/pA)[Y]/(Y^D)\) with \(Y\) the image of \(\pi\), which has the single maximal ideal \((Y)\). So \(A[\xi]\) is local with maximal ideal \((p,\pi)\). Moreover, \(\Phi_{p^k}(1+\pi)=0\) and \(\Phi_{p^k}(1+Y)=Y^D+p\bigl(1+\sum_{j=1}^{D-1}c_jY^j\bigr)\) with \(c_j\in\mathbb Z\), so \(\pi^D=-p\,u_\pi\) with \(u_\pi=1+\sum c_j\pi^j\), a unit since its residue is \(1\). So \(p\in\pi A[\xi]\), the maximal ideal is \(\pi A[\xi]\), and \(p=-u_\pi^{-1}\pi^D\).

(c) The first statement is Lemma 1.2 with \(t=\pi\). Since \(p^\ell\) is a unit times \(\pi^{\ell D}\), \(v(z)\ge\ell\) iff \(z\in\pi^{\ell D}A[\xi]=p^\ell A[\xi]\). By (a), \(p^\ell A[\xi]=\bigoplus_{j<D}p^\ell A\,\xi^j\), which gives the coefficient statement. \(\square\)

If \(A=\mathbb Z_{(p)}\) (that is, \(L=1\)), complex conjugation maps \(A[\xi]\) to itself, and \(\overline\pi=\xi^{-1}-1=-\xi^{-1}\pi\) is an associate of \(\pi\); so \(v(\overline z)=v(z)\).

4. The order restriction

Proposition 4.1 (Turyn). If a circulant Hadamard matrix of order \(n>1\) exists, then \(n=4u^2\) for an odd integer \(u\ge1\).

Proof. Let \(h\in\mathbb Z[C_n]\) be the first row, with \(hh^*=n\) (Lemma 2.1 of Circulant Hadamard matrices and group rings). By Corollary 2.2 there, \(n=2^{2s}u^2\) with \(s\ge1\) and \(u\) odd. Suppose \(s\ge2\). Project \(C_n\) onto its quotient of order \(2^{2s}\); the image \(T\) of \(h\) has odd coefficients (each a sum of \(u^2\) signs) and \(TT^*=2^{2s}u^2\).

One step. Let \(T\in\mathbb Z[C_{2^k}]\), \(k\ge1\), have odd coefficients \(t_0,\dots,t_{2^k-1}\) and \(TT^*=2^{2t}u^2\) with \(t\ge2\). Use Lemma 3.1 with \(p=2\), \(A=\mathbb Z_{(2)}\) and a primitive \(2^k\)-th root \(\xi\). Evaluating, \(T(\xi)\overline{T(\xi)}=2^{2t}u^2\) with \(u^2\) a unit of \(A\); since \(v(\overline{T(\xi)})=v(T(\xi))\), \(v(T(\xi))=t\), so \(T(\xi)\in2^tA[\xi]\). Modulo \(\Phi_{2^k}(X)=1+X^{2^{k-1}}\), \(T\) reduces to \(\sum_{j<2^{k-1}}(t_j-t_{j+2^{k-1}})X^j\), so by Lemma 3.1(c) each difference \(t_j-t_{j+2^{k-1}}\) is divisible by \(2^t\). Let \(T'\in\mathbb Z[C_{2^{k-1}}]\) be the projection of \(T\) divided by \(2\): its coefficients \(\frac12(t_j+t_{j+2^{k-1}})=t_j+\frac12(t_{j+2^{k-1}}-t_j)\) are odd integers, as \(4\) divides the difference. Projection commutes with \(*\), so \(T'T'^*=2^{2(t-1)}u^2\).

Descent. Start with \(k=2s\), \(t=s\), and apply the step \(s-1\) times; each time \(t\ge2\) and \(k\ge s+2\ge1\). The result has \(2^{s+1}\) odd coefficients and \(T'T'^*=4u^2\). The coefficient of the identity in \(T'T'^*\) is the sum of the squares of the coefficients; odd squares are \(1\) modulo \(8\) and \(8\mid2^{s+1}\), so this sum is \(0\) modulo \(8\), while \(4u^2\equiv4\pmod8\). Hence \(s=1\). \(\square\)

5. Exercises

Exercise 5.1 (easy). For \(p=2\), \(k=2\), \(A=\mathbb Z_{(2)}\) and \(\xi=i\), show that \(2=i(i-1)^2\) and that \(A[i]\) has basis \(1,i\).

Exercise 5.2 (easy). Show that \(\Phi_{p^k}(1)=p\) and that \(\Phi_{p^k}(X)\) is the product of \(X-\zeta\) over the primitive \(p^k\)-th roots of unity \(\zeta\).

Exercise 5.3 (medium). Run the descent of Proposition 4.1 for \(n=16\) and \(n=64\): which contradiction is reached, and after how many steps?

Exercise 5.4 (medium). Show by an example that Lemma 2.1 fails when \(p\mid L\): take \(p=2\), \(L=4\), \(\eta=i\), and show that the maximal ideal of \(\mathbb Z[i]_{(1+i)}\) is not generated by \(2\).

6. Solutions

5.1. \((i-1)^2=-2i\), so \(i(i-1)^2=2\). \(\Phi_4=1+X^2\) has degree \(2\).

5.2. \(\Phi_{p^k}(1)=\sum_{j<p}1=p\). The roots of \(X^{p^k}-1\) not among those of \(X^{p^{k-1}}-1\) are exactly the primitive \(p^k\)-th roots, all simple.

5.3. For \(n=16\): \(s=2\), \(u=1\), and one step takes \(T\in\mathbb Z[C_{16}]\) (\(k=4\), \(t=2\)) to \(T'\in\mathbb Z[C_8]\) with eight odd coefficients and \(T'T'^*=4\); the sum of eight odd squares is \(0\) modulo \(8\), not \(4\). For \(n=64\): \(s=3\), and two steps end with sixteen odd coefficients and \(T'T'^*=4\), the same contradiction.

5.4. In \(\mathbb Z[i]\), \(2=-i(1+i)^2\), and \(1+i\) generates the only maximal ideal containing \(2\). In the localization, \((2)=(1+i)^2\) is strictly smaller than \((1+i)\): otherwise \(1+i=2c\), so \((1-i)c=1\) and \(1-i=-i(1+i)\) would be a unit.

References