Proof. The ‘if’ direction is immediate. For the converse, if the $\omega _j$ are not all equal, then the triangle inequality gives $|\mu |<1$ and, similarly, each Galois conjugate $\mu '$ satisfies $|\mu '|\le 1$ with at least one strict inequality. The product of all conjugates of $\mu $ is an algebraic integer of absolute value $<1$ ; hence $0$ , so $\mu $ itself must be $0$ .

Proof. For each $b\in \mathbb {Z}_n$ , Lemma 2.1 applied to the multiset $\{\omega ^{\,f(x)+bx}:x\in \mathbb {Z}_n\}$ says that either the sum $\sum _x\omega ^{\,f(x)+bx}$ vanishes or all terms are equal, that is, $f(x)+bx$ is constant modulo n. Suppose that no b makes $f(x)+bx$ constant modulo n. Then, for every b,

$$ \begin{align*} \sum_{x=0}^{n-1}\omega^{\,f(x)+bx}=0. \end{align*} $$

Summing these equalities over all $b\in \mathbb {Z}_n$ and reversing the order of summation gives

$$ \begin{align*} \sum_{b=0}^{n-1}\sum_{x=0}^{n-1}\omega^{\,f(x)+bx} =\sum_{x=0}^{n-1}\omega^{\,f(x)}\!\sum_{b=0}^{n-1}\omega^{\,bx} = n\,\omega^{\,f(0)}\neq 0, \end{align*} $$

which is a contradiction. Hence, there exists $b_0$ and c with $f(x)+b_0x\equiv c$ for all x. Thus, $f(x)\equiv \alpha x+\beta \pmod n$ as required.

Proof. By (4.1), it suffices to show $\mu \in (\mathcal {O}_F)_{\mathfrak {q}}$ for every nonzero prime $\mathfrak {q}\subset \mathcal {O}_F$ .

Case 1: $\mathfrak {q}\nmid \mathfrak {p}$ . In the discrete valuation ring $(\mathcal {O}_F)_{\mathfrak {q}}$ , we have $v_{\mathfrak {q}}(p)=0$ , so $p^r$ is a unit. Since $S\in \mathcal {O}_F\subset (\mathcal {O}_F)_{\mathfrak {q}}$ , it follows that

$$ \begin{align*} v_{\mathfrak{q}}(\mu)=v_{\mathfrak{q}}(S)-v_{\mathfrak{q}}(p^r)\ge 0-0=0. \end{align*} $$

Hence, $\mu \in (\mathcal {O}_F)_{\mathfrak {q}}$ .

Case 2: $\mathfrak {q}=\mathfrak {p}$ . By hypothesis, $\mu \in \mathcal {O}_{F_{\mathfrak {p}}}$ . Since $(\mathcal {O}_F)_{\mathfrak {p}}=\mathcal {O}_{F_{\mathfrak {p}}}\cap F$ (as subrings of $F_{\mathfrak {p}}$ ), we conclude $\mu \in (\mathcal {O}_F)_{\mathfrak {p}}$ .

Therefore, $\mu \in (\mathcal {O}_F)_{\mathfrak {q}}$ for all nonzero primes $\mathfrak {q}$ and (4.1) yields $\mu \in \mathcal {O}_F$ .

Proof. By Theorem 4.2, the hypothesis implies $S/p^r\in \mathcal {O}_F$ . Applying the roots-of-unity average criterion (Lemma 2.1) to the multiset of terms, we see that the average $S/p^r$ is an algebraic integer only in the two cases stated.

Proof. $(\Rightarrow )$ Suppose f yields a perfect isometry. By condition (i), for every $a,b\in \mathbb {Z}_n$ ,

$$ \begin{align*} \frac{1}{n}\sum_{j=0}^{n-1}\omega^{a f(j)+bj} \in \mathbb{Z}_p[\omega]. \end{align*} $$

Fix $a=1$ . Then, for all b,

$$ \begin{align*} \frac{1}{n}\sum_{j=0}^{n-1}\omega^{f(j)+bj} \in \mathbb{Z}_p[\omega]. \end{align*} $$

By Theorem 4.2, this average is an algebraic integer, so Theorem 3.1 applies and gives $\alpha ,\beta \in \mathbb {Z}_n$ with

$$ \begin{align*} f(j)\equiv \alpha j + \beta\, \pmod n \quad\text{for all }j. \end{align*} $$

Since f is bijective, multiplication by $\alpha $ must be invertible modulo n. It follows that $\gcd (\alpha ,n)=1$ .

$(\Leftarrow )$ Conversely, assume $f(x)\equiv \alpha x+\beta $ with $\gcd (\alpha ,n)=1$ . Then,

$$ \begin{align*} \mu_I(g^a,g^b) = \sum_{j=0}^{n-1}\omega^{a(\alpha j+\beta)+bj} = \omega^{a\beta}\sum_{j=0}^{n-1}\omega^{(a\alpha+b)j}. \end{align*} $$

The inner sum equals n if $a\alpha +b\equiv 0\pmod n$ and $0$ otherwise, so

$$ \begin{align*} \frac{\mu_I(g^a,g^b)}{n} = \begin{cases} \omega^{a\beta}\in \mathbb{Z}_p[\omega] & \text{if }b\equiv -a\alpha\, \pmod n, \\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

Thus, condition (i) holds. For condition (ii), if $\mu _I(g^a,g^b)\ne 0$ , then $b\equiv -a\alpha \pmod n$ and because $\alpha $ is a unit, we have $a=0$ if and only if $b=0$ . Hence, $g^a$ and $g^b$ are simultaneously the identity or both nonidentity elements, satisfying the separation property.