Proof. \((1)\Rightarrow (2)\): Suppose that \(\alpha =\sum\limits_k n_k[a_k]\in \mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]\) and \(\overline{\alpha}=0\). Then \[\overline{\alpha}=\sum_{1\leq a\leq d} n_a [\overline{a}],\] where \(n_a=\sum\limits_{\substack{k:\, a_k\equiv a\mod d}} n_k\). Because \(\overline{\alpha}=0\), \(n_a=0\), for all \(1\leq a\leq d\). We have \[\alpha= \sum\limits_k n_k[a_k] =\sum_{1\leq a\leq d }\sum\limits_{k:\,a_k\equiv a} n_k[a_k]=\sum_{1\leq a\leq d }\sum\limits_{k:\,a_k\equiv a} n_k([a_k]-[a]).\] This implies that \(\alpha\equiv 0\mod {[d]}\).

\((2)\Rightarrow (1)\): Suppose that \(\alpha\equiv 0\mod [d]\). Then \(\alpha=\sum n_k([a_k]-[b_k])\), where \(a_k\equiv b_k\mod d\) for every \(k\). Clearly, \(\bar{\alpha}=\phi_d(\alpha)=\sum n_k([\bar{a_k}]-[\bar{b_k}])=0\) in \(\mathop{\mathrm{\mathbb{Z}}}[(\mathop{\mathrm{\mathbb{Z}}}/d\mathop{\mathrm{\mathbb{Z}}})^\circ]\).

Now we assume further that \(\alpha\in \mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{N}}}^\circ]\).

\((2)\Rightarrow (3)\): Suppose that \(\alpha\equiv 0\mod [d]\) and \(f(x)\) is a polynomial in \(\mathop{\mathrm{\mathbb{Z}}}[x]\). We can write \(\alpha =\sum_{k} n_k([a_k]-[b_k])\), where \(n_k\in \mathop{\mathrm{\mathbb{Z}}}\) and \(a_k\equiv b_k\mod d\). We have \[\alpha f(x)=\sum_k n_k (f(x^{a_k})-f(x^{b_k})).\] Since \(a_k\equiv b_k\mod d\), this implies that \(f(x^{a_k})-f(x^{b_k})\) is divisible by \(x^d-1\). Hence \(x^d-1\) divides \(\alpha f(x)\).

\((3)\Rightarrow (4)\): Clearly, \((3)\) implies \((4)\).

\((4)\Rightarrow (1)\): Suppose that \(x^d-1\) divides \(\alpha x\). We can write \[\alpha x= (x^d-1)(\sum_{k=0}^n n_k x^{b_k}),\] for some \(n_k\in \mathop{\mathrm{\mathbb{Z}}}\), \(b_k\in \mathop{\mathrm{\mathbb{N}}}\). We have \[\alpha x = \sum_{k=0}^n n_k (x^{b_k+d}-x^{b_k}) =\left( \sum_{k=0}^n n_k ([b_k+d]-[b_k])\right ) x.\] Hence \(\alpha=\sum_{k=0}^n n_k ([b_k+d]-[b_k])\). Thus \(\overline{\alpha}=0\). ◻

Proof. The proof is straightforward. ◻

Proof. Write \(\alpha=\sum_{m\in \mathop{\mathrm{{\rm Supp}}}(\alpha) } a_m[m]\). Since \(\phi_d(\alpha)=0\), for each residue class \(r\mod d\), we have \[\sum_{\substack{m\equiv r\mod d\\ m\in \mathop{\mathrm{{\rm Supp}}}(\alpha)} } a_m=0.\] Therefore \[\alpha f(\zeta_d) = \sum_{m\in \mathop{\mathrm{{\rm Supp}}}(\alpha) } a_m f(\zeta_d^m) =\sum_{r\mod d} \left(\sum_{\substack{m\equiv r\mod d\\ m\in \mathop{\mathrm{{\rm Supp}}}(\alpha)} } a_m\right) f(\zeta_d^r)=0. \qedhere\] ◻

Proof. \((1)\Rightarrow (2)\): This is trivial.

\((2)\Rightarrow (3)\): Suppose that the set of divisors of \(N\mid n\) can be paired as \(\{a_1,b_1\},\ldots, \{a_l,b_l\}\) such that \(\mu(a_i)+\mu(b_i)=0\) and \(a\equiv b\pmod d\) for every \(i\). We have \[\begin{align} \psi_N &= \sum_{m \mid N} \mu(m) [m] = \sum_{i=1}^{l} \mu(a_i)[a_i]+ \mu(b_i)[b_i] \\ &= \sum_{i=1}^{l} \mu(a_i) ([a_i]-[b_i]). \end{align}\] Since \(d \mid a_i-b_i,\) for all \(i \in \{1,2,\ldots,l\},\) we have \(\phi_d(\psi_N)= 0.\) Because \(n\) is squarefree, \(\psi_n = \psi_N \psi_{n/N}\) and hence \(\phi_d(\psi_n)=\phi_d(\psi_N)\phi_d(\psi_{n/N})=0\).

\((3)\Rightarrow (1)\): For each residue class \(r\), let \[S_r=\{a\mid n : a\equiv r\pmod d\},\] and decompose it according to the sign of \(\mu\): \[S_r^{+}=\{a\in S_r : \mu(a)=1\}, \qquad S_r^{-}=\{a\in S_r : \mu(a)=-1\}.\] Set \(p_r = |S_r^{+}|\) and \(q_r = |S_r^{-}|\).

Since \(\phi_d(\psi_n)=0,\) we have \(\sum_{a\mid n}\mu(a)[\overline{a}]=0\) in \(\mathbb{Z}[(\mathbb{Z}/d\mathbb{Z})^\circ]\). Equating the coefficient of each residue class \(r\) gives \(p_r-q_r=0,\) that is, \(p_r=q_r\) for every residue class \(r\). Hence, for each residue class \(r\), the sets \(S_r^+\) and \(S_r^-\) have the same cardinality. We may therefore pair each element of \(S_r^+\) with a unique element of \(S_r^-\). Thus the set of divisors of \(n\) can be partitioned into pairs \[\{a_1,b_1\},\ldots,\{a_\ell,b_\ell\}\] such that \(a_i\equiv b_i \pmod d\) and \(\mu(a_i)+\mu(b_i)=0\) for every \(i\). ◻

Proof. Since \(\beta\in {\rm Ann}_R(f)\), \(\beta f=0\) and hence \(\alpha f=(\alpha-\beta )f\) in \(\mathop{\mathrm{\mathbb{Q}}}(x)\). Note that \(\mathop{\mathrm{{\rm Supp}}}(\alpha-\beta)\subset \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup \mathop{\mathrm{{\rm Supp}}}(\beta)\). Hence \(f\) is defined at \(\zeta_d^a\) for each \(a\in \mathop{\mathrm{{\rm Supp}}}(\alpha-\beta)\). By Proposition 7, \(\Phi_d\) is a factor of \((\alpha-\beta )f=\alpha f\). ◻

Proof. Clearly, we have \(\phi_d(\alpha)=\sum c_r[r]\). On the other hand, \(\phi_d(\langle [1] + [-1] \rangle)= \langle [1] + [-1] \rangle \subset \mathop{\mathrm{\mathbb{Z}}}[(\mathop{\mathrm{\mathbb{Z}}}/d)^\circ]\) and an arbitrary element in \(\phi_d(\langle [1] + [-1] \rangle)\) is of the form \[\left(\sum_{r} m_r [r]\right) ([1] + [-1]) = \sum_{r} m_r ([r] + [-r]).\] From this we see that \(\phi_d(\alpha)\in \phi_d(\langle [1] + [-1] \rangle)\) if and only if \(c_r=c_{-r}\) for \(r\not\equiv -r\pmod d\), and \(c_r\) is even for \(r\equiv -r\pmod d\). Note that \(r\equiv -r\pmod d\) if and only if \(2r\equiv 0\pmod d\) if and only if \(r=0\) or \(r=d/2\) if \(d\) is even. ◻

Proof. Write \(\alpha=\sum\limits_{a\in \mathop{\mathrm{{\rm Supp}}}(\alpha)} n_a[a]\). For each residue class \(r\in (\mathop{\mathrm{\mathbb{Z}}}/d)^\circ\), define \(c_r=\sum\limits_{\substack{a\equiv r\mod d\\ a\in \mathop{\mathrm{{\rm Supp}}}(\alpha)}}n_a\). By the previous lemma, \(c_r=c_{-r}\) for \(r\not\equiv -r\pmod d\) and \(c_r\) is even for \(r\equiv -r\pmod d\). Now for each pair of non-self-inverse classes \(\{r,-r\}\) such that \(c_r\not=0\), we choose a representative \(a_r\in \mathop{\mathrm{{\rm Supp}}}(\alpha)\) and set \(\beta_{\{r,-r\}}= c_r([a_r]+[-a_r])\). For class \(r\) with \(r\equiv-r\pmod d\), we have \(c_r=2 c^\prime_ r\), and if \(c_r\not=0\) then we choose a representative \(a_r\in \mathop{\mathrm{{\rm Supp}}}(\alpha)\) and set \(\beta^\prime_{r}= c^\prime_r ([a_r]+[-a_r])\). Let \(\beta\) be the sum of those \(\beta_{\{r,-r\}}\) and \(\beta^\prime_r\). Then clearly \(\beta\) is in \(\langle [1] + [-1] \rangle\), and \(\mathop{\mathrm{{\rm Supp}}}(\beta)\subseteq \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha))\) and \(\phi_d(\beta)=\phi_d(\alpha)\). ◻

Proof. Since \(I:=\langle [1] + [-1] \rangle \subset {\rm Ann}_{\mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^{\circ}]}(f)\), one has \(([1]+[-1])f=0\), i.e., \(f(x)+f(1/x)=0\). From this we see that if \(f\) is defined at \(\zeta_d^a\) then \(f\) is also defined at \(\zeta_d^{-a}\). Hence \(f\) is defined at \(\zeta_d^a\) for each \(a\in \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha))\).

By the previous lemma, there exists \(\beta\in I\) such that \(\phi_d(\beta)=\phi_d(\alpha)\) and \(\mathop{\mathrm{{\rm Supp}}}(\beta)\subseteq \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha)).\) Clearly, \(\mathop{\mathrm{{\rm Supp}}}(\alpha)\cup\mathop{\mathrm{{\rm Supp}}}(\beta) \subseteq \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha)).\) Hence \(f\) is defined at \(\zeta_d^a\) for each \(a\in \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup \mathop{\mathrm{{\rm Supp}}}(\beta)\). By Theorem 13, \(\Phi_d\) is a factor of \(\alpha f\). ◻

Proof. \((1)\Rightarrow (2):\) Take \(N := n.\) The conclusion follows immediately.

\((2)\Rightarrow (3):\) Suppose that the set of divisors of \(N\) can be partitioned into pairs \[\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_\ell,b_\ell\},\] such that \(d\mid \mu(a_i)a_i+\mu(b_i)b_i,\) for every \(i\). Let \[\alpha_1=\sum_{\mu(a_i)=\mu(b_i)}(\mu(a_i)[a_i]+\mu(b_i)[b_i]),\quad\text{ and } \alpha_2=\sum\limits_{\mu(a_i)=-\mu(b_i)}(\mu(a_i)[a_i]+\mu(b_i)[b_i]).\] We have \(\psi_N=\sum_{a\mid N}\mu(a)[a]=\alpha_1+\alpha_2\).

For each pair \((a_i,b_i),\) we have \(d\mid \mu(a_i)a_i+\mu(b_i)b_i.\) If \(\mu(a_i)=\mu(b_i),\) then \(d\mid(a_i+b_i)\) and \(b_i\equiv -a_i\pmod d\). Hence \[\phi_d(\alpha_1) =\sum_i\mu(a_i)\bigl([\overline{a_i}]+[\overline{-a_i}]\bigr) \in\phi_d(I).\] If \(\mu(a_i)=-\mu(b_i),\) then \(d\mid(a_i-b_i)\) and \(\phi_d(\alpha_2)=\sum\limits_{\mu(a_i)=-\mu(b_i)}(\mu(a_i)[\overline{a_i}]+\mu(b_i)[\overline{b_i}]) =0.\) Thus \[\phi_d(\psi_N)=\phi_d(\alpha_1+\alpha_2)=\phi_d(\alpha_1)\in\phi_d(I).\] Since \(\phi_d\) is a surjective ring homomorphism, \(\phi_d(I)\) is an ideal of \(\mathop{\mathrm{\mathbb{Z}}}[(\mathop{\mathrm{\mathbb{Z}}}/d)^\circ]\). Since \(n\) is squarefree and \(N\mid n\), \(\gcd(N,n/N)=1\) and \(\psi_n = \psi_N \psi_{n/N}.\) Hence \[\phi_d(\psi_n)=\phi_d(\psi_N)\phi_d(\psi_{n/N})\in \phi_d(I).\]

\((3)\Rightarrow (1):\) For each residue class \(r\), let \[S_r=\{a\mid n : a\equiv r\pmod d\},\] and decompose it according to the sign of \(\mu\): \[S_r^{+}=\{a\in S_r : \mu(a)=1\}, \qquad S_r^{-}=\{a\in S_r : \mu(a)=-1\}.\] Set \(p_r = |S_r^{+}|\) and \(q_r = |S_r^{-}|\).

The condition for \(\phi_d(\psi_n)\in \phi_d(I)\) is equivalent to \[p_r - q_r = p_{-r} - q_{-r},\] for all non-self-inverse residue classes \(r\not=-r\), and \[p_r - q_r\equiv 0\pmod 2,\] for self-inverse residue classes \(r=-r\).

Consider a non-self-inverse class \(r\) (\(r\not=-r\)). Without loss of generality, we may assume that \(p_r \geq p_{-r}\). Then \(s:=q_r-q_{-r}=p_{r}-p_{-r}\geq0\). We pair \(p_{-r}\) elements in \(S_r^+\) with \(p_{-r}\) elements in \(S_{-r}^+\). We pair \(q_{-r}\) elements in \(S_r^-\) with \(q_{-r}\) elements in \(S_{-r}^-\). Clearly, for such a pair \(\{a,b\}\) we have \(a\equiv -b\pmod d\) and \(\mu(a)=\mu(b)\). The remaining \(s\) elements in \(S_r^{+}\) are paired with \(s\) elements in \(S_{r}^{-}\). For such a latter pair \(\{a,b\}\) we have \(a\equiv b\equiv r\pmod d\) and \(\mu(a)=-\mu(b)\). In any case, for each obtained pair \(\{a_i,b_i\}\), \[d \mid \mu(a_i)a_i+\mu(b_i)b_i.\]

Now consider a self-inverse class \(r\) (\(r=-r\)). We have \(|S_r|=p_r+q_r\equiv p_r-q_r\equiv 0\pmod 2\). Hence \(S_r\) has an even number of elements. We pair the elements of \(S_r\) arbitrarily. For such a pair \(\{a,b\}\) we have \(a\equiv b\equiv r\equiv -b\pmod d\). Hence \(d\mid \mu(a)a+\mu(b)b\). ◻

Proof. We recall the following identity [15]. For any positive integer \(n\), it is well known that \[\varphi(n) = \sum_{e \mid n} \mu(e)\,\frac{n}{e}.\] In particular, when \(n\) is squarefree, this identity reduces to \[\varphi(n) = \mu(n) \sum_{e \mid n} \mu(e)\, e.\]

By Proposition 21, there exists a partition of the set of all divisors of \(n\) into pairs \[\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_\ell,b_\ell\},\] such that \(d\mid \mu(a_i)a_i+\mu(b_i)b_i\) for every \(i\). Summing over all pairs gives \[d\mid \sum_{i=1}^\ell \mu(a_i)a_i+\mu(b_i)b_i = \sum_{e\mid n} \mu(e)e=\mu(n)\varphi(n).\] Hence \(d\mid\varphi(n)\). ◻

Proof. Suppose that \(\phi_d(\psi_n)\in \phi_d(\langle [1] + [-1] \rangle)\). Then \(d \mid \varphi(n)=\prod_{p \mid n}(p-1).\) Since \(d\) is a prime number, there exists \(p\) such that \(d \mid p-1.\) Conversely, suppose that \(d \mid p-1\) for some prime divisor \(p\) of \(n.\) Then, we can apply the second criterion in 21 for \(N=p\). ◻

Proof. \((1)\Rightarrow (2):\) Take \(N := n.\) The conclusion follows immediately.

\((2)\Rightarrow (3):\) Suppose that the set of divisors of \(N\) can be partitioned into pairs \[\{a_1,b_1\},\{a_2,b_2\},\ldots,\{a_\ell,b_\ell\},\] such that \(\mu(a_i)+\mu(b_i)=0\) and either \(d\mid a_i-b_i\) or \(d\mid a_i+b_i\), for every \(i\). Let \(I=\langle[1]-[-1]\rangle\). Let \[\alpha_1=\sum_{d\mid a_i+b_i}(\mu(a_i)[a_i]+\mu(b_i)[b_i]),\quad\text{ and } \quad \alpha_2=\sum\limits_{d\mid a_i-b_i}(\mu(a_i)[a_i]+\mu(b_i)[b_i]).\] Clearly, \(\psi_N=\sum_{a\mid N}\mu(a)[a]=\alpha_1+\alpha_2\).

We have \[\phi_d(\alpha_1) =\sum_{d\mid (a_i+b_i)}\mu(a_i)\bigl([\overline{a_i}]-[\overline{-a_i}]\bigr) \in\phi_d(I),\] and \(\phi_d(\alpha_2)=\sum\limits_{d\mid a_i-b_i}(\mu(a_i)[\overline{a_i}]-\mu(b_i)[\overline{b_i}]) =0.\) Thus \[\phi_d(\psi_N)=\phi_d(\alpha_1+\alpha_2)=\phi_d(\alpha_1)\in\phi_d(I).\] Since \(\phi_d\) is a surjective ring homomorphism, \(\phi_d(I)\) is an ideal of \(\mathop{\mathrm{\mathbb{Z}}}[(\mathop{\mathrm{\mathbb{Z}}}/d)^\circ]\). Since \(n\) is squarefree and \(N\mid n\), \(\gcd(N,n/N)=1\) and \(\psi_n = \psi_N \psi_{n/N}.\) Hence \[\phi_d(\psi_n)=\phi_d(\psi_N)\phi_d(\psi_{n/N})\in \phi_d(I).\]

\((3)\Rightarrow (1):\) For each residue class \(r\), let \[S_r=\{a\mid n : a\equiv r\pmod d\},\] and decompose it according to the sign of \(\mu\): \[S_r^{+}=\{a\in S_r : \mu(a)=1\}, \qquad S_r^{-}=\{a\in S_r : \mu(a)=-1\}.\] Set \(p_r = |S_r^{+}|\) and \(q_r = |S_r^{-}|\).

The condition for \(\phi_d(\psi_n)\in \phi_d(I)\) is equivalent to \[p_r - q_r =-(p_{-r} - q_{-r}),\] all non-self-inverse residue classes \(r\not=-r\), and \[p_r - q_r= 0,\] for self-inverse residue classes \(r=-r\).

Consider a non-self-inverse class \(r\) (\(r\not=-r\)). Without loss of generality, we may assume that \(p_r \geq q_{-r}\). Then \(s:=p_r-q_{-r}=q_{r}-p_{-r}\geq0\). We pair \(q_{-r}\) elements in \(S_r^+\) with \(q_{-r}\) elements in \(S_{-r}^-\). We pair \(p_{-r}\) elements in \(S_{-r}^+\) with \(p_{-r}\) elements in \(S_{r}^-\). Clearly, for such a pair \(\{a,b\}\) we have \(a\equiv -b\pmod d\) and \(\mu(a)=-\mu(b)\). The remaining \(s\) elements in \(S_r^{+}\) are paired with the remaining \(s\) elements in \(S_{r}^{-}\). For such a latter pair \(\{a,b\}\) we have \(a\equiv b\equiv r\pmod d\) and \(\mu(a)=-\mu(b)\).

Now consider a self-inverse class \(r\) (\(r=-r\)). In this case \(|S_r^+|=|S_r^-|\). We pair each element in \(S_r^+\) with an element in \(S_r^-\). For such a pair \(\{a,b\}\) we have \(\mu(a)=-\mu(b)\) and \(a\equiv b\equiv r\pmod d\). ◻

Proof. By 26, the divisors of \(n\) can be paired so that \[\mu(a)+\mu(b)=0 \qquad\text{and}\qquad a\equiv \pm b \pmod d.\] Hence \[\mu(a)a^2+\mu(b)b^2 =\mu(a)(a^2-b^2) \equiv 0 \pmod d\] for every pair. Summing over all pairs yields \[\sum_{e\mid n}\mu(e)e^2\equiv 0 \pmod d.\] Since \(n\) is squarefree, \[\sum_{e\mid n}\mu(e)e^2 = \prod_{p\mid n}(1-p^2) = (-1)^{\omega(n)}J_2(n).\] Therefore \(d\mid J_2(n)\). ◻

Proof. Let \(m,n \in \mathop{\mathrm{\mathbb{Z}}}^{{\rm odd}}.\) We will show that \[\theta_d((-1,m)) \theta_d((-1, n))= \theta_d((1, mn)),\] and \[\theta_d((-1,m)(1, n))= \theta_d((-1, mn)) .\] For the first equality, we have \[\theta_d((-1,m)) \theta_d((-1,n))=\overline{(d_0+m)(d_0+n)} = \overline{d_0^2 +d_0(m+n) + mn}.\] Since \(d_0\) is even, \(d_0^2 \equiv 0 \pmod{d}.\) Additionally, since \(m,n\) are both odd, \(d_0(m+n) \equiv 0 \pmod{d}.\) We conclude that \[\theta_d((-1,m)) \theta_d((-1,n)) = \overline{mn} = \theta_d((1,mn)).\] For the second equality, we have \[\theta_d((-1,m)) \theta_d((1,n))= \overline{(d_0+m)n} = \overline{d_0n +mn}.\] Since \(n\) is odd, \(d_0n\equiv d_0\pmod d\). Hence \(d_0n+mn\equiv d_0+mn\pmod d\) and we conclude that \[\theta_d((-1,m)) \theta_d((1,n)) = \overline{d_0+mn} = \theta_d((-1,mn)). \qedhere\] ◻

Proof. Let \(\zeta=\zeta_d\). We may write \[\alpha = \sum_{(1,m)\in S_1} c_m[(1, m)] +\sum_{(-1,n)\in S_2} d_n [(-1, n)],\] where \(c_m, d_n \in \mathbb{Z},\; m, n \in \mathbb{Z}^{\mathrm{odd}}\), and \(\mathop{\mathrm{{\rm Supp}}}(\alpha)=S_1\sqcup S_2\).

Define \[\gamma := \sum_{(1,m)\in S_1} c_m [m] + \sum_{(-1,n)\in S_2} d_n \left[n + \frac{d}{2}\right] \in \mathbb{Z}[\mathbb{Z}^{\circ}].\] Since \(d\) is even, we have \(\zeta^{\,d/2} = -1\). Note also that \(f\) is defined at \(\zeta^a\), for each \(a\in \mathop{\mathrm{{\rm Supp}}}(\gamma)\). If \(a=m\in \mathop{\mathrm{{\rm Supp}}}(\gamma)\), then definedness follows from the hypothesis on \((1,m)\in S_1\). If \(a=n+d/2\in \mathop{\mathrm{{\rm Supp}}}(\gamma)\), then \(\zeta^a =-\zeta^n\), and the definedness follows from the hypothesis on \((-1,n)\in S_2\).

We have \[\begin{align} \alpha f(\zeta) &= \left( \sum_{(1,m)\in S_1} c_m[(1, m)] +\sum_{(-1,n)\in S_2} d_n [(-1, n)] \right) f(\zeta) \\ &= \sum_{(1,m)\in S_1} c_m f(\zeta^m) +\sum_{(-1,n)\in S_2} d_n f(-\zeta^n)\\ &= \sum_{(1,m)\in S_1} c_m f(\zeta^m) +\sum_{(-1,n)\in S_2} d_n f(\zeta^{n+d/2}) \\ &= \left( \sum_{(1,m)\in S_1} c_m [m] + \sum_{(-1,n)\in S_2} d_n \left[n + \frac{d}{2}\right] \right) f(\zeta) \\ &= \gamma f(\zeta). \end{align}\]

Moreover, in \(\mathop{\mathrm{\mathbb{Z}}}[(\mathop{\mathrm{\mathbb{Z}}}/d)^\circ]\) we have \[0=\theta_d (\alpha) =\sum_{(1,m)\in S_1} c_m [m] + \sum_{(-1,n)\in S_2} d_n [n+d/2]=\phi_d(\gamma),\] where \(\phi_d\) is the natural homomorphism \(\phi_d\colon \mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]\to \mathop{\mathrm{\mathbb{Z}}}[(\mathop{\mathrm{\mathbb{Z}}}/d)^\circ]\). By Proposition 7, it follows that \(\gamma f(\zeta) = 0\). ◻

Proof. Since \(\beta\in {\rm Ann}_{\mathop{\mathrm{\mathbb{Z}}}[\{\pm 1\}\times\mathop{\mathrm{\mathbb{Z}}}^{{\rm odd}}]}(f)\), \(\beta f=0\) and hence \(\alpha f=(\alpha-\beta )f\) in \(\mathop{\mathrm{\mathbb{Q}}}(x)\). Note that \(\mathop{\mathrm{{\rm Supp}}}(\alpha-\beta)\subset \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup \mathop{\mathrm{{\rm Supp}}}(\beta)\). Hence \(f\) is defined at \(\zeta_d^{(\epsilon,m)}\) for each \((\epsilon,m) \in \mathop{\mathrm{{\rm Supp}}}(\alpha-\beta)\). By Proposition 31, \(\Phi_d\) is a factor of \((\alpha-\beta )f=\alpha f\). ◻

Proof. Clearly, we have \(\theta_d(\alpha)=\sum c_r[r]\). Note also that for any \(\beta\in R^\pm\), \(\phi_d(\beta)= \sum_{r} m_r[r]\) is supported on odd residue classes. (Since \(d\) is even, the notion of odd residue classes makes sense.) Since \(\phi_d([(1,1)]+[(-1,1)] )=[1]+[d_0+1]\) an arbitrary element in \(\phi_d( \langle [(1,1)]+[(-1,1)] \rangle)\) is of the form \[\left(\sum_{r} m_r [r]\right) ([1] + [d_0+1]) = \sum_{r} m_r ([r] + [rd_0+r]) = \sum_{r} m_r ([r] + [d_0+r]).\] Here we note that for \(r\) is odd then \(rd_0+r\equiv d_0+r\pmod d\). From this we see that \(\phi_d(\alpha)\in \phi_d(\langle [(1,1)]+[(-1,1)] \rangle)\) if and only if \(c_r=c_{r+d_0}\) for every residue class \(r\). ◻

Proof. Write \(\alpha=\sum\limits_{(\epsilon,m)\in \mathop{\mathrm{\text{Supp}}}(\alpha)} a_{\epsilon,m}[(\epsilon,m)].\) For each residue class \(r\), set \(c_r=\sum\limits_{\theta_d(\epsilon,m)=r}a_{\epsilon,m}.\) By the previous lemma, \(c_r=c_{r+d_0}.\)

For each pair of residue classes \(\{r,r+d_0\}\) with \(c_r\neq 0\), we choose a representative \((\epsilon_r,m_r)\in \mathop{\mathrm{\text{Supp}}}(\alpha)\) such that \(\theta_d(\epsilon_r,m_r)=r.\) Define \(\beta_r =c_r\left([(\epsilon_r,m_r)]+[(-\epsilon_r,m_r)]\right).\) Then \(\beta_r\in \langle [(1,1)]+[(-1,1)] \rangle\), and \(\theta_d(\beta_r) = c_r\big([r]+[r+d_0]\big).\)

Now set \(\beta=\sum\limits_{\{r,r+d_0\}}\beta_r.\) Then \(\beta\in \langle [(1,1)]+[(-1,1)] \rangle\) and \(\theta_d(\beta)=\theta_d(\alpha).\) Moreover, by construction, \(\mathop{\mathrm{{\rm Supp}}}(\beta)\subseteq \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha))\). ◻

Proof. Since \(I:=\langle [(1,1)]+[(-1,1)] \rangle \subset {\rm Ann}_{R^\pm}(f)\), one has \(([(1,1)]+[(-1,1)])f=0\), i.e., \(f(x)+f(-x)=0\). From this we see that if \(f\) is defined at \(\pm\zeta_d^m\) then \(f\) is also defined at \(\mp\zeta_d^{m}\). Hence \(f\) is defined at \(\zeta_d^{(\epsilon,a)}\) for each \((\epsilon,a)\in \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha))\).

By the previous lemma, there exists \(\beta\in I\) such that \(\theta_d(\beta)=\theta_d(\alpha)\) and \(\mathop{\mathrm{{\rm Supp}}}(\beta)\subseteq \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha)).\) Clearly, \(\mathop{\mathrm{{\rm Supp}}}(\alpha)\cup\mathop{\mathrm{{\rm Supp}}}(\beta) \subseteq \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup(-\mathop{\mathrm{{\rm Supp}}}(\alpha)).\) Hence \(f\) is defined at \(\zeta_d^{(\epsilon,a)}\) for each \((\epsilon,a)\in \mathop{\mathrm{{\rm Supp}}}(\alpha)\cup \mathop{\mathrm{{\rm Supp}}}(\beta)\). By Theorem 33, \(\Phi_d\) is a factor of \(\alpha f\). ◻

Proof. (1) Let \(\zeta=\zeta_{24}\). We have \[\begin{align} \tilde{F}_{21}(\zeta)&=\frac{\zeta}{1-\zeta} -\frac{\zeta^3}{1-\zeta^3}-\frac{\zeta^7}{1-\zeta^7}+\frac{\zeta^{21}}{1-\zeta^{21}} =\frac{\zeta(1+\zeta+\zeta^2)-\zeta^3}{1-\zeta^3} -\frac{\zeta^7(1+\zeta^7+\zeta^{14})-\zeta^{21}}{1-\zeta^{21}}\\ &=\frac{\zeta+\zeta^2}{1-\zeta^3} - \frac{\zeta^3(\zeta^7+\zeta^{14})}{\zeta^3(1-\zeta^{21})} =\frac{\zeta+\zeta^2+\zeta^{10}+\zeta^{17}}{1-\zeta^3} =\frac{\zeta^{-2}(\zeta^3+\zeta^{12})+\zeta^2(1+\zeta^{15})}{1-\zeta^3}\\ &=\frac{\zeta^{-2}(\zeta^3-1)+\zeta^2(1-\zeta^{3})}{1-\zeta^3}=\zeta^2-\zeta^{-2}=\zeta^6. \end{align}\] (Since \(\Phi_{24}(x)=x^8-x^4+1\), one has \(\zeta^2-\zeta^{-2}=\zeta^6\).)

Hence \(\tilde{F}_{21p}(\zeta)=\tilde{F}_{21}(\zeta)-\tilde{F}_{21}(\zeta^p)=\zeta^6-\zeta^{6p}=\zeta^6-\zeta^6=0\).

(2) Let \(\zeta=\zeta_{20}\). Note that \(\zeta^{10}=-1\). We have \[\begin{align} \tilde{F}_{15}(\zeta)&=\frac{\zeta}{1-\zeta} -\frac{\zeta^3}{1-\zeta^3}-\frac{\zeta^5}{1-\zeta^5}+\frac{\zeta^{15}}{1-\zeta^{15}}\\ &=\frac{\zeta(1+\zeta+\zeta^2+\zeta^3+\zeta^4)-\zeta^5}{1-\zeta^5} -\frac{\zeta^3(1+\zeta^3+\zeta^{6}+\zeta^9+\zeta^{12})-\zeta^{15}}{1-\zeta^{15}}\\ &=\frac{\zeta+\zeta^2+\zeta^3+\zeta^4}{1-\zeta^3} - \frac{\zeta^5(\zeta^3+\zeta^6+\zeta^9+\zeta^{12})}{\zeta^5(1-\zeta^{15})}\\ &=\frac{\zeta+\zeta^2+\zeta^3+\zeta^4+\zeta^8+\zeta^{11}+\zeta^{14}+\zeta^{17}}{1-\zeta^5}\\ &=(\frac{\zeta+\zeta^{11})+(\zeta^4+\zeta^{14})+( \zeta^3+ \zeta^8)+(\zeta^2+ \zeta^{17})}{1-\zeta^5}\\ &=\frac{\zeta^{-2}(\zeta^5+\zeta^{10})+\zeta^2(1+\zeta^{15})}{1-\zeta^{5}}=\zeta^2-\zeta^{-2} = \zeta_5^2 - \zeta_5^3. \end{align}\]

Hence \(\tilde{F}_{15p}(\zeta)=\tilde{F}_{15}(\zeta)-\tilde{F}_{15}(\zeta^p)=\zeta^2-\zeta^{2p}-(\zeta^{-2}-\zeta^{-2p})=0\). ◻

Proof. Write \(f(x) = \frac{g(x)}{h(x)}\) with \(g(x), h(x) \in \mathbb{Q}[x]\) and \(\gcd(g,h)=1\). By assumption, \(h(\zeta_d^a) \neq 0\).

Let \(t\) be an integer with \(\gcd(t,d)=1.\) Then the map \(\zeta_d \mapsto \zeta_d^t\) extends to a field automorphism \(\sigma_t\) of \(\mathbb{Q}(\zeta_d).\) Hence, we have \[h(\zeta_d^{at}) = \sigma_t\big(h(\zeta_d^a)\big).\] Since \(h(\zeta_d^a) \neq 0\), it follows that \(h(\zeta_d^{at}) \neq 0\), and thus \(f\) is defined at \(\zeta_d^{at}\).

Now let \(q \mid a\) with \(\gcd(q,d)=1\). Choose an integer \(l\) such that \(lq \equiv 1 \pmod{d}\). Then \(a/q\equiv al\pmod d\) and hence \(\zeta_d^{a/q} = \zeta_d^{al}\). Since \(\gcd(l,d)=1\), the previous argument shows that \(h(\zeta_d^{a/q}) \neq 0\). Therefore, \(f\) is defined at \(\zeta_d^{a/q}\) as desired. ◻

Proof. Let \(\zeta=\zeta_{24}\). By our assumption, \(M_{3\times7}(f)\) is defined at \(\zeta\). We will show that \(M_{3\times 7}(f)(\zeta)\in \mathop{\mathrm{\mathbb{Q}}}(\zeta_8)\).

We know that \(G=\mathop{\mathrm{\text{Gal}}}(\mathop{\mathrm{\mathbb{Q}}}(\zeta_{24})/\mathop{\mathrm{\mathbb{Q}}})=(\mathop{\mathrm{\mathbb{Z}}}/24\mathop{\mathrm{\mathbb{Z}}})^\times=\{\overline{1},\overline{5},\overline{7},\overline{11},\overline{13},\overline{17},\overline{19},\overline{23}\}\). We denote the Galois action of \(\overline{a}\in G\) on an element \(u\in \mathop{\mathrm{\mathbb{Q}}}(\zeta)\) by \(\overline{a}\cdot u\). Recall that this action is given by \(\overline{a}\cdot \zeta=\zeta^a\). Let \(H=\{\overline{1},\overline{17}\}\) be a subgroup of \(G\). We have \[\begin{align} \overline{17}\cdot M_{21}(f)(\zeta)&=M_{21}(f)(\zeta^{17})=f(\zeta^{17})-f(\zeta^{3\cdot 17})-f(\zeta^{7\cdot 17})+f(\zeta^{21\cdot 17})\\ &=f(\zeta^{17})-f(\zeta^{3})-f(\zeta^{23})+f(\zeta^{21})\\ &=f(\zeta^{1})-f(\zeta^{3})-f(\zeta^{7})+f(\zeta^{21}). \end{align}\] The last equality follows from \(f(\zeta^{17})+f(\zeta^{7})=f(\zeta^{23})+f(\zeta^{1})\). (Note also that by Lemma 40\(, f\) is defined at \(\zeta^{17a}\), for \(a=1,3,7,21\).) Thus \(M_{21}(f)(\zeta)\in \mathop{\mathrm{\mathbb{Q}}}(\zeta)^H= \mathop{\mathrm{\mathbb{Q}}}(\zeta_{8})\).

(a) We can write \(M_{21}(f)(\zeta)= P(\zeta_8)\), for some \(P(x)\in \mathop{\mathrm{\mathbb{Q}}}[x]\). Suppose that \(p\equiv 1 \pmod 8\) is a prime. Then by Lemma 40, \(M_{21}(f)\) is defined at \(\zeta^p\), and we have \[M_{21}f(\zeta^p)=\overline{p} \cdot M_{21}(f)(\zeta) = \overline{p}\cdot P(\zeta_8)=P(\zeta_8^p)=P(\zeta_8).\] Hence \(M_{21p}(f)(\zeta)=M_{21}(f)(\zeta)-M_{21}(f)(\zeta^p)=0\).

(b) Since \(M_{3\times 7}(g)(\zeta_{24})=0\), we see that \[M_{21}(f)(\zeta)=M_{21}(f)(-\zeta)=M_{21}(f)(\zeta^{13})= \overline{13} \cdot M_{21}(f)(\zeta).\] Hence \(M_{21}(f)(\zeta)\) is also fixed by \(\overline{13}\in G=\mathop{\mathrm{\text{Gal}}}(\mathop{\mathrm{\mathbb{Q}}}(\zeta_{24})/\mathop{\mathrm{\mathbb{Q}}})\). Thus \(M_{21}(f)(\zeta)\) is fixed by the subgroup \(H'=\langle \overline{13},\overline{17}\rangle =\{\overline{1},\overline{5},\overline{13},\overline{17}\}\). This implies that \(M_{21}(f)(\zeta)\in Q(\zeta)^{H'}=\mathop{\mathrm{\mathbb{Q}}}(\zeta_4)\). We can write \(M_{21}(f)(\zeta)= P(\zeta_4)\), for some \(P(x)\in \mathop{\mathrm{\mathbb{Q}}}[x]\). Suppose that \(p\equiv 1 \pmod 4\) is a prime. Then by Lemma 40, \(M_{21}(f)\) is defined at \(\zeta^p\), and we have \[M_{21}f(\zeta^p)=\overline{p} \cdot M_{21}(f)(\zeta) = \overline{p}\cdot P(\zeta_4)=P(\zeta_4^p)=P(\zeta_4).\] Hence \(M_{21p}(f)(\zeta)=M_{21}(f)(\zeta)-M_{21}(f)(\zeta^p)=0\). ◻

Proof. Let \(\zeta=\zeta_d\). By our assumption, \(\alpha_1 f\) is defined at \(\zeta\). We will show that \(\alpha_1 f(\zeta)\in \mathop{\mathrm{\mathbb{Q}}}(\zeta_{d_1})\).

We know that \(G=\mathop{\mathrm{\text{Gal}}}(\mathop{\mathrm{\mathbb{Q}}}(\zeta)/\mathop{\mathrm{\mathbb{Q}}})=(\mathop{\mathrm{\mathbb{Z}}}/d\mathop{\mathrm{\mathbb{Z}}})^\times = \{\overline{a} \mid 1\leq a < d , \gcd(a,d) =1\}.\) We denote the Galois action of \(\overline{a}\in G\) on an element \(u\in \mathop{\mathrm{\mathbb{Q}}}(\zeta)\) by \(\overline{a}\cdot u\). Recall that this action is given by \(\overline{a}\cdot \zeta =\zeta^a\). Here, we make an important remark about the relationship between the Galois action and the Mahler action. Namely; if \(\gcd(a, m)=1\), and \(p \in \mathop{\mathrm{\mathbb{Q}}}(x)\) which is defined at \(\zeta_m\), then these actions are compatible. More precisely, we have \(\bar{a} p(\zeta_m) =([a] \cdot p )(\zeta_m)\) since they are both equal to \(p(\zeta_m^a).\)

By the assumption that \(f\) is defined at \(\zeta^a\) for each \(a\in \mathop{\mathrm{{\rm Supp}}}(\alpha_1)\) and by 40, we see that \(f\) is defined at each \(\zeta^{b}\) for each \(b \in \mathop{\mathrm{\text{Supp}}}(([h]-[1])\alpha_1), \overline{h} \in H.\) For each \(\overline{h}\in H\), we have \[\overline{h}\cdot \alpha_1 (f)(\zeta)=\alpha_1 (f)(\zeta^{h}) =([h]\alpha_1) (f)(\zeta) =\alpha_1 (f)(\zeta).\] The last equality follows from \(([h]-[1])\alpha_1 f(\zeta) = 0,\) by 17. Thus \(\alpha_1(f)(\zeta)\in \mathop{\mathrm{\mathbb{Q}}}(\zeta)^{H}= \mathop{\mathrm{\mathbb{Q}}}(\zeta_{d_1})\). Indeed, \(H\) is precisely the subgroup fixing \(\zeta_{d_1}=\zeta^{d/d_1}\).

(a) We can write \(\alpha_1(f)(\zeta)= P(\zeta_{d_1})\), for some \(P(x)\in \mathop{\mathrm{\mathbb{Q}}}[x]\). Write \(\alpha_2=\sum\limits_{a\in S} n_a[a]\), where \(S\) is the support of \(\alpha_2\). By the assumption that \(\gcd(a,d)=1\) for each \(a\in S\) and by 40, we see that \(f\) is defined at each \(\zeta^{b}\) for each \(b\in \mathop{\mathrm{{\rm Supp}}}(\alpha_2\alpha_1)\). Then \[\begin{align} \alpha f(\zeta) &= \alpha_2 \alpha_1f(\zeta)= \sum_{a \in S} n_a[a]\alpha_1f(\zeta) = \sum_{a \in S} n_a\alpha_1 f(\zeta^{a})=\sum_{a \in S} n_a \overline{a}\cdot \alpha_1 f(\zeta)\\ &=\sum_{a \in S} n_a \overline{a} \cdot P(\zeta_{d_1}) = \sum_{a \in S} n_a [a]P(\zeta_{d_1}) =\alpha_2 P(\zeta_{d_1}). \end{align}\] Since \(\alpha_2\equiv 0\mod {[d_1]}\), by Proposition 7 we have \(\alpha_2 P(\zeta_{d_1})=0\). Hence \(\alpha f(\zeta)=0\) and we are done.

(b) Now we use the additional hypothesis involving \(g(x)=f(x)-f(-x).\) Since \(d\) is divisible by \(4\), the integer \(d/2+1\) is coprime to \(d\), and \(-\zeta=\zeta^{d/2+1}.\) By assumption, \(\alpha_1 g(\zeta)=0.\) Thus \[0 = \alpha_1 f(\zeta)-\alpha_1 f(-\zeta) = \alpha_1 f(\zeta)-\alpha_1 f(\zeta^{d/2+1}) = \alpha_1 f(\zeta) -\overline{(\frac{d}{2}+1)} \cdot \alpha_1 f(\zeta).\] Therefore \(\alpha_1 f(\zeta)\) is fixed by the Galois automorphism \(\zeta\longmapsto \zeta^{d/2+1}.\) Consequently, \[\alpha_1 f(\zeta) \in \mathbb{Q}(\zeta_{d_1})\cap \mathbb{Q}(\zeta_d)^{\langle d/2+1\rangle}.\] Since the automorphism \(\zeta_d\mapsto -\zeta_d\) restricts on \(\mathbb{Q}(\zeta_{4d_0})\) to the automorphism \[\zeta_{4d_0}\longmapsto -\zeta_{4d_0},\] its fixed field inside \(\mathbb{Q}(\zeta_{4d_0})\) is \(\mathbb{Q}(\zeta_{2d_0}).\) Hence \[\alpha_1 f(\zeta)\in \mathbb{Q}(\zeta_{2d_0}).\] Thus there exists a polynomial \(P(x)\in \mathbb{Q}[x]\) such that \(\alpha_1 f(\zeta)=P(\zeta_{2d_0}).\)

Now write \(\alpha_2=\sum\limits_{a\in \operatorname{Supp}(\alpha_2)} n_a[a].\) By the assumption that \(\gcd(a,d)=1\) for each \(a\in S\) and by 40, we see that \(f\) is defined at each \(\zeta^{b}\) for each \(b\in \mathop{\mathrm{{\rm Supp}}}(\alpha_2\alpha_1)\). Then \[\begin{align} \alpha f(\zeta) &= \alpha_2 \alpha_1f(\zeta)= \sum_{a \in S} n_a[a]\alpha_1f(\zeta) = \sum_{a \in S} n_a\alpha_1 f(\zeta^{a})=\sum_{a \in S} n_a \overline{a}\cdot \alpha_1 f(\zeta)\\ &=\sum_{a \in S} n_a \overline{a} \cdot P(\zeta_{2d_0}) = \sum_{a \in S} n_a [a]P(\zeta_{2d_0}) =\alpha_2 P(\zeta_{2d_0}). \end{align}\] Since \(\alpha_2\equiv 0\mod {[2d_0]}\), by Proposition 7 we have \(\alpha_2 P(\zeta_{2d_0})=0\). Hence \(\alpha f(\zeta)=0\) and we are done. ◻

Proof. First, we remark that \(\mathop{\mathrm{\mathbb{Z}}}^\circ\). \(\mathop{\mathrm{\mathbb{Z}}}^\circ = \mathop{\mathrm{\mathbb{Z}}}~\backslash~\{0\}\) can be identified as a direct product of two simpler monoids:

• The group \(G = \{1, -1\}\);

• The monoid of positive integers \(\mathop{\mathrm{\mathbb{N}}}= \mathop{\mathrm{\mathbb{Z}}}^+ = \{1, 2, 3, \dots\}\).

Every \(m \in \mathop{\mathrm{\mathbb{Z}}}^\circ\) has a unique representation as \(m = s \times p\) with \(s \in G\) and \(p \in N\). Thus, we have a monoid isomorphism \(\mathop{\mathrm{\mathbb{Z}}}^\circ = G \times N\). Hence, we can rewrite \[R = \mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ] \cong \mathop{\mathrm{\mathbb{Z}}}[G \times N] \cong (\mathop{\mathrm{\mathbb{Z}}}[N])[G].\] Now we define \(S = \mathop{\mathrm{\mathbb{Z}}}[N] = \mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^+]\), then our ring \(R\) is isomorphic to the group ring \(S[G] = S[\{-1, 1\}]\). Note that since the monoid \(N = \mathop{\mathrm{\mathbb{Z}}}^+\) is cancellative (that is if \(ab = ac\) then \(b = c\)) and torsion-free (that is the only element \(n\) with \(n^k = 1\) is \(n = 1\)). A result by Gilmer [16] stated that if a ring \(U\) is an integral domain and a monoid \(V\) is cancellative and torsion-free, then the monoid ring \(U[V]\) is also an integral domain. Since our \(\mathop{\mathrm{\mathbb{Z}}}\) is an integral domain and \(N\) is cancellative and torsion-free, \(\mathop{\mathrm{\mathbb{Z}}}[N]\) is an integral domain, thus \(\mathop{\mathrm{\mathbb{Z}}}[N]\) has no zero divisors.
For the group ring \(R\), let \(g\) be the element of \(S[G]\) corresponding to \([-1]\), thus \(g^2 = [1]\). Any \(\alpha \in R\) can be uniquely written as \(\alpha = A + Bg\) with \(A, B \in S\) and \(A\) represents the part of \(\alpha\) with positive indices while \(B\) represents the part with negative indices. Specifically, for \(\alpha = \sum_{m \in Z^\circ}a_m[m]\), then:

• \(A = \sum_{p \in \mathop{\mathrm{\mathbb{Z}}}^+}a_p[p]\);

• \(B = \sum_{p \in \mathop{\mathrm{\mathbb{Z}}}^+}a_{-p}[p]\).

An element \(\alpha = A + Bg\) is a zero divisor if there exists a non-zero element \(\beta = C + Dg\), with \(C, D \in S\) such that \(\alpha\beta = 0\). This is equivalent to \[(A + Bg)(C + Dg) = (AC + BD) + (AD + BC)g = 0.\] Since we are calculating in the ring \(S[G]\), then it gives us the system of equations \[\begin{cases} AC + BD = 0\\ AD + BC = 0.\\ \end{cases}\] For this system to have non-zero solution \(C, D\), the determinant of the coefficient matrix must be zero: \[\det\left(\begin{matrix} A & B \\ B & A \end{matrix}\right) = A^2 - B^2 = 0.\] \(S\) is an integral domain, then \(A^2 - B^2 = 0\) implies \((A - B)(A + B) = 0\), which implies that either \(A - B = 0\) or \(A + B = 0\). We now consider cases:

1. \(A = B\), then the element \(\alpha = A(1 + g)\), with \(A\) is non-zero. This is a zero divisor because it is annihilated by the non-zero element \(1 - g\): \(\alpha(1 - g) = A(1 + g)(1 - g) = A(1 - g^2) = A(1 - 1) = 0\). Since \(A = B\), it means \(\sum_{p \in \mathop{\mathrm{\mathbb{Z}}}^+}a_p[p] = \sum_{p \in \mathop{\mathrm{\mathbb{Z}}}^+}a_{-p}[p]\). Since the elements \([p]\) form a basis for \(S\), the coefficients must be equal for each \(p\), that is \(a_p = a_{-p}\) for all \(p \in \mathop{\mathrm{\mathbb{Z}}}^+\). We call these “even” elements.

2. \(A = -B\), then the element \(\alpha = A(1 - g)\), with \(A\) is non-zero. This is a zero divisor because it is annihilated by the non-zero element \(1 + g\): \(\alpha(1 + g) = A(1 - g)(1 + g) = A(1 - g^2) = A(1 - 1) = 0\). Since \(A = -B\), it means \(\sum_{p \in \mathop{\mathrm{\mathbb{Z}}}^+}a_p[p] = -\sum_{p \in \mathop{\mathrm{\mathbb{Z}}}^+}a_{-p}[p]\). Since the elements \([p]\) form a basis for \(S\), the coefficients must be equal for each \(p\), that is \(a_p = -a_{-p}\) for all \(p \in \mathop{\mathrm{\mathbb{Z}}}^+\). We call these “odd” elements

Thus, the set of zero divisors in \(\mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]\) is the union of two sets of non-zero elements: those whose coefficients have even symmetry (\(a_p = a_{-p}\)) and those whose coefficients have odd symmetry (\(a_p = -a_{-p}\)). Furthermore, this also means the set of all zero divisors is the union of the two principal ideals \(([1] + [-1])\) and \(([1] - [-1])\), excluding the zero element. ◻

Proof. The given sequence is exact if and only if \(\operatorname{Im}(\alpha) = \ker(\beta)\) and \(\operatorname{Im}(\beta) = \ker(\alpha)\). We only prove that \(\operatorname{Im}(\alpha) = \ker(\beta)\). The other set equality can be proved in a similar way. First, we will identify \(\operatorname{Im}(\alpha)\) and \(\ker(\beta)\).

1. \(\operatorname{Im}(\alpha)\): The image of \(\alpha\) is the set of all \(g(x) \in M\) such that \(g(x) = \alpha \cdot h(x)\) for some \(h(x) \in M\):

   • \(\alpha \cdot h(x) = ([1] - [-1])\cdot h(x) = h(x^1) - h(x^{-1}) = h(x) - h(1/x)\).

   • Then, \(\operatorname{Im}(\alpha) = \{h(x) - h(1/x): h(x) \in \mathop{\mathrm{\mathbb{Q}}}(x)\}\)

2. \(\ker(\beta)\): The kernel of \(\beta\) is the set of all \(f(x)\) such that \(\beta \cdot f(x) = 0\).

   • \(\beta \cdot f(x) = ([1] + [-1])\cdot f(x) = f(x^1) + f(x^{-1}) = f(x) + f(1/x)\).

   • Then, \(\ker(\beta) = \{f(x) \in \mathop{\mathrm{\mathbb{Q}}}(x): f(x) + f(1/x) = 0\}\)

We will show that \(\operatorname{Im}(\alpha) = \ker(\beta)\):

1. \(\operatorname{Im}(\alpha) \subseteq \ker(\beta)\): Consider \(g(x) \in \operatorname{Im}(\alpha)\), then \(g(x) = h(x) - h(1/x)\) for \(h(x) \in \mathop{\mathrm{\mathbb{Q}}}(x)\). We will check if \(\beta \cdot g(x) = 0\). \(\beta \cdot g(x) = g(x) + g(1/x)\). Note that we also have \(g(1/x) = h(1/x) - h(x)\), then \(\beta \cdot g(x) = (h(x) - h(1/x)) + (h(1/x) - h(x)) = 0\), or \(g(x) \in \ker(\beta)\).

2. \(\ker(\beta) \subseteq \operatorname{Im}(\alpha)\): Consider \(g(x) \in \ker(\beta)\), then \(g(x) + g(1/x) = 0\). Take \(h(x) = \frac{g(x)}{2}\), then \(h(x) - h(1/x) = \frac{g(x)}{2} - \frac{g(1/x)}{2}\). Since \(g(x) + g(1/x) = 0\), it means \(g(x) = -g(1/x)\), thus \(h(x) - h(1/x) = \frac{g(x)}{2} + \frac{g(x)}{2} = g(x)\), thus \(g(x) \in \operatorname{Im}(\alpha)\), then, \(g(x) \in \operatorname{Im}(\alpha)\).

These implies \(\operatorname{Im}(\alpha) \subseteq \ker(\beta)\) and \(\ker(\beta) \subseteq \operatorname{Im}(\alpha)\), that is \(\operatorname{Im}(\alpha) = \ker(\beta)\). ◻

Proof. Write \(\alpha=\sum_{m\in S} a_m[m],\) where \(S=\mathop{\mathrm{{\rm Supp}}}(\alpha)\subset \mathbb{N}^\circ\) is finite and \(a_m\neq 0\) for all \(m\in S\). We will show that \(\alpha f=0\) implies \(S=\varnothing\), hence \(\alpha=0\).

Write \(f=P/Q\), where \(P,Q\in \mathbb{Q}[x]\) are coprime and \(Q\neq 0\). Since \(f\notin \mathbb{Q}\), at least one of \(P,Q\) has positive degree.

Suppose \[0=\alpha f(x)=\sum_{m\in S} a_m f(x^m)=\sum_{m\in S} a_m \frac{P(x^m)}{Q(x^m)}.\] Multiplying by \(\prod\limits_{m\in S} Q(x^m),\) we get \[\sum_{m\in S} a_m P(x^m)\prod_{\substack{n\in S\\ n\neq m}} Q(x^n)=0.\]

The summand corresponding to \(m\) has a degree \[d_m:=m\deg P+\sum_{\substack{n\in S\\ n\neq m}} n\deg Q =\sum_{n\in S} n\deg Q + m(\deg P-\deg Q).\] Hence for \(m,m'\in S\), we have \[d_m-d_{m'}=(m-m')(\deg P-\deg Q).\] Suppose \(\deg P\not=\deg Q\) then all degrees \(d_m\) are distinct. This forces \(S\) is empty and \(\alpha=0\).

Assume now that \(\deg P=\deg Q\). Let \(c=\dfrac{\operatorname{lc}(P)}{\operatorname{lc}(Q)}\) and set \(R=P-cQ.\) Then \(f=c+\frac{R}{Q}=c+g,\) and \(\deg R<\deg Q.\).

Since \(\alpha f=0\), we have \(\alpha\!\left(\frac{R}{Q}\right) = -c\,\alpha(1).\) The right-hand side is a constant, denoted \(C\in \mathop{\mathrm{\mathbb{Q}}}\). We obtain \[\sum_{m\in S} a_m \frac{R(x^m)}{Q(x^m)}=C.\]

Multiplying by \(\prod_{m\in S}Q(x^m)\) yields \[\sum_{m\in S} a_m R(x^m) \prod_{\substack{n\in S\\ n\neq m}} Q(x^n) = C\prod_{m\in S}Q(x^m).\]

If \(C\not=0\) then the degree of the right-hand side is \(\sum\limits_{m\in S} m\deg Q.\) On the other hand, each summand on the left has degree \[m\deg R+ \sum_{\substack{n\in S\\ n\neq m}} n\deg Q = \sum_{n\in S} n\deg Q - m(\deg Q-\deg R).\] Since \(\deg R<\deg Q\), every summand on the left has degree strictly less than \(\sum_{n\in S} n\deg Q\). Therefore \(C=0\), and hence \(\alpha g=0\), where \(g=\dfrac{R}{Q}\) with \(\deg R<\deg Q\). Note that since \(f\) is not a constant \(g\) is also not a constant. Thus we reduce to the previous case and hence \(\alpha=0\). ◻

Proof. Write \(\alpha=A+[-1]B,\) where \(A,B\in \mathbb{Z}[\mathbb{N}^\circ]\).

Define \[f_{+}(x)=\dfrac{f(x)+f(1/x)}{2}, \qquad f_{-}(x)=\dfrac{f(x)-f(1/x)}{2}.\] Note that \(f_{+}(1/x)=f_+(x)\) and \(f_{-}(1/x)=-f_{-}(x)\), i.e, \([-1]f_{+}=f_{+}\), and \([-1]f_{-}=-f_{-}\).

We have \[0=\alpha f=\alpha f_{+}+\alpha f_{-}=(A+[-1]B)f_{+} +(A+[-1]B)f_{-}=(A+B)f_{+}+(A-B)f_{-}.\]

Applying \([-1]\) to this equation gives \[(A+B)f_{+}-(A-B)f_{-}=0.\] Adding and subtracting the two equations yields \[(A+B)f_{+}=0, \qquad (A-B)f_{-}=0.\]

Since \(f\notin \mathbb{Q}\), at least one of \(f_{+}\) and \(f_{-}\) is non-constant. By Lemma A.4, if \(f_{+}\) is non-constant, then \(A+B=0\), and therefore \[\alpha=A+[-1]B=A-[-1]A=A([1]-[-1]).\] Similarly, if \(f_{-}\) is non-constant, then \(A-B=0\), and therefore \[\alpha=A+[-1]A=A([1]+[-1]).\]

Thus \(\alpha\) is a multiple of either \([1]-[-1]\) or \([1]+[-1]\). ◻

Proof. Let \(\alpha\in \mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]\) and write \(\alpha=A+[-1]B,\) where \(A,B\in \mathbb{Z}[\mathbb{N}^\circ]\). As in the proof of Corollary A.5, we see that \(\alpha \in {\rm Ann}_{\mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]}(f)\) if and only if \[(A+B)f_+=(A-B)f_- =0.\]

(1) Suppose \(f_+=0\). This implies that \(f_-\) is nonconstant. By Lemma A.4, the condition \((A-B)f_-=0\) is equivalent to \(A-B=0\), i.e, \(\alpha=A([1]+[-1])\). Hence \({\rm Ann}_{\mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]}(f)=\langle [1]+[-1]\rangle.\)

(2) Suppose \(f_+=c\), where \(c\) is a nonzero constant. This implies that \(f_-\) is nonconstant. By Lemma A.4, the condition \((A-B)f_-=0\) is equivalent to \(A-B=0\). And the condition \(0=(A+B)f_+=2A(c)=2c\epsilon(A)\) is equivalent to \(A\in \ker\epsilon\). Hence \({\rm Ann}_{\mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]}(f)=\ker(\epsilon)\langle [1]+[-1]\rangle.\)

Case (3) and Case (4) can be proved by similar arguments as in case (1) and case (2).

(5) Suppose \(f_+\) and \(f_-\) are both nonconstant. By Lemma A.4, the condition \((A+B)f+=(A-B)f_- = 0\) is equivalent to \(A+B=A-B=0\). This latter condition is equivalent to \(A=B=0\). Hence \({\rm Ann}_{\mathop{\mathrm{\mathbb{Z}}}[\mathop{\mathrm{\mathbb{Z}}}^\circ]}(f)=0.\) ◻