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Section 4.1 Proof of Theorem 4.1

We fix an odd prime \(p\text{.}\) For \(\alpha\geq0\text{,}\) recall that \(F_\alpha\) is the unique subfield of \(K_\alpha=\Q(\mu_{p^{\alpha+1}})\text{,}\) that corresponds to the unique cyclic subgroup \(H\cong\Z/(p-1)\Z\subset(\Z/p^{\alpha+1}\Z)^\times\) via the Galois correspondence [28, Chapter VI, Theorem 1.1]. Let \(\chi\) be a character of \(\Gal(F_\alpha/\Q)\cong\Gamma_\alpha\text{.}\) Therefore \(\chi\) is of the second kind, i.e., \(\psi=\chi\) and \(\theta=1\) in the decomposition (3.12). Define \(\chi_n\coloneqq\chi\omega^n\text{,}\) where \(\omega\) is the Teichmüller character as defined on page 30. Therefore, if we decompose \(\chi_n=\theta_n\psi_n\text{,}\) we see that \(\psi_n=\chi\text{,}\) \(\theta_n=\omega^n\text{,}\) and \(\zeta_{\chi_n}=\chi_n(1+p)^{-1}=\chi(1+p)^{-1}=\zeta_\chi\text{.}\) As a result, by Proposition 3.6, \(f(T,\theta_n)\in\Z_p\llbracket T\rrbracket\text{.}\) Now, we define
\begin{gather} F_n(T)\coloneqq f((1+T)(1+p)^{1-n}-1,\omega^n).\tag{4.3} \end{gather}
For any \(n\geq1\) such that \(p-1\nmid n\text{,}\) \(\theta_n=\omega^n\neq1\text{;}\) hence \(f(T,\theta_n)\in\Z_p\llbracket T\rrbracket\) and \(1+p\in1+p\Z_p\) is a unit in \(\Z_p\text{.}\) Consequently, \(F_n(T)\in\Z_p\llbracket T\rrbracket\text{.}\) Moreover, by Theorem 3.9 and equation (4.3) we obtain
\begin{align} F_n(\zeta_\chi-1)=L_p(1-n,\chi_n)\amp=(1-\chi(p)p^{n-1})L(1-n,\chi)\notag\\ \amp=\left\{\begin{matrix} L(1-n,\chi)\amp\chi\neq1\\(1-p^{n-1})\zeta(1-n)\amp\chi=1 \end{matrix}\right.\tag{4.4} \end{align}
as \(\chi\) is a character of conductor dividing \(p^\alpha\) and hence \(\chi(p)=0\) for \(\chi\neq1\) and \(\chi(p)=1\) for the trivial character.
To proceed further, we need to analyze the leading coefficient \(a_0\coloneqq F_n(0)\) of the power series \(F_n(T)\in\Z_p\llbracket T\rrbracket\text{.}\) Note that, for the trivial character \(\chi\text{,}\) \(\zeta_{\chi_n}=\zeta_\chi=\chi(1+p)^{-1}=1\) and hence
\begin{align} a_0=F_n(0)\amp=F_n(\zeta_\chi(1+p)^{1-n}-1,\omega^n)\notag\\ \amp=(1-p^{n-1})\zeta(1-n)=-(1-p^{n-1})B_n/n.\tag{4.5} \end{align}
We know that \(B_n/n\) is a \(p\)–adic integer whenever \(p-1\nmid n\) [19, Proposition 15.2.4]. A prime \(p\) is called regular if \(B_{2i}\) is in \(\Z_p^\times\) for all \(1\leq i\leq (p-3)/2\text{.}\) For every positive even integer \(n\) there exists unique \(m\in\{2,4,\ldots,p-3\}\) such that \(n\equiv m\pmod{p-1}\text{.}\) As a result of Kummer congruences (Theorem A), \(B_n/n\) is in \(\Z_p^\times\) if and only if \(B_m/m\) is in \(\Z_p^\times\text{.}\) If \(p\) is regular, \(p\nmid B_m\) and hence \(p\nmid B_m/m\) as \(0\lt m\lt p\text{,}\) for all \(m\in\{2,4,\ldots,p-3\}\) and hence \(p\) is regular if and only if \(B_n/n\) is in \(\Z_p^{\times}\) for all even positive integers \(n\) not divisible by \(p-1\text{.}\) Equivalently, \(p\) is regular if and only if \(F_n(0)\) is in \(\Z_p^\times\) for all even positive integers \(n\) not divisible by \(p-1\text{.}\) But, if \(p\) is irregular, then it’s not necessarily true that \(p\mid B_n/n\) for all even positive integers \(n\) is not divisible by \(p-1\text{.}\) So, there are two cases we have to consider: (1) when \(p\) does not divide \(B_n/n\text{,}\) \(a_0=F_n(0)\) is in \(\Z_p^\times\text{,}\) and (2) when \(p\mid B_n/n\text{,}\) \(a_0=F_n(0)\in p\Z_p\text{,}\) for a fixed even positive integer \(n\) not divisible by \(p-1\text{.}\)
For a positive integer \(n\text{,}\) let \(R_n\) denote the set of all \(n\)-th roots of unity and \(P_n\) denote the set of all primitive \(n\)-th roots of unity. Then \(\abs{R_n}=n\) and \(\abs{P_n}=\varphi(n)\text{,}\) where \(\varphi\) is the Euler totient function. It is worth noting that as \(\chi\) runs over the characters of \(\Gamma_\alpha\text{,}\) the values \(\zeta_\chi=\chi(1+p)^{-1}\) runs over all \(p^{\alpha}\)–th roots of unity. We will use these notations and facts in the proof of Theorem 4.1, especially, in Lemma 4.8.
For \(\alpha\geq0\text{,}\) the local field \(\Q_p(\mu_{p^{\alpha+1}})\) is totally ramified of degree \(\varphi(p^{\alpha+1})=p^\alpha(p-1)\text{,}\) its ring of integers \(\mathscr{O}_{\Q_p(\mu_{p^{\alpha+1}})}\) is \(\Z_p[\mu_{p^{\alpha}}]\text{,}\) the unique maximal ideal of \(\Z_p[\mu_{p^{\alpha+1}}]\) is generated by \(\xi-1\) for a primitive \(p^{\alpha+1}\)–th root of unity \(\xi\text{,}\) and \(\Q_p(\mu_{p^{\alpha+1}})/\Q_p\) is a Galois extension with \(\Gal(\Q_p(\mu_{p^{\alpha+1}})/\Q_p)\cong(\Z/p^{\alpha+1}\Z)^\times\) [34, Chapter II, Proposition 7.13].
Before we proceed, we need a few Lemmas.

Proof.
  1. Let \(N_\alpha\) be the norm \(N_{\Q_p(\mu_{p^{\alpha+1}})/\Q_p}\) and \(\xi\) be a primitive \(p^{\alpha+1}\)-th root of unity. It is easy to see that for any \(\sigma\in\Gal(\Q_p(\xi)/\Q)\text{,}\) \(\sigma(F_n(\xi-1))=F_n(\sigma(\xi)-1)\) as \(F_n(T)\in\Z_p\llbracket T\rrbracket\text{.}\) As \(\sigma\) varies in \(\Gal(\Q_p(\xi)/\Q)\text{,}\) \(\sigma(\xi)\) varies over all the primitive \(p^{\alpha+1}\)-th roots of unity. Therefore,
    \begin{align} N_\alpha(F_n(\xi-1))\amp=\prod_{\sigma\in\Gal(\Q_p(\xi)/\Q_p)}\sigma(F_n(\xi-1))\notag\\ \amp=\prod_{\sigma\in\Gal(\Q_p(\xi)/\Q_p)}F_n(\sigma(\xi)-1)\notag\\ \amp=\prod_{\zeta\in P_{p^{\alpha+1}}}F_n(\zeta-1)=H_{\alpha,n}.\tag{4.6} \end{align}
    and hence \(H_{\alpha,n}\in\Z_p\text{.}\) Now, by equation (4.5), \(p\mid a_0\text{;}\) hence \(F_n(\xi-1)\in\ideal{\xi-1}=\mf{p}\text{.}\) Therefore, \(H_{\alpha,n}=N_\alpha(F_n(\xi-1))\in p\Z_p\text{,}\) since \(\Z_p\cap\mf{p}=p\Z_p\text{.}\)
  2. Recall that, for the trivial character \(\chi\text{,}\) \(\zeta_{\chi_n}=\zeta_\chi=\chi(1+p)^{-1}=1\text{;}\) hence, from equation (4.5), \(a_0=F_n(0)=-(1-p^{n-1})B_n/n\text{.}\) Since \(p\nmid B_n/n\) and \(n-1\geq1\text{,}\) \(a_0=-(1-p^{n-1})B_n/n\) is in \(\Z_p^{\times}\text{.}\) Now, \(F_n(T)\) can be written as \(\sum_{n=0}^\infty a_nT^n\text{,}\) where \(a_n\in\Z_p\) and \(F_n(0)=a_0\) is a unit in \(\Z_p\text{.}\) Therefore,
    \begin{gather} F_n(\zeta_\chi-1)\equiv a_0\pmod{\zeta_\chi-1}.\tag{4.7} \end{gather}
    For a primitive \(p^{\alpha+1}\)-th root of unity \(\xi\text{,}\) \(\ideal{\xi-1}=\mf{p}\text{.}\) Consequently, we can write congruence (4.7) as
    \begin{gather*} F_n(\xi-1)\equiv a_0\pmod{\mf{p}}. \end{gather*}
    Since \(a_0\in\Z_p^\times\text{,}\) \(a_0^{p^\alpha(p-1)}\equiv1\pmod{p}\) and hence taking the product over all primitive \(p^{\alpha+1}\)–th roots of unity \(\zeta\text{,}\) we obtain
    \begin{gather} H_{\alpha,n}=\prod_{\zeta\in P_{p^\alpha}}F_n(\zeta-1)\equiv a_0^{p^{\alpha}(p-1)}\equiv1\pmod{\mf{p}}.\tag{4.8} \end{gather}
    Since \(H_{\alpha,n}\) is the norm of \(F_n(\xi-1)\) and \(F_n(\xi-1)\in\Z_p[\xi]^\times\text{,}\) we conclude that \(H_{\alpha,n}\in\Z_p^\times\text{.}\) Therefore, congruence (4.8) updates itself to \(H_{\alpha,n}\equiv1\pmod{p}\text{,}\) since \(\Z_p\cap\mf{p}=p\Z_p\text{.}\)
This finishes the proof.

Proof.
Fix an odd prime \(p\) and an even positive integer \(n\) not divisible by \(p-1\text{.}\) We will proceed via induction on \(\alpha\text{.}\) The base case is \(\alpha=0\) and hence \(\zeta_{F_0}=\zeta\) is the Riemann Zeta function. Therefore, \(\zeta_{F_0}(1-n)=-B_n/n\) is a \(p\)–adic integer by [19, Proposition 15.2.4]. Suppose that \(\zeta_{F_{\alpha}}(1-n)\) is a \(p\)–adic integer for some \(\alpha\geq0\text{.}\) Then, by Theorem 2.16,
\begin{align} \zeta_{F_{\alpha+1}}(1-n)\amp=\prod_{\chi\in\widehat{\Gamma_{\alpha+1}}}L(1-n,\chi)\notag\\ \amp=(1-p^{n-1})^{-1}F_n(0)\prod_{\substack{\chi \in\widehat{\Gamma_{\alpha+1}}\\\chi\neq1}}F_n(\zeta_\chi-1)\notag\\ \amp=(1-p^{n-1})^{-1}\prod_{\zeta\in R_{p^{\alpha+1}}}F_n(\zeta-1).\tag{4.9} \end{align}
Note that \(P_{p^{\alpha+1}}\) consists of those elements of \(R_{p^{\alpha+1}}\) which are not elements of \(R_{p^{\alpha}}\text{.}\) Therefore, equation (4.9) shows that
\begin{align} \zeta_{F_{\alpha+1}}(1-n)\amp=(1-p^{n-1})^{-1}F_n(0)\prod_{\substack{\zeta\in R^{\alpha}\\\zeta\neq1}}F_n(\zeta-1)\prod_{\zeta\in P_{p^{\alpha+1}}}F_n(\zeta-1)\notag\\ \amp=(1-p^{n-1})^{-1}F_n(0)\prod_{\substack{\chi\in\widehat{\Gamma_\alpha}\\\chi\neq1}}F_n(\zeta_\chi-1)H_{\alpha,n}\notag\\ \amp=\zeta_{F_\alpha}(1-n)H_{\alpha,n},\tag{4.10} \end{align}
where the last equality follows from Theorem 2.16 and equation (4.4). As shown in the proof of Lemma 4.7, \(H_{\alpha,n}\text{,}\) being the norm of an element \(F_n(\xi-1)\in\Z_p[\xi]\text{,}\) is a \(p\)–adic integer. Therefore, by induction hypothesis, we conclude that \(\zeta_{F_{\alpha+1}}(1-n)\) is a \(p\)–adic integer as well. This finishes the induction step and hence the proof.

Proof of Theorem 4.1.
Recall, from equation (4.4) that
\begin{align*} F_n(\zeta_\chi-1)\amp=\left\{\begin{matrix} L(1-n,\chi)\amp\chi\neq1\\(1-p^{n-1})\zeta(1-n)\amp\chi=1 \end{matrix}\right. \end{align*}
Consequently, by Theorem 2.16, and as computed in the previous Lemma, equation (4.10)
\begin{gather} \zeta_{F_{\alpha+1}}(1-n)=\zeta_{F_\alpha}(1-n)H_{\alpha,n}.\tag{4.11} \end{gather}
As discussed above, before Lemma 4.7, we divide the proof in two cases:
Case 1: When \(v_p(B_n/n)\geq1\text{,}\) where \(v_p\) is the normalized \(p\)–adic valuation on \(\Q_p\) such that \(v_p(p)=1\text{.}\) We claim that \(v_p(\zeta_{F_\alpha}(1-n))\geq\alpha+1\) for all \(\alpha\geq0\text{.}\) We proceed via induction on \(\alpha\text{.}\) The base case is \(\alpha=0\text{.}\) Note that \(\zeta_{F_0}(1-n)=-B_n/n\) is divisible by \(p^{\alpha+1}=p\text{;}\) the base case is true, by assumption. Suppose that \(v_p(\zeta_{F_{\alpha}}(1-n))\geq\alpha+1\) for some \(\alpha\geq0\text{.}\) Then, by Lemma 4.7, equation (4.10), and the induction hypothesis,
\begin{align*} v_p(\zeta_{F_{\alpha+1}}(1-n))\amp=v_p(\zeta_{F_\alpha}(1-n)H_{\alpha,n})\\ \amp=v_p(\zeta_{F_\alpha}(1-n))+v_p(H_{\alpha,n})\\ \amp\geq \alpha+1+1\\ \amp=\alpha+2. \end{align*}
This finishes the induction step and hence the proof. This, together with Lemma 4.8, implies that
\begin{gather*} \zeta_{F_{\alpha+1}}(1-n)\equiv\zeta_{F_\alpha}(1-n)\pmod{p^{\alpha+1}\Z_p}. \end{gather*}
Case 2: When \(p\) does not divide \(\zeta(1-n)=-B_n/n\text{,}\) by Lemma 4.7 part (2), \(H_{\alpha,n}\equiv1\pmod{p}\text{.}\) As we have computed in the proof of Lemma 4.8,
\begin{gather*} \zeta_{F_{\alpha+1}}(1-n)-\zeta_{F_\alpha}(1-n)=\zeta_{F_\alpha}(1-n)(H_{\alpha,n}-1). \end{gather*}
Thus, it’s enough to prove \(H_{\alpha,n}\equiv1\pmod{p^{\alpha+1}}\text{.}\) Let \(U^{(n)}\) denote the unit group \(U^{(n)}=1+p^{n}\Z_p\text{,}\) for \(n\geq1\text{.}\) Local Class Field Theory [34, Chapter V] allows us to compute the norm group \(N_\alpha(\Q_p(\xi)^\times)\) [34, Chapter V, Proposition 1.8, p. 323]:
\begin{gather*} N_\alpha(\Q_p(\xi)^\times)=\ideal{p}\times U^{(\alpha+1)}. \end{gather*}
Therefore, \(H_{\alpha,n}\text{,}\) being the norm \(N_\alpha(F_n(\xi-1))\) of \(F_n(\xi-1)\in\Z_p[\xi]^\times\subset\Q_p(\xi)^\times\text{;}\) hence an element of \(N_\alpha(\Q_p(\xi)^\times)\text{.}\) But \(v_p(H_{\alpha,n})=0\text{;}\) hence \(H_{\alpha,n}\) belongs to \(U^{(\alpha+1)}\text{.}\) Equivalently,
\begin{gather*} H_{\alpha,n}\equiv1\pmod{p^{\alpha+1}\Z_p}. \end{gather*}
This concludes the proof of our first main result.
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