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I am reading about $p$-adic $L$-functions and I have one question in mind.

To start with, I will write a proof I've learned of a congruence of $L$-values:

Theorem: Let $p\geq5$ be a prime, $\alpha\geq1$ be any integer, and $F_{\alpha}$ be the subextension of $\Bbb Q(p^{\alpha+1})$ such that ${\rm Gal}(F_{\alpha}/\Bbb Q)=\Bbb Z/p^{\alpha}\Bbb Z$, which exists by Galois theory since the Galois group of $\Bbb Q(\zeta_{p^{\alpha+1}})$ is cyclic of order $p^{\alpha}(p-1)$. Then $\zeta_{F_{\alpha}}(-1)\equiv\zeta_{F_{\alpha+1}}(-1)\pmod{p}$.

Proof: Let $\Lambda=\Bbb Z_p[\![T]\!]$ be the Iwasawa algebra, $\omega:(\Bbb Z/p\Bbb Z)^{\times}\longrightarrow\Bbb Q(\zeta_p)\subset\Bbb C_p$ be the $p$-adic cyclotomic character. There exists a function $g\in\Lambda$ (sometimes denoted as $g(T,\omega^{-1})$) such that \begin{align*} g((1+p)^{-1}-1)=(1-p)\zeta(-1)=\frac{1}{12}(p-1) \end{align*}For every Dirichlet character $\chi\neq1$, the conductor $p^{\alpha}$ is the $p$-primary order and hence associated to $F_{\alpha}$ such that \begin{align*} g(\zeta(1+p)^{-1}-1)=L(-1,\chi) \end{align*} where $\zeta$ is the value of $\chi$ on the topological generator $(1+p)$ of $1+p\Bbb Z_p$. We can also write in the following way: put \begin{align*} T=(1+p)^{-1}(1+X)-1=-\frac{p}{p+1}+\frac{X}{p+1} \end{align*} Therefore, $g(T)=f(X)$ for $f\in\Lambda$ now verifies \begin{align*} f(\zeta-1)=g((1+p^{-1})\zeta-1)=L(-1,\chi) \end{align*} For $\chi=1$, we can write $f(0)=L(-1,1)=\zeta(-1)$. Therefore, we have \begin{align*} \prod_{\zeta}f(\zeta-1)=\prod_{\chi}L(-1,\chi)=(1-p)\zeta_{F_{\alpha}}(-1) \end{align*} Here we have used the well-known Artin formalism for complex $L$-functions.(Example: For $K=\Bbb Q(\zeta_n)$ a cyclotomic field, we have $\zeta_K(s)=\prod_{\chi}L(s,\chi)$, product taken over the Dirichlet characters modulo $n$.) Therefore, \begin{align*} \zeta_{F_{\alpha}}(-1)-\zeta_{F_{\alpha+1}}(-1) &=(1-p)\zeta_{F_{\alpha}}(-1)H_{\alpha+1} \end{align*} where \begin{align*} H_{\alpha+1}=\prod_{\substack{\zeta\\\zeta\text{ primitive}\\\text{of order $p^{\alpha+1}$}}}f(\zeta-1)-1 \end{align*} There are total $\varphi(p^{\alpha})=(p-1)p^{\alpha}$ such primitive roots of unity. Also, there is a unit $a_0$ such that \begin{align*} f(\zeta-1)\equiv a_0\pmod{\zeta-1} \end{align*} It is a well-known fact that \begin{align*} \prod_{\substack{\zeta\\\zeta\text{ primitive}\\\text{of order $p^{\alpha+1}$}}}(\zeta-1)=\Phi_{p^{\alpha+1}}(1)=p \end{align*} Therefore, \begin{align*} H_{\alpha+1} &=\prod_{\substack{\zeta\\\zeta\text{ primitive}\\\text{of order $p^{\alpha+1}$}}}f(\zeta-1)\\ &\equiv a_0^{(p-1)p^{\alpha}}-1\pmod{p}\\\implies H_{\alpha+1} &\equiv 0\pmod{p}\\\tag{since $a_0$ is a unit in $\Bbb Z/p\Bbb Z$} \end{align*} This completes the proof.

Please let me know if this proof is correct!

Here as we can see the proof uses the factorization formula in a crucial way. I am thinking about similar factorization for $p$-adic $L$-functions. I know that in his "Lectures on $p$-adic $L$-functions" Iwasawa defined: if $K$ is a real abelian extension then the $p$-adic zeta function for $K$ is given by $\zeta_{K,p}(s)=\prod_{\chi\in\widehat{{\rm Gal}(K/\Bbb Q)}}L_p(s,\chi)$. This is clearly motivated by the Artin formalism for complex $L$-functions.

My question is: Can we define a $p$-adic analog of $\zeta_{F_{\alpha}}$ and rephrase the proof of the Theorem above in terms of a factorization of $p$-adic zeta function $\zeta_{F_{\alpha},p}$ into $p$-adic Dirichlet $L$-functions?

Can we obtain simpler proof of other congruences of complex $L$-values using $p$-adic $L$-function/$p$-adic Artin formalism?

Edit/Addition: The Galois group of the subextension $F_{\alpha}$ is inside the totally real subfield of $\Bbb Q(\zeta_{\alpha})$. Hence following Iwasawa/Leopoldt, we may define their $p$-adic $L$-function as \begin{align*} \zeta_{F_{\alpha},p}(1-n) &=\prod_{\chi}L_p(1-n,\chi) \end{align*} where the product is taken over the characters $\chi$ of ${\rm Gal}(F_{\alpha}/\Bbb Q)\cong\Bbb Z/p^{\alpha}\Bbb Z$.

Also, we know by the interpolation formula that \begin{align*} L_p(1-n,\chi)=(1-\chi(p)\omega^{-n}(p)p^{n-1})L(1-n,\chi\omega^{-n})\equiv L(1-n,\chi\omega^{-n})\pmod{p} \end{align*} ($\omega$ is the $p$-adic cyclotomic character)which gives us \begin{align*} \zeta_{F_{\alpha},p}(-1)\equiv\prod_{\chi}L(-1,\chi\omega^{-2})\pmod{p} \end{align*}

From this can we say the that the Theorem above gives us a congruence for $p$-adic L-functions as follows?

\begin{align*} \zeta_{F_{\alpha},p}(-1)\equiv\zeta_{F_{\alpha+1},p}(-1)\pmod{p} \end{align*}

Thanks in advance for any helpful comments/answers!

PS. This is not so well-thought or well-written, so apologies for posting here.

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    $\begingroup$ Please define $F_{\alpha}$: specifying its Galois group is far from sufficient. $\endgroup$ Feb 13, 2023 at 21:35
  • $\begingroup$ @HenriCohen Since the Galois group of $\Bbb Q(p^{\alpha+1})$ is cyclic of order $p^{\alpha}(p-1)$, it has a subextension whose Galois group is cyclic of order $p^{\alpha}$. So, I meant that subextension. $\endgroup$
    – ShBh
    Feb 13, 2023 at 22:00
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    $\begingroup$ I thought as much. You should edit accordingly. $\endgroup$ Feb 13, 2023 at 22:03
  • $\begingroup$ @HenriCohen I have edited! Thanks for the suggestion! $\endgroup$
    – ShBh
    Feb 13, 2023 at 22:11

1 Answer 1

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I have not checked your proof, but did some numerical experiments, which should be easy to prove:

For $p=3$ your claim is definitely false: for $\alpha=1$, $2$, $3$ the values of the zeta function are -1/9, -373312/27, -bignumber/81, so probably always integer/$3^{\alpha+1}$.

For $p=5$ and $p=7$, the zeta values are integers/3, and for $\alpha=1$ they are indeed congruent modulo $p$, and even modulo $p^2$, so perhaps modulo $p^{\alpha+1}$ in general.

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  • $\begingroup$ I am really sorry! My proof actually works for primes $p\geq 5$. I have edited. Do you mean this congruence is true for the $p$-adic Zeta functions as well that I mentioned at the end? $\endgroup$
    – ShBh
    Feb 14, 2023 at 11:27
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    $\begingroup$ There is a good reason why $p = 3$ does not work: for $p = 3$, you have $-1 = 1 \bmod {p-1}$. Throughout this theory, one needs to split into cases according to the value of $s$ modulo $p-1$ (or mod 4 if $p = 2$); and the values $s = 1 \bmod {p-1}$ are bad, because you are in the same congruence class as $s = 1$ where the zeta function has a pole -- meaning you get growing denominators, just as in this example. $\endgroup$ Feb 14, 2023 at 21:24
  • $\begingroup$ @DavidLoeffler Yes! That's very illuminating! Thanks for the comment. Do you think that the p-adic L-function congruence is true at the end? $\endgroup$
    – ShBh
    Feb 15, 2023 at 2:03

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