# $p$-adic $L$-functions and congruence of $L$-values

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*}

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

• Please define $F_{\alpha}$: specifying its Galois group is far from sufficient. Feb 13, 2023 at 21:35
• @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.
– ShBh
Feb 13, 2023 at 22:00
• I thought as much. You should edit accordingly. Feb 13, 2023 at 22:03
• @HenriCohen I have edited! Thanks for the suggestion!
– ShBh
Feb 13, 2023 at 22:11

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.

• 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?
– ShBh
Feb 14, 2023 at 11:27
• 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. Feb 14, 2023 at 21:24
• @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?
– ShBh
Feb 15, 2023 at 2:03