On some congruence in cyclotomic field Let $k = \mathbb{Q}(\zeta_5)$  the $5^{th}$ cyclotomic field and $p$ a prime number verify $p\equiv 1 \pmod 5$.
Let $\lambda = 1-\zeta_5$ the unique prime of $k$ above $5$, precisely $5 = (1-\zeta_5)^4 = \lambda ^4$. We have that $p$ splits in $k$ as $p=\pi_1\pi_2\pi_3\pi_4$ with $\pi_i$ are primes in $k_0$.
I need to know what is the exact congruence of $\pi_i$ modulo $\lambda^5$ ? is it  $\pi_i \equiv\,\, 1  \pmod {\lambda^5}$ or $\pi_i \not\equiv \,\,1\,\, \pmod {\lambda^5}$.
In the case when  $p\equiv 1 \pmod {25}$ we see that $\pi_1\pi_2\pi_3\pi_4 \equiv\,\, 1  \pmod  {\lambda^5}$ so we should have $\pi_i \equiv\,\, 1 \pmod {\lambda^5}$ in order to have the congruence ? and what happen if $p \not\equiv 1 \pmod {25}$
 A: I think it will depend on $p$ due to class field theory and Chebotarev.
Let $H$ be the image of $O_k^\times$ in $O_k/(\lambda^5)^\times$ and $G=(O_k/(\lambda^5)^\times) / H$ which is a non-trivial group.

$G$ is non-trivial because $O_k/(\lambda^5)^\times= O_k/(\lambda^5) - (\lambda)/(\lambda^5)$ has
$N(\lambda^5)-N(\lambda^4)=5^5-5^4$ elements,
$O_k^\times$ is generated by $-\zeta_5$ and $1+\zeta_5$ and $1+\zeta_5$ has order $20$ in $O_k/(\lambda^5)^\times$ so $H$ has at most $200$ elements.

It is known that $O_k$ is a PID.
For a (fractional) ideal $\alpha O_k$ coprime with $\lambda$ let $f$ be the well-defined homomorphism sending $\alpha O_k$ to $\alpha\in G$.
$f$ is clearly surjective.
There is a non-trivial abelian extension $L/k$ whose Artin map is given by $f$.
If $f(\alpha)=1$ then for all $\sigma\in Gal(k/\Bbb{Q})$ we have $f(\sigma(\alpha))=1$.
Whence $p$ splits completely in $L$ iff $p \equiv 1\bmod 5$ and $f(\pi)=1$ for some prime $(\pi)$ above $p$.
If $p\equiv 1\bmod 5$ implies that $f(\pi)=1$ then every $p\equiv 1\bmod 5$ will split completely in $L$.
This is a set of primes of density $1/4$, so (by Chebotarev) it must be that $[L:\Bbb{Q}]\le 4$ ie. $L=k$, a contradiction.

The problem of understanding the $p\equiv 1\bmod 5$ with $f(\pi)=1$ will probably become clearer if someone finds $Gal(L/\Bbb{Q})$ and a field generator of $L/k$

