prove an equivalence mod $p$ in a ring of cyclotomic field with $p$ prime Let $p\ge 5$ be an odd prime in $\Bbb Z$.
Let $R=\Bbb Z[\xi]$ with $\xi$ a $p$-th root of unity.
Let $b\in R$ such that $b=\sum_{i=0}^{p-2}\alpha_i\xi^i$.
I am trying to prove that if $a+b\xi\equiv u b^p$ for some unit $u$ then: $a+b\xi\equiv (a+b\xi^{-1})\xi^k \bmod\ p$ for some $0\le k<p$ and that actually $k=1$ if $p\nmid ab$.
What I did:
I proved that $b^p\equiv c \bmod p$ for $c=\sum_{i=0}^{p-2}\alpha_i^p$.
We have: $a+b\xi^{-1}=\overline {a+b\xi} \equiv \overline {u\alpha  ^{p}}\bmod p \equiv \overline {u}c\bmod p$
and we have also by Kummer's lemma since $u$ is a unit: $u=\overline  u \xi^k$ for some $0\le k<p$
Then:
$a+b\xi \equiv u\alpha  ^{p}\mod p \equiv uc\mod p \equiv \overline uc\ 
\xi^k \mod p$
from the above we deduce that $a+b\xi\equiv (a+b\xi^{-1})\xi^k\mod p$.


*

*Is this part correct? 

*I don't have any clues on why $k=1$?
Thank you for your help.
 A: I am trying to restate, so that the statement becomes natural, and the ring supporing the computations becomes simpler.
The given relation $a+b\xi=ub^p$ modulo $p$, and the 
condition $p\not| ab$ are both relations modulo $p$. So it is natural to consider them in the ring $S=R/p$, which has a simpler structure:
$$
\begin{aligned}
S 
&:= R/p\\
&=\Bbb Z[\xi]/p\\
&=\Big(\ \Bbb Z[X]/(X^{p-1}+\dots+X+1)\ \Big)\ /\ p\\
&=\Bbb Z[X]\ / (\ X^{p-1}+\dots+X+1\ ,\ p\ )\\
&=\Big(\ \Bbb Z[X]/\ p\ \Big)\ /\ (X^{p-1}+\dots+X+1)\\
&=\Bbb F_p [X]\ /\ (X^{p-1}+\dots+X+1)\\
&=\Bbb F_p [X]\ /\ (1-X)^{p-1}\\
&=\Bbb F_p [Y]\ /\ Y^{p-1}\\
\end{aligned}
$$
Isomorphisms were written as equalities. The last passage corresponds to $1+X=Y$.
(More exactly, the last $=$ corresponds to two maps, the one map $\to$ brings
$X\to (1+Y)$, the inverse map $\leftarrow$ then $Y\to 1-X$.) 
Let us denote by $x$ the value of $X$ taken modulo $(1-X)^{p-1}$, and by $y$ the value of $Y$ taken modulo $Y^{p-1}$.
Then we have formally (elements mapped in each other by isomorphisms, with arrows denoted by the equal sign):
$$
(\xi \text{ modulo }p) \ =\ 
x \ =\  1+y\ .
$$
Now the given framework of the OP is as follows:

Given data: Let $p\ge 5$ be prime, let 
  $S$ be the ring
  $$ S=\Bbb F_p[y]=\Bbb F_p[Y]/(Y^{p-1})\ . $$
  An element $a\in S$ can then be uniquely written as $a=a_0+a_1y+\dots +
a_{p-2}y^{p-2}$, and we tacitly use this notation for the coefficients of
  the basis $1,y,\dots,y^{p-2}$ of $S$ over the field $\Bbb F_p$ with $p$ elements.
In particular, $a$ is invertible in the ring $S$, in notation $a\in S^\times$, iff $a_0\in\Bbb F_p^\times$, i.e.
  $a_0\ne 0$. 
Consider $a,b\in S$, $u\in S^\times$, so that a relation of the shape is valid in $R$:
  $$ a+b(1+y)=ub^p\ .$$

In this framework we ask the

Question: Under which circumstances there is an integer power $k$ such that 
  $$ a+b(1+y) = (a+b(1+y)^{-1})(1+y)^k \ ? $$

Discussion:
We write $b=b_0+y(\dots)$, so $b^p=b_0^p+y^p(\dots)^p=b_0^p=b_0$.
In the case of $b_0=0$ we start thus with the relation $a+b(1+y)=0$. So also $a_0=0$, and $b$ is of the shape $y^L\beta$, where $\beta$ is a unit. We obtain a formula for $a$, which is linear in $\beta$, and we want to show an other formula, which is linear in $\beta$. So we may and do assume $\beta=1$. (Or minus one.) It is further enough to take $L=1$. This gives a counterexample:
$$
b=-y\ ,\qquad   a=-b(1+y)=y(1+y)\ ,
$$
because $a+b(1+y)^{-1}=y(1+y)-y(1+y)^{-1}=y(1+y)^{-1}\Big(\ (1+y)^2-1\ \Big)
=y^2(1+y)^{-1}(2+y)\ne 0$. (Ironically, we need $p>3$ at the last step.)
In the OP, i could not figure out why we can write $a+b\xi^{-1}=\overline{a+b\xi}$.
For the case $p=5$ we can for instance easily compute in sage:
sage: K.<c> = CyclotomicField(5)
sage: R = K.OK()
sage: y = R(c)-1
sage: y.is_unit()
False
sage: y.norm()
5
sage: a, b = y*(1+y), -y  
sage: b^5
-15*c^3 + 5*c^2 - 10*c - 5
sage: a+b*c, a+b/c
(0, -c^3 - 2*c - 2)
sage: (a+b*c).norm(), (a+b/c).norm()
(0, 25)

In the case $b_0\ne 0$, we start with a relation which is linear in $b_0$, and need one which is also linear in $b_0$. So we may and do assume $b_0=1$. But here we have even less constraints, on the R.H.S of $a+b\xi=u b^p(=u)$ we have too much freedom of choice.

I suppose that $a,b$ may be / should have been integers. (Then the argument using the fact that $a+b\xi^{-1}$ and $a+b\xi$ are conjugated is valid.) 
A: This is a proof by elimination to $k=1$ but I wonder if there is a more elegant proof.
Suppose $k=0$. then:
$a+b\xi\equiv a+b\xi^{-1} mod\ p \Rightarrow b(\xi-\xi^{-1})\equiv 0 mod\ p $.
$\Rightarrow \xi^{-1} b(\xi^2-1)\equiv 0 mod\ p $.
$\Rightarrow \xi^{-1} b(\xi-1)(\xi+1)\equiv 0 mod\ p $.
but $(\xi-1)$ and $(\xi+1)$ and $\xi^{-1}$ are units in $R$ so multiplying by their respective inverses in $R$ we get: $b\equiv 0\ mod\ p$ which contradicts $p\nmid ab$. 
Suppose $k\ge 2$.
Then $a+b\xi\equiv a\xi^{k}+b\xi^{k-1}\ mod\ p$.
Take $\delta =\sum_{i=0}^{p-2}\delta_i\xi^i$ such that $a+b\xi\equiv a\xi^{k}+b\xi^{k-1}+\delta p$
by identifying right and left sides in the power basis of $R$, we have $a=\delta_0 p$. If $\delta_0=0$ then $a=0$ then $p|ab$ which is a contradiction.
Therefore $\delta_0\ne0$ and $p|a$ which gives $p|ab$ which is also a contradiction.
Finally $k=1$.
I found no use here of the condition $p\ge 5$ so either it is superfluous or my proof above is wrong.
