Let $p$ be an odd prime in $\Bbb Z$.
Let $R$ be the $p$-th cyclotomic ring: $R=\Bbb Z[\xi]$ 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 2 facts:
1) $p|b$ $\Leftrightarrow$ $\forall i, p|\alpha_i$
2) if $p\nmid b$ then $b^p$ is a unit in $R$. [is that even true?]
What I did:
1) $\Leftarrow$) is obvious but I have no clue for the $\Rightarrow $)
2) we have $b^p\equiv \sum_{i=0}^{p-2} \alpha_i^p\ mod\ p$,if $p\nmid b$ then $p\nmid \alpha_i$ for some family $(\alpha_j)$. these $\alpha_j$ are units as they are prime to $p$, but how to go from there?
Thank you for your help.
For the $\implies$ in (1), let $c= \sum_{i=0}^{p-2} \beta_i \xi^i$ be such that $pc=b$. Then matching coefficients says $p\beta_i= \alpha_i$, i.e. $p \mid \alpha_i$. You can match coefficients because the powers of $\xi$ are a basis for $\mathbb{Z}[\xi]$.
To elaborate on why 2 is false, take $p=3$, so $\mathbb{Z}[\xi]$ is generated by $\frac{-1+\sqrt{-3}}{2}$. By a norms argument, if $\frac{a+b\sqrt{-3}}{2}$ is a unit, then $$a^2+3b^2=4,$$ from which you can exhaust a finite number of possibilities, and see that the only units are $\{\pm 1, \pm \xi, \pm \xi^2 = \pm (-1-\xi)\}$. Hence, for example, $8=2^3$ is not a unit.