Cyclotomic fields and valuation.

Question

Consider $L=\mathbb{Q}[\zeta]$ the $p^n$-th cyclotomic field. The ideal above $p$ is generated by $\pi_L=1-\zeta$. My question is: how can I compute the following valuation:

$$ v_{\pi_L}(1-\zeta^{a-1})$$

with $(a,p)=1$? Sorry if it is a silly question, but I am trying from hours to find a value.

Answer 1

Edit: this answer addresses the case $n=1$.

If $a\equiv 1$ (mod $p$) then $1-\zeta^{a-1}=0$. Otherwise $b=a-1$ will be prime to $p$.

It turns out that if $(b,p)=1$ then $1-\zeta$ and $1-\zeta^b$ are associates. Indeed, we have $$ \frac{1-\zeta^b}{1-\zeta}=1+\zeta+\dots+\zeta^{b-1}\in\mathbb{Z}[\zeta]$$ and since $(b,p)=1$ there is an integer $c$ such that $bc\equiv 1$ (mod $p$), hence $$ \frac{1-\zeta}{1-\zeta^b}=\frac{1-\zeta^{bc}}{1-\zeta^b}=1+\zeta^b+\dots+\zeta^{b(c-1)}\in\mathbb{Z}[\zeta]$$

Therefore $1-\zeta^b=u(1-\zeta)$ for a unit $u$, so $v_{\pi_{L}}(1-\zeta^b)=1$.

Answer 2

For a general $n$. I call $\zeta_{p^k}$ the $p^k$-th roots of unity.

Take $(a,p^n)=1$ we have:

$$ \begin{cases} 1-\zeta_{p^n}^{a-1}=0 & \ \ \text{if} \ \ a\equiv 1 \ \ (\mathrm{mod} \: p^n) \\ 1-\zeta_{p^n}^{a-1}=u(1-\zeta_{p^n}), \ u\in\left(\mathbb{Z}[\zeta_{p^n}]\right)^{\times} & \ \ \text{if} \ \ (a-1,p^n)=1 \\ 1-\zeta_{p^n}^{a-1}=1-\zeta_{p^{n-t}}^s=u(1-\zeta_{p^{n-t}}) & \ \ \text{if} \ \ a-1\equiv 0 \ \ (\mathrm{mod} \: p^t) \end{cases}$$

with $1\leq t\leq n-1, \: (s,p^t)=1,\:u\in\left(\mathbb{Z}[\zeta_{p^n}]\right)^{\times}$. We have obtained the equality in the second and third case using the reasonament of carmichael561. In the first case valuation is $\infty$, in the second we have

$$ v_{\pi}\left(1-\zeta_{p^n}^{a-1}\right)=v_{\pi}\left(1-\zeta_{p^n}\right)=1$$

For the third notice that for every $t$ we have the Galois extension $\mathbb{Q}[\zeta_{p^n}]/\mathbb{Q}[\zeta_{p^{n-t}}]$ with degree $[\mathbb{Q}[\zeta_{p^n}]:\mathbb{Q}[\zeta_{p^{n-t}}]]=p^{t}$. We have just one prime ideal $P$ above $p$ in $\mathbb{Q}[\zeta_{p^{n-t}}]$, since $p$ totally ramify in $\mathbb{Q}[\zeta_{p^n}]$. In particular $P$ is generated by $1-\zeta_{p^{n-t}}$. Again, since $p$ totally ramify in the big extension also $1-\zeta_{p^{n-t}}$ totally ramify in $\mathbb{Q}[\zeta_{p^n}]$. The ideal above $p$ in large extension is generated by $\pi=1-\zeta_{p^n}$ so we have

$$ (1-\zeta_{p^{n-t}})\mathbb{Z}[\zeta_{p^n}]=(\pi)^{p^{t}}$$

Therefore

$$ v_{\pi}(1-\zeta_{p^{n-t}})= p^{t} $$