Let $p$ be an odd prime and $\zeta_p$ a primitive $p$th root of unity, that is a $p$th root of unity other than 1. I need to show that $p$ and $(1-\zeta_p)^{p-1}$ are associates in $\mathbb{Z}[\zeta_p]$.

I don't know what I can do about this. Should I try factorizing $p$ into polynomials in $\zeta_p$?


Recall that: $$ \text{Nm}(1-\zeta)=(1-\zeta)\cdots(1-\zeta^{p-1})=\Phi(1)=p, $$ where $\Phi$ is the $p$-th cyclotomic polynomial. So the ideal generated by $p$ is: $$ (p)=(1-\zeta)\cdots(1-\zeta^{p-1}). $$

Lemma. Let $r,s\in\mathbb{Z}$ such that $\text{gcd}(p,rs)=1$. Then: $$ \frac{\zeta^r-1}{\zeta^s-1}\in(\mathbb{Z}[\zeta])^\times $$

Now $1-\zeta$ and $1-\zeta^i$ are easily shown to be associate if $\text{gcd}(p,i)=1$ by the above lemma, so they generate the same ideal, and: $$ (1-\zeta)^{p-1}=(p). $$ So $(1-\zeta)^{p-1}$ and $p$ are associate.

  • 1
    $\begingroup$ As a direct method of proving the Lemma: Since the minimal polynomial of $\zeta_p$ over $Q$ is $f(x)=(x^p-1)/(x-1)$, and as $f(x+1)=\Sigma_{i=0}^{p-1} \binom{p}{i+1}x^{î}$ is a polynomial of Eisenstein type, we know that $(p)$ totally ramifies in $Q(\zeta_p)$. But $(p)=(1-\zeta)\ldots(1-\zeta^{p-1})$, and hence every appearing term should be the same ideal, i.e. they are all associates. $\endgroup$
    – awllower
    May 9 '13 at 16:49

Let $f(x)=(x^p-1)/(x-1)$. By considering $f(1)$, show that $\prod(1-\zeta_p^k)=p$, the product being over $1\le k\le p-1$. Then show that the numbers $1-\zeta_p^k$ are all associates, by showing that both $(1-\zeta_p^k)/(1-\zeta_p)$ and its inverse are algebraic integers.


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