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I am trying to prove the following:

Let $l$ be a prime and let $\zeta$ be a $l$th root of unity. Show that, in the order $\{ 1, \zeta, \ldots, \zeta^{l-2} \}$ of the field $\mathbb{Q}(\zeta)$, if a product $\alpha \beta$ is divisible by $1-\zeta$, then $\alpha$ or $\beta$ must be divisible by $1-\zeta$.

I know that $1-\zeta$ is irreducible in the maximal order $\mathbb{Z}[ \zeta ]$, and I am trying to mimic the proof of that statement, but I'm stuck.

Can someone please help?

I see that all the products I am dealing with are of the form $\zeta^k$.

Also does the order contain $\zeta^{l-1}$?

I have a feeling that "the order ${ 1, \zeta, \ldots, \zeta^{l-2} }$" is actually just $\mathbb{Z}[ \zeta ]$. Is this true?

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Doesn't this come for free once you know that $1-\zeta$ is prime in the maximal order? If $1-\zeta$ had factors in the smaller order then that factorization would automatically lift to the maximal order and disprove its primality... –  Steven Stadnicki Feb 4 '11 at 19:27
    
Sorry, I should have said that I know that $1-\zeta$ is irreducible. –  AWC Feb 4 '11 at 20:23
    
@AWC: Irreducibility is not enough in any case... And Steven's argument is incorrect as stated. $6$ is irreducible in the ring $\mathbb{Z}[1/2]$, but not in the smaller ring $\mathbb{Z}$. You would need to argue no element of the order becomes a unit in the maximal order. –  Arturo Magidin Feb 4 '11 at 20:27
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@AWC: Since $\zeta$ is a root of $1+x+x^2+\cdots+x^{l-1}$, then it follows that $$\zeta^{l-1} = -1 - \zeta - \zeta^2 -\cdots - \zeta^{l-2}.$$ So, certainly, $\zeta^{l-1}$ is contained in the order generated by $1,\zeta,\ldots,\zeta^{l-2}$. And yes, the order is $\mathbb{Z}[\zeta]$. –  Arturo Magidin Feb 4 '11 at 20:30
    
@Steven: Primality implies irreducibility, but not the other way around. And in general, a nontrivial factorization in a smaller ring may become trivial in the larger one if one of the factors is invertible in the larger ring but not the smaller one. You would need to argue this does not occur with this order. –  Arturo Magidin Feb 4 '11 at 20:32

3 Answers 3

First, the ring you are talking about is just the ring of integers $\mathbb{Z}[\zeta]$, as Arturo has remarked in the comments. Actually, to see that $\mathbb{Z}[\zeta]$ is precisely the ring of integers is not entirely trivial, but I will assume this. One reference that spontaneously comes to my mind is Washington's book on Cyclotomic Fields, but I'm sure there are more elementary ones.

Here is one easy way to see that $1-\zeta$ is prime, avoiding questions like "if norms of elements divide each other, do the elements divide each other?" :

Step 1. Show that an element $x$ of an integral domain is prime if and only if the ideal $(x)$ is prime.

Step 2. Show that if an ideal has prime norm, then it is prime. Recall that the norm of an ideal is its index in the ring. (Hint: any maximal ideal is prime. Prove it!)

Step 3. Show that in the case of rings of integers of number fields, the norm of the ideal $(x)$ is equal to the absolute norm of $x$. Now, what is $\text{Norm}_{\mathbb{Q}(\zeta)/\mathbb{Q}}(1-\zeta)$? (Hint: the norm of an element of a number field is the constant coefficient of its minimal polynomial. Why?)

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Let $\alpha$ be an element of $\mathbb{Q}(\zeta)$. We denote by $N(\alpha)$ the norm of $\alpha$ with respect to $\mathbb{Q}(\zeta)/\mathbb{Q}$.

Let $\alpha$ and $\beta$ be elements of $\mathbb{Z}(\zeta)$. We denote $\alpha \equiv \beta$ (mod $(1 - \zeta)$) if $\alpha - \beta$ belongs to $(1 - \zeta)\mathbb{Z}(\zeta)$.

Lemma 1 $N(1 - \zeta) = l$

Proof: This is immediate by the following formula(replacing $X$ by $1$). $1 + X + \cdots + X^{l-1} = (X - \zeta)(X - \zeta^2)\cdots (X - \zeta^{l-1})$.

Lemma 2 Let $\alpha = f(\zeta)$ be an element of $\mathbb{Z}(\zeta)$, where $f(X)$ is a polynomial in $\mathbb{Z}[X]$. Then $\alpha \equiv 0$ (mod $(1 - \zeta)$) if and only if $f(1) \equiv 0$ (mod $l$).

Proof: Since $\zeta \equiv 1$ (mod $(1 - \zeta))$, $f(\zeta) \equiv f(1)$ (mod $(1 - \zeta))$.

Suppose $\alpha \equiv 0$ (mod $(1 - \zeta)$). Then $f(1) \equiv 0$ (mod $(1 - \zeta)$) by the above congruence. Taking norms of $f(1)$ and $1 - \zeta$, we get $f(1)^{l-1} \equiv 0$ (mod $l$) by the Lemma 1. Hence $f(1) \equiv 0$ (mod $l$)

Conversely suppose $f(1) \equiv 0$ (mod $l$). By the Lemma 1, $f(1) \equiv 0$ (mod $(1 - \zeta)$). Hence $\alpha \equiv 0$ (mod $(1 - \zeta)$) by the above congruence. This completes the proof.

Proposition Let $\alpha$ and $\beta$ be elements of $\mathbb{Z}(\zeta)$. Suppose $\alpha\beta \equiv 0$ (mod $(1 - \zeta)$). Then $\alpha \equiv 0$ (mod $(1 - \zeta)$) or $\beta \equiv 0$ (mod $(1 - \zeta)$).

Proof: This follows immediately from Lemma 2

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Here's a more direct way not relying on norms, but is phrased slighly differently than Makoto's answer.

Note that $\ell\in (\zeta-1)$ since if $f(T)=1+\cdots+T^{\ell-1}$ then $f(T+1)$ annihilates $\zeta-1$, but

$$\displaystyle f(T+1)=\ell +\sum_{j=1}^{\ell-1}a_j T^j\qquad(\ast)$$

for some $a_j$ with $\ell\mid a_j$ for $j\leqslant 1<\ell-1$. So,

$$\displaystyle 0=f((\zeta-1)+1)=\ell+\sum_{j=1}^{\ell-1}a_j T$$

so $\ell\in (\zeta-\ell)$ as desired.

Note then, note that

$$\begin{aligned}\mathbb{Z}[\zeta]/(\zeta-1) &=\mathbb{Z}[\zeta]/(\ell,\zeta-1)\\ &=\mathbb{Z}[T]/(f(T),\ell,T-1)\\ &= (\mathbb{Z}/\ell\mathbb{Z})[T]/(\overline{f(T)},\overline{T-1})\\ &= (\mathbb{Z}/\ell\mathbb{Z})[T]/(\overline{T-1})\\ &\cong \mathbb{Z}/\ell\mathbb{Z}\end{aligned}$$

Where the bar denotes taking the coefficients modulo $\ell$ and the fact that $(\overline{f(T)},\overline{T-1})=(\overline{T-1})$ follows from $(\ast)$.

EDIT: I just noticed this is like a two year old question..

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