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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?

  • 1
    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...2011-02-04
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    Sorry, I should have said that I know that $1-\zeta$ is irreducible.2011-02-04
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    @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.2011-02-04
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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 $\mathbb{Z}[\zeta]$.2011-02-04
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    @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.2011-02-04
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    Ahhh! Mea culpa - it's been quite a while since I poked at this stuff (and I never had a whole lot of ring/field theory) so my recollection is embarrassingly fuzzy.2011-02-04

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