# Why $(1-\zeta)$ unit where $\zeta$ is a primitive nth and n divisible by two primes

From Chapter VII of Lang's Algebra.

The question asks if $n\geq 6$ and $n$ is divisible by at least two primes, show that $1-\zeta$ is a unit in the ring $\mathbb{Z}[\zeta]$

I am having a hard time understanding why this is true. This is in the integral dependence chapter, but that has not given me any inspiration. I have also tried using cyclotonic polynomial to no avail

Thanks for any direction.

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Looking at concrete examples often helps. $n=6$ is small enough that maybe it's not too hard to find an explicit inverse, assuming one exists. I bet finding the inverse will suggest a general form. On a different note, my first instinct is to find the norm. –  Hurkyl Oct 18 '12 at 2:36

Here is a more theoretical way to think about it; I've never read Lang's book, so I don't know how well it fits with the material of the chapter (but it is a standard argument in number theory):

Write $n = p^k m$ with $p \not\mid m$. Note $(1-\zeta)^{p^k} \equiv 1 - \zeta^{p^k} \bmod p.$

Now $\zeta^{p^k}$ is a primitive $m$th root of $1$, where $p \not \mid m$.

Assuming $m \neq 1$, can you use this to prove that $1 - \zeta^{p^k}$ is a unit mod $p$? And hence that $1 - \zeta$ is a unit mod $p$?

Now find another prime $q$ so that $1 - \zeta$ is also a unit mod $q$.
Once you've done this, you're done. Do you see why?

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In the question both assumptions $n>=6$ and $n$ divisible by two primes were made. If these primes are assumed distinct, then automatically $n>=6=2*3$ would follow. So is there a similar proof for the case $n=p^k$ assuming we're not looking at $n=4$? (Of course in the $n=4$ case $1-i$ isn't a unit.) –  coffeemath Oct 18 '12 at 19:06
@coffeemath: Dear coffeemath, the statement is false unless $n$ is divisible by two distinct primes. Regards, –  Matt E Oct 18 '12 at 20:25