TITLE: Why $(1-\zeta)$ unit where $\zeta$ is a primitive nth and n divisible by two primes
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: Hurkyl's advice in the comments is sensible.  
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?