Can $1-w$ divide an integer in $Z[w] $? I'm trying to determine whether the following is true.  Let $w$ be a primitive $p-th$ root of unity, $p$ prime not equal to 2.   In $Z[w]$ is it possible for $1-w$ to divide a nonzero integer? 
 A: Sure. Any non-zero algebraic number divides some non-zero rational integer. Look at the norm. (Multiply this by an integer if necessary.)
A: The norm of $1-\omega$ is $p$ (exercise), so $1-\omega$ always divides $p$ in $\mathbb{Z}[\omega]$, which is the ring of integers of $\mathbb{Q}[\omega]$, and no other rational prime.
A: HINT $\ $ No knowledge of norms is needed - this follows from high-school polynomial arithmetic. Namely, every algebraic number $\rm\:w\:$ divides the constant term of its minimal polynomial, i.e.
$$\rm\ a\ w^n +\:\cdots\: + b\ w + c\ =\ 0\ \ \Rightarrow\ \ (a\ w^{n-1}+\:\cdots\:+b)\ w\ =\: -c\ \ \Rightarrow\ \ w\ |\ c\:\ in\:\ \mathbb Z[w]$$
Note that the same argument works for any algebraic element that is not a zero-divisor, for then we easily infer that $\rm\:c\ne0\:$ in a minimal polynomial. On the other hand, it fails for zero-divisors, e.g. $\:$ in $\rm\ \mathbb Z[w]/(2\:w)\:,\:$ if $\rm\ f(w)\: w\: =\: n\ $ then, multiplying by $2\:,$ we infer $\rm\ 2\:n\ = 0\:,\:$ so $\rm\: n = 0\:.$
