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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?

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    You mean, a nonzero *rational* integer?2011-05-26

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Sure. Any non-zero algebraic number divides some non-zero rational integer. Look at the norm. (Multiply this by an integer if necessary.)

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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.

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    Thanks. Can you tell me if there exists a non zero element in Z[w] that has minimal norm?2011-05-26
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    @Myke: The question needs to change somewhat, since $1$ has minimal non-zero norm.2011-05-26
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    Okay so every non zero element has norm no smaller than 1?2011-05-26
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    @Myke: The norm of x is the product of the Galois conjugates of x, and so is pretty clearly an integer (being a sum of products of integers times powers of w, it must be an integer combination of powers of w, but it is fixed by the galois group and so lies in Q, so it is an integer).2011-05-26
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    Never mind stupid question2011-05-26
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    Alternatively, the norm is ± the constant term of its minimal polynomial.2011-05-26
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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 $\rm\: n = 0\:.$