polynomial with integer coefficients Question: Let $\Pi_{j=1}^n (z-z_j)$ be a polynomial with integer coefficients. Is also $\Pi_{j=1}^n (z-z_j^k)$ for $k=1,2,3,\dots$ a polynomial with integer coefficients?
In fact, this is a question that someone asked 3 days ago, but the answer is not clear (I think)
For this, I think that it is enough to show that $z_J$ and $z_j^k$ have same minimal polynomial, since there is a  $\mathbb{Z}$-automorphism($\mathbb{Z}$ left fixed) which map an element into its conjugate. 
But this is impossible to show, really.
How can I solve this problem?
Separately, Who knows why I can't write a comment in some questions??
 A: Yes, they are integers.
Note that the $z_j^k$ are algebraic integers, since the $z_j$ are and thus so are the coefficients of $\displaystyle \prod_j(z-z_j^k)$ it suffices to prove that the coefficients are rational. 
To do this, let $K=\mathbb{Q}(\{z_j\})$. Note that $K$ is the splitting field of $\displaystyle \prod_j(z-z_j)$ a polynomial with $\mathbb{Q}$-coefficients, and thus $K/\mathbb{Q}$ is Galois.
Let $\sigma\in\text{Gal}(K/\mathbb{Q})$ and consider any of the coefficients of $\displaystyle \prod_j (z-z_j^k)$ which are elementary symmetric polynomials $e_i(z_1^k,\ldots,z_n^k)$. Then, since $\sigma$ permutes the set $\{z_1,\ldots,z_n\}$ we have that 
$$\sigma(e_i(z_1^k,\ldots,z_n^k))=e_1(\sigma(z_1)^k,\ldots,\sigma(z_n)^k)=e_i(z_1^k,\ldots,z_n^k)$$
where we used the fact that $e_i$ is the elementary symmetric polynomial.
Thus, since $K/\mathbb{Q}$ is Galois, we have that $e_i(z_1^k,\ldots,z_n^k)\in K^{\text{Gal}(K/\mathbb{Q})}=\mathbb{Q}$. Thus, from previous comment it follows that $e_i(z_1^k,\ldots,z_n^k)\in\mathbb{Z}$ as desired.
A: Yes. Obviously $\prod (z - z_j^k)$ is invariant under permutation of $z_j$-s, so if you multiply it out, at each power of $z$ you'll get a polynomial in variables $z_1, \ldots, z_n$ that's invariant under permutation of variables. A fundamental theorem of symmetric polynomials says that each such polynomials is actually a polynomial in elementary symmetric polynomials with integer coefficients, but the values of elementary symmetric polynomials taken on $(z_1, \ldots, z_n)$ are just the coefficients of $\prod(z - z_j)$ which are integers, so the coefficients of $\prod(z - z_j^k)$ are values of polynomials with integer coefficients on integer tuples, thus they're integers.
