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??

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    $\begingroup$ Because you do not have 50 reputation points yet, you can only comment on your own questions and answers. The "add comment" button will only appear for you once you gain 50 points. Here is an explanation of reputation points. $\endgroup$ – Zev Chonoles Oct 10 '13 at 5:21
  • $\begingroup$ oh.. I see.. Thanks!:) $\endgroup$ – NNNN Oct 10 '13 at 5:31
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    $\begingroup$ Please use **text** to make bold text, unless the text is actually part of the equation/math. $\endgroup$ – dfeuer Oct 10 '13 at 5:39
  • $\begingroup$ @dfeuer why??? Is any difference between them?? $\endgroup$ – NNNN Oct 10 '13 at 5:55
  • $\begingroup$ @JeongNam-ho, yes. Aside from being slightly different fonts, the MathJax version wastes time running Javascript for no reason. This sort of thing actually does matter for people using smartphones and and tablets. $\endgroup$ – dfeuer Oct 10 '13 at 6:29

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


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.

  • $\begingroup$ Um.....Is'elementary symmetric polynomial' a concept that is learned in graduate school?? $\endgroup$ – NNNN Oct 10 '13 at 5:48
  • $\begingroup$ @JeongNam-ho No. If THAT is the part of my answer you're confused about, then I am confused. $\endgroup$ – Alex Youcis Oct 10 '13 at 5:49
  • $\begingroup$ Because I have learned the elementary symmetric polynomial during undergraduate, I should study it first, to understand your explain...... Anyway, Thank you very much~ $\endgroup$ – NNNN Oct 10 '13 at 5:53
  • $\begingroup$ @JeongNam-ho Yeah, if you don't know it, look it up. If you understand Galois theory, you should have no issue understanding elementary symmetric polynomials. $\endgroup$ – Alex Youcis Oct 10 '13 at 6:05
  • $\begingroup$ You are correct. Thank you for your help:) $\endgroup$ – NNNN Oct 10 '13 at 6:11

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.


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