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In the proof of Kronecker's theorem here. Can someone explain to me why the statement "Since $\alpha$ is an algebraic integer then $a_{k,n-1},\ldots,a_{k,0}\in\mathbb{Z}$" is true? I know that if $\alpha$ is an algebraic integer then the minimal polynomial of $\alpha$ is an integer coefficient polynomial. I also know that $\alpha^k$ is an algebraic integer. But does that mean $f_k$ is the minimal polynomial of $\alpha^k$?

Thanks in advance!

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+1 for showing your own thinking –  Jyrki Lahtonen Feb 8 '13 at 19:53
    
Sorry, I lost internet the last few days. –  user44322 Feb 13 '13 at 2:59

2 Answers 2

up vote 4 down vote accepted

For the benefit of the readers who don't want to check out the source let me recap. Here $\alpha$ is an algebraic integer, an $\alpha_1=\alpha,\alpha_2,\ldots,\alpha_n$ are its conjugates. The question is about the coefficients of the polynomial $$ f_k(x)=\prod_i(x-\alpha_i^k)=x^n+a_{k,n-1}x^{k-1}+\cdots+a_{k,0}. $$ The coefficients of the polynomial are symmetric polynomials in the unknowns $\alpha_i$ with integer coefficients. By basic results on symmetric polynomials they are thus values of some polynomials $g(s_1,s_2,\ldots,s_n)$ with integer coefficients evaluated at the elementary symmetric polynomials $s_i$ of $\alpha_1,\alpha_2,\ldots,\alpha_n$. Because $\alpha$ is an algebraic integer, the elementary symmetric polynomials are all rational integers (they are up to sign the coefficients of the minimal polynomial of $\alpha$). The claim follows from this.

Note that $f_k(x)$ is not necessarily the minimal polynomial of $\alpha^k$. Indeed, when $\alpha$ is a root of unity (the case in the text), there will be some repetitions among the roots $\alpha_1^k,\alpha_2^k,\ldots,\alpha_n^k$ of $f_k(x)$. At least for some exponents $k$.

Edit:

On second thought. Even though $f_k(x)$ is not necessarily the minimal polynomial of $\alpha_k$ in the problem scenario outlined in the previous paragraph, it is always a power of the minimal polynomial of $\alpha^k$. This is because by basic Galois theory each distinct conjugate of $\alpha^k$ appears in the list $\alpha_1^k,\alpha_2^k,\ldots,\alpha_n^k$ the same number of times. The fact that $f_k(x)\in\mathbb{Z}[x]$ follows also from this.

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This is a nice answer but I have never really learned symmetric polynomials. Do you have a method that doesn't involve symmetric polynomials? –  user44322 Feb 8 '13 at 20:41
    
@user44322: Does the addition of the last paragraph help you? –  Jyrki Lahtonen Feb 8 '13 at 21:16
    
An elementray symmetric polynomials with $n$ indeterminates $y_1,..y_n$ are: $$s_0=1,\ s_1=\sum_k y_k,\ s_2=\sum_{j<k} y_jy_k, \dots,\ s_n=y_1y_2\dots y_n$$ All symmetric polynomials contains these as building blocks, in the sense that it can be written as a polynomial $g(s_0,\dots,s_n)$ of these –  Berci Feb 8 '13 at 21:33

Let $K/L$ be a finite Galois extension of fields with $G$ its Galois group. Let $\alpha \in K$ and consider

$$f_\alpha(t)=\prod_{\sigma \in G}(x-\sigma(\alpha)),$$

notice that for any $\tau \in G$ we have that

$$\tau(f_\alpha(t))=\tau\left(\prod_{\sigma \in G}(x-\sigma(\alpha))\right)= \prod_{\sigma \in G}(x-\tau(\sigma(\alpha))=\prod_{\phi \in \tau G}(x-\phi(\alpha))=f_\alpha(t)$$

since $\tau G=G$. Hence $f_\alpha(t)$ is fixed by $G$ and so $f_\alpha(t) \in L[t]$ thereby $\min_L(\alpha,t) \mid f_\alpha(t)$. We can do a little better though. In particular the set $\{\sigma(\alpha) : \sigma \in G\}$ contains precisely the roots of $\min_L(\alpha,t)$, since these are the conjugates of $\alpha$. Thereby $f_\alpha(t)$ has the same roots as an irreducible polynomial and so must be a power of the irreducible polynomial.

In your instance this gives us that $f_k(t)$ is a power of the minimum polynomial of $\alpha^k$ which has $\mathbb Z$-coefficients because it is an algebraic integer.

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