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Let $f \in \mathbb{Q}[x]$ of degreee $d$ be irreducible, with roots $\alpha_1,\ldots, \alpha_d$. One particular basis for the field extension of $\mathbb{Q}$ obtained by adjoining the roots of $f$ is given by $\{\alpha_j\}_{j=1}^d$. For which $k$ does the set $\{\alpha_j^k\}_{j=1}^d$ also provide a basis?

If this is in general a difficult question, could I at least show that I obtain a basis for infinitely many $k$?

Example: Consider the polynomial $x^2-2$. The $k$th powers of the roots $\{\pm \sqrt{2}\}$ provide a basis for my field extension precisely when $k$ is odd.

Edit: As Gerry Myerson points out, and as my example indicates, I am actually more interested in when the $k$th powers of the roots generate the extension over $\mathbb{Q}$. Calling this a $\mathbb{Q}$-basis was erroneous.

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As your own example shows, the roots of a polynomial do not need to be linearly independent, and in particular do not form a basis in general. –  Alex B. Oct 25 '11 at 8:31
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@Alex, I suspect what OP means by "basis for the field extension" is really generating set (in the field sense, not the vector space sense). –  Gerry Myerson Oct 25 '11 at 8:45
    
My guess is that for infinitely many $k$ you will have ${\bf Q}(\alpha_1^k,\dots,\alpha_d^k)={\bf Q}(\alpha_1,\dots,\alpha_d)$ (which, I think, is the question you are asking), but I don't see how to prove it. –  Gerry Myerson Oct 25 '11 at 8:49
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If $[\mathbb{Q}(\alpha_1,\ldots,\alpha_d):\mathbb{Q}]$ is odd, then $k=2^r$ will work for any $r$: just note that replacing a single $\alpha_i$ with $\alpha_i^2$ will give you an intermediate field of index at most $2$ in $\mathbb{Q}(\alpha_1,\ldots,\alpha_d)$, but the index must divide the degree of the extension. Lather, rinse, repeat. –  Arturo Magidin Oct 25 '11 at 17:11

1 Answer 1

up vote 3 down vote accepted

Let $L$ be a field of characteristic $0$ and $K$ a finite simple separable extension, with primitive element $\alpha \in K$. Let $E$ be the galois closure of $K$ over $L$. So we have a tower of fields $L \subset L(\alpha)=K \subset E$. Now if for some $k \in \mathbb N$ we have that $L(\alpha^k) \subsetneq K$ then there exists some $\sigma \in \mathrm{Gal}(E,L)$ fixing $\alpha^k$ that does not fix $\alpha$. Now $\sigma(\alpha^k)=\sigma(\alpha)^k=\alpha^k$ in particular $\sigma(\alpha)$ is a solution to $x^k-\alpha^k$ and thereby $\sigma(\alpha)=\zeta \alpha$ where $\zeta$ is a $k$-th root of unity. We see that there are finitely many automorphisms of $E$ over $L$ and since there are infinitely many $p$-th roots of unity, we have that $L(\alpha^p)=L(\alpha)$ for all but finitely many primes $p$. This then extends naturally to the case of $n$ generators by taking intersections.

In the case that the base field is $L=\mathbb Q$ we can make this slightly more effective by noting that if $\varphi(p)>[\mathbb Q(\alpha): \mathbb Q]$ then $\alpha^p$ is still a generator. Which can then be extended to composite numbers by imposing the same constraint on the prime factors. But this doesn't necessarily tell the whole story. For most primitive generators I would expect that for every $k \in \mathbb N$ $\alpha^k$ generates the field. Implicit in my previous argument is the observation that $\alpha^k$ does not generate $L(\alpha)$ if and only if $\min_L(\alpha) \mid x^k-\alpha^k$. Which seems to me the exceptional case.

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