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Assume we're working in $\mathbb{Q}[x]$. So I know that for a polynomial like $f(x) = x^3 - 2$, the roots are found by realizing that $\sqrt[3]{2}$ is obviously a root, and then having the other 2 roots be $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$ where $\omega$ is the third root of unity. Can someone generalize the situations where I can do this, like just multiplying a root by powers of roots of unity? I'm assuming we can't do this for third degree equations with 3 real roots: can someone explain why not, and how to detect such situations?

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Consider the polynomial $x^n-a\in \mathbb Q[x]$, with $a>0$. One root of this polynomial is $\alpha$, the $n$-th root of $a$. Now let $\omega\in \mathbb C$ be a primitive $n$-th root of unity. That means that $\omega ^n=1$ and $\omega^k\ne 1$ for all $0<k<n$. So, the (typically complex) numbers $\beta_i=\omega ^i\alpha $, where $0\le i < n$, are distinct and all satisfy: $\beta _i^n=(\omega ^i\alpha )^n=(\omega ^n)^i\alpha ^n=a$. So, this shows that the polynomial has the $n$ distinct roots $\beta _0,\cdots ,\beta _n$.

(Partly) Conversely, if a polynomial $p(x)\in \mathbb Q[x]$ is of degree $n$, and $\alpha$ is a root of $p(x)$ such that for some primitive $n$-th root of unity $\omega $, it holds that $\omega^i\alpha$ is a root of $p(x)$ for all $i$, then $p(x)$ splits as $(x-\alpha)(x-\omega \alpha)(x-\omega ^2 \alpha)\cdots (x-\omega ^{n-1}\alpha)$ which is the splitting of the polynomial $x^n-(\alpha^n)$. Writing $a=\alpha ^n$ this shows that $p(x)=x^n-a$.

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  • $\begingroup$ Thank you. I understand it works for this particular kind of polynomial, but I was wondering if there are more such cases where this would work, or if that's all. $\endgroup$ – tyur43 Apr 26 '13 at 22:51
  • $\begingroup$ I think the OP knows that. Rather, he's asking if it generalizes to other types of polynomials. $\endgroup$ – Math Gems Apr 26 '13 at 22:51
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    $\begingroup$ ok, that wasn't clear to me from the question. So, I added some partly converse result. $\endgroup$ – Ittay Weiss Apr 26 '13 at 22:59

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