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This should be a fairly trivial question, to which I have nevertheless found no satisfactory answer.

I am interested in effective algorithmic computation of minimal polynomials of some particularly simple algebraic numbers, especially the complex $n$-th roots of rational numbers: $$\alpha = \sqrt[n]{q}e^{i2k\pi/n}.$$ Each such $\alpha$ is a root of the polynomial $x^n - q$, but the trouble is that this polynomial might not be irreducible over $\mathbb{Q}$. I know that the irreducible factors can be found in polynomial time, e.g., using the LLL algorithm. But how to determine the correct factor? This can usually be done easily by hand, but what about algorithmic computation? The only way of evaluating a polynomial at $\alpha$ that I am aware of already assumes that the minimal polynomial of $\alpha$ is known. Or should I use a completely different approach?

Moreover, since $q$ can be assumed to be an integer, an algorithm for computing minimal polynomials of algebraic integers would be sufficient. This MathOverflow answer suggest that computing minimal polynomials of algebraic integers is easy, but I was unable to find out why. Any reference would be helpful here.

Thank you in advance.

Jára Cimrman
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    See this post. $x^n-q$ is irreducible, except for the case where $q=p^m$ for some prime $p$ dividing $n$. – Dietrich Burde Aug 15 '21 at 08:08
  • If you have the irreducible factors, you can determine the correct factor simply by inserting the given algebraic number. That is an algorithm. Whether it is "easy" to determine the minimal polynomial of an algebraic number in general, depends what is meant with "easy". – Peter Aug 15 '21 at 08:20
  • @DietrichBurde I seem to understand. However, what about the "exceptional" case? For instance, if $q = 1$, one would hope to obtain the correct cyclotomic polynomial... I believe there should be a general algorithm here. – Jára Cimrman Aug 15 '21 at 08:40
  • @Peter Thanks for a comment, but my question was largely about how can one just "insert" the given algebraic number. To my knowledge, you need to work in an appropriate algebraic extension when performing computations involving the algebraic number. However, in order to do that, you already need to know the minimal polynomial. – Jára Cimrman Aug 15 '21 at 08:41
  • See here. The general proof is a bit trickier. When $n$ happens to be a prime the proof is easier. – Jyrki Lahtonen Aug 15 '21 at 09:00
  • @JyrkiLahtonen OK, but this is still just concerned with the question when the polynomial is reducible, which is not what I have asked. – Jára Cimrman Aug 15 '21 at 09:05
  • Sorry, but I quite do not understand a reason for closing this question... – Jára Cimrman Aug 15 '21 at 09:06
  • You were not asking about irreducibility of $x^n-a$? Ok. I'll reopen for a while, so that you can clarify in peace :-) – Jyrki Lahtonen Aug 15 '21 at 09:09
  • So $x^n-q$ is irreducible unless $q$ is a $p$th power of a rational number for some prime $p\mid n$. Or $4\mid n$ and $q=-4r^4$ for some rational number $r$. When $q=r^p$, $p\mid n$, then $x^{n/p}-r$ is a factor. In the other exceptional case we have factorizations modelled after $$(x^4+4)=(x^4+4x^2+4)-4x^2=(x^2+2x+2)(x^2-2x+2).$$ – Jyrki Lahtonen Aug 15 '21 at 09:17
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    Hmm. There is a subtlety I overlooked... – Jyrki Lahtonen Aug 15 '21 at 09:25
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    @JyrkiLahtonen Nevertheless, thanks for your comments. I seem to understand how the problem could be solved for positive real algebraic numbers, but I am still quite unsure about the complex numbers... – Jára Cimrman Aug 15 '21 at 09:28
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    With the aid of the minimal polynomial of $|q|^{1/n}$ and the roots of unity you get a field where you can work. Galois theory then tells us how to find the conjugates of $\alpha$. There are subtleties when $q$ is negative (see the second exceptional case). – Jyrki Lahtonen Aug 15 '21 at 10:10
  • @JyrkiLahtonen Thank you, I will take a look. – Jára Cimrman Aug 15 '21 at 10:12

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