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What is the minimal degree of a polynomial with integer coefficients with $a^{1/b}$ as a root where a and b are integers?

Assume it can't be simplified, as in, there are no integer $x$ and $y<b$ such that $a^{1/b}=x^{1/y}$.

It feels like the answer really should be $b$. In some cases it can be shown using Eisenstein, and for $b=3$ we can show it with the general second degree solution. I can't see a proof that it should be $b$ generally, though.

Alice Ryhl
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1 Answers1

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By Karpilovsky's theorem (see Bill Dubuque's answer here), the minimal polynomial is $X^b-a$ whenever $b$ is not a multiple of $4$, or $a$ is not $-4$ times a fourth power.

If, on the other hand, we are in the "nasty" case when $b$ is a multiple of $4$ (say $b=4\beta$) and $a=-4c^4$, the theorem only tells us that the minimal polynomial will have degree strictly less than $b$. We have

$$ X^b-a=X^{4\beta}+4c^4=(X^{2\beta}-2cX^{\beta}+2c^2)(X^{2\beta}+2cX^{\beta}+2c^2) $$ So we have two "ex aequo" candidates of degree $2\beta=\frac{b}{2}$. It is not clear if they are irreducible however.

Ewan Delanoy
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  • But .. as far as I can see in the nasty case we have $(-4c^4)^{1/4\beta} = i (2c^2)^{1/2\beta}$, which means it's simplifiable with the definition I gave? Does this mean that the answer to my question is that "yes, the minimal degree is, in fact always $b$, when it is not simplifiable"? – Alice Ryhl May 04 '17 at 16:00
  • And what happened to things like $x^3 - 27 = (x-3)(x^2 + 3x + 9)$? – Alice Ryhl May 04 '17 at 16:04
  • @AliceRyhl I don't get your $x^3-27$ example. Isn't $27^{\frac{1}{3}}=3^{\frac{1}{1}}$ simplifiable by your definition ? – Ewan Delanoy May 04 '17 at 17:21
  • @AliceRyhl In the nasty case we have $(-4c^4)^{\frac{1}{4\beta}}=x^{\frac{1}{y}}$ with $x=2ic^2$ and $y=2\beta$, but the problem is that $x$ is not an integer (unless you wish to include Gaussian integers such as $i$ in the definition). – Ewan Delanoy May 04 '17 at 17:24
  • Ah, it is simplifiable, but I just don't see why the theorem you mentioned doesn't show that $x^3 - 27$ should also be irreducible. – Alice Ryhl May 05 '17 at 18:17
  • @AliceRyhl : read the theorem : it says that $X^b-a$ is irreducible when $a$ is not a $p$-power for any prime divisor of $b$, except when $b$ is a multiple of $4$ and $a$ is $-4$ times a fourth power.. – Ewan Delanoy May 05 '17 at 18:54
  • Sorry, I was having some trouble interpreting the theorem, but I get it. Thanks for the help. – Alice Ryhl May 05 '17 at 18:57