When $X^n-a$ is irreducible over F?

Question

Let $F$ be a field, let $\omega$ be a primitive $n$th root of unity in an algebraic closure of $F$. If $a \in F$ is not an $m$th power in $F(\omega)$ for any $m\gt 1$ that divides $n$, how to show that $x^n -a$ is irreducible over $F$?

This breaks down for $n = 4$. In ${\mathbf Q}[x]$, the polynomial $x^4 + 4$ has $a = -4$, which is not a square or 4th power in ${\mathbf Q}$, but this polynomial is reducible over ${\mathbf Q}$: it is $(x^2+2x-2)(x^2-2x-2)$. This counterexample is essentially the only kind of counterexample, in light of Bill Dubuque's answer. — KCd, Apr 19 '12 at 00:19
Closing this as a duplicate of a "newer" question. This is an admittedly debatable action. I want to emphasize that this thread stands on its own merits. But the other one is more self-contained in the sense that Bill Dubuque included an excerpt from Karpilovsky in his answer to the companion thread. Therefore A) the viewers (without access to a decent math library) will benefit more from the material in the other thread, B) I will under no circumstances support the idea of deleting either. The main motivation is to strongly link these two, and guide the readers. — Jyrki Lahtonen, Jul 04 '25 at 06:47
(cont'd) For the record, I have upvoted all the answers to both these threads. And I have referred readers of several other questions to THIS thread simply because it is older, and hence made its way to the list of common questions in meta. — Jyrki Lahtonen, Jul 04 '25 at 06:49
This question about the special case of a prime degree is also worth a link. Even though that case is a step in the proof of the general result. — Jyrki Lahtonen, Jul 04 '25 at 06:52

Bill Dubuque · Answer 1 · 2022-05-27T08:11:38.477

22

Below is a classical result:

Theorem $\ $ Suppose $\,c\in F\,$ a field, and $\,0 < n\in\mathbb Z$.

$\ \ \ x^n - c\ $ is irreducible over $F\! \iff c \not\in F^p\,$ for all primes $\,p\mid n,\,$ and $\ c\not\in -4F^4$ when $\, 4\mid n$

A proof is in many Field Theory textbooks, e.g. Karpilovsky, Topics in Field Theory, Theorem 8.1.6.

edited May 27 '22 at 08:11

answered Apr 18 '12 at 19:44

Bill Dubuque

285,713

10

See also Lang's Algebra. There is a whole section about the polynomial $x^n - a$ in the chapter on Galois theory. – KCd Apr 19 '12 at 00:20

score 9 · Answer 2 · edited Nov 26 '25 at 16:58

9

I will assume "$m \geq 1$", since otherwise $a \in F(\omega)$, but $F(\omega)$ is $(n-1)$th extension and not $n$th extension, so $x^n-a$ must have been reducible.

Let $b^n=a$ (from the algebraic closure of $F$).

$x^n-a$ is irreducible even over $F(\omega)$. Otherwise $$f= \prod_{k=0}^{n - 1} (x-\omega^k b) = (x^p + \cdots + \omega^o b^p)(x^{n-p} + \cdots + \omega^ó b^{n-p}),$$ so $b^p$ and $b^{n-p}$ are in $F(\omega)$. Consequenty $b^{\gcd(p,n-p)}$ is in $F(\omega)$, but $\gcd(p,n-p)$ divides $n$, so $(b^{\gcd})^\frac{n}{\gcd} = a$, a contradiction.

edited Nov 26 '25 at 16:58

hdecristo

1,277

answered Apr 18 '12 at 19:41

savick01

4,587

I don't understand how this proves it. Wouldn't you want to show if $f$ is reducible then a is an $m$th power. It looks like you showed when $f$ is reducible that $a$ is an $n$th power. Also minor thing, the upper index of your product should be $n-1$. – Wyatt Kuehster Jul 22 '19 at 01:40
@WyattKuehster note it has proved that $b^{gcd}\in F $ so that proved the claim. – mathemather Nov 19 '19 at 14:58
Your first sentence doesn't make sense. (But the rest does make sense.) – mr_e_man Aug 02 '24 at 02:38

When $X^n-a$ is irreducible over F?

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