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

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 0:19
• 1. You should have: $(x^2 + 2x + 2)(x^2 - 2x + 2)$. 2. This is not a counterexample to the original question, because $-4 = (2i)^2$ and $2i \in Q(i)$ and of course $2$ divides $4$. (Anyway, I've given a proof.) – savick01 Apr 19 '12 at 19:25

Here is a classic result:

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

$\rm\quad x^n - c\$ is irreducible over $\rm\:F \iff c \not\in F^p\:$ for all primes $\rm\:p\: |\: n\:$ and $\rm\ c\not\in -4\:F^4\:$ when $\rm\: 4\ |\ n\:.$

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

• 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 0:20
• I'm apparently at my dullest. Knowing that you cite your sources well, and searching for "Karpilovsky", made it easy. Fixing a few links :-) – Jyrki Lahtonen Aug 11 '19 at 16:31

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 $$f= \prod_{k=0}^n (x-\omega^k b) = (x^p + \cdots + \omega^o b^p)(x^{n-p} + \cdots + \omega^ó b^{n-p}),$$ 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.

• 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 1:40
• @WyattKuehster note it has proved that $b^{gcd}\in F$ so that proved the claim. – mathemather Nov 19 '19 at 14:58
• @savick01 I did not understand the opening comment. – mathemather Nov 19 '19 at 15:11