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 $F$?
When is $X^n-a$ is irreducible over F?
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galois-theory
irreducible-polynomials
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0This 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. – 2012-04-19
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01. 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.) – 2012-04-19