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Fix a field $K$, and $a\in K$. Consider the polynomial $x^m-a$, where $m$ is prime to the characteristic of $K$, and suppose $a$ does not have any $d$-th roots in $K$ for any $d|m$. Is $x^m-a$ irreducible over $K$?

The answer seems to be yes under the a priori stronger assumption that $a^{m/d}$ is not an $m$-th power, by adapting the arguments here: Irreducibility of a polynomial if it has no root (Capelli). (Proof: if $[K(\sqrt[m]{a}):K]=d'<m$, then $N(\sqrt[m]{a})^m=N(a)=a^{d'}$, so $a^{d'}\in (K^{\times})^m$, hence $a^{\gcd(d',m)}\in (K^{\times})^m$, contradiction.)

However, I don't know how to prove this with the weaker hypothesis.

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Not necessarily. The number $-2$ has no real square roots. Yet $x^4+2$ is not irreducible over $\Bbb{R}$ for the simple reason that $\Bbb{R}$ only has a single algebraic extension, and that is of degree two.

For a different example, let $K=\Bbb{F}_3$, and $L$ be its quadratic extension. Because $-1$ is not a quadratic residue modulo $3$ it has no square root in $K$. Yet, the group $L^*$ is cyclic of order $8$, so $K^*$ consists of the fourth powers of elements of $L$. Therefore $p(x)=x^4+1$ has a root in $L$, and consequently $p(x)$ has a quadratic factor over $K$.

Jyrki Lahtonen
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The "correct" hypothesis should be: $m$ is coprime to the characteric of $K$, $a\in K$ is non null and has no $p$-th root in $K$ for any prime $p$ dividing $m$, and if $4$ divides $m$, then $a\notin -4K^4$. Then $X^m - a$ is irreducible over $K$. See e.g. Lang's "Algebra", chapter 8, §9, thm. 16.

nguyen quang do
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