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The question:

Can one characterize all fields $K$ over which there exists an irreducible polynomial $f(x)$ that does not divide a polynomial of the form $x^n + a$?

Examples:

  • Such a polynomial clearly exists over $\mathbb{Q}$.
  • It also exists over $\mathbb{R}$ (the polynomial with roots $(3 + 4i)/5$ and $(3 - 4i)/5$).
  • As far as I understand, such polynomials do not exist over finite fields.
  • Apart from the trivial case of $K$ algebraically closed, this is the only negative example I know. I would thus also appreciate any example of an infinite field over which there is no polynomial with the described properties.

Possibly related:

I have found some papers dealing with possible multiplicative relations between roots of polynomials over certain fields (for example, this one). However, these do not seem to directly address my question. I suspect that my question should be much easier (hopefully not completely trivial).

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    $\begingroup$ Separably closed fields are also examples: if $L/K$ is purely inseparable, then for any $a\in L$, you have $a^{p^n}\in K$ for some $n$. $\endgroup$
    – Wojowu
    Commented Oct 10, 2021 at 16:38
  • $\begingroup$ I think the Galois group of a splitting field for $x^n+a$ is abelian (right?) so I would guess that a sufficient condition is that the absolute Galois group of $K$ be non-abelian. $\endgroup$
    – Paul Levy
    Commented Oct 10, 2021 at 19:14
  • $\begingroup$ @PaulLevy No, that's not the case - $x^4-2$ has nonabelian Galois group. It is, however, solvable, so unsolvability of the absolute Galois group is a sufficient condition. $\endgroup$
    – Wojowu
    Commented Oct 10, 2021 at 20:37
  • $\begingroup$ @Wojowu Ok, the absolute Galois group of the field obtained from $K$ by adjoining all roots of unity should be non-abelian... $\endgroup$
    – Paul Levy
    Commented Oct 10, 2021 at 20:59

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