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My attempt

$f(x)$ is irreducible by Eisenstein’s criterion with $p = 2$.

$f’(x)= 5x^4 -10x$ has two real roots.

$f’’(x)= 20x^3$ has a single real root.

Therefore, $f$ has exactly 3 real solutions and two complex ones. Therefore its Galois group is a transitive subgroup of $S_5$, which contains a transposition, so it must be the whole of $S_5$, which is not solvable, so $f$ is not solvable by radicals over $\mathbb Q$.

Is it correct/complete? It seems very simple, but I don't know how to improve.

Thank's for any help.

Arturo Magidin
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Bruno
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  • The Galois group of $x^5-10x+2$ is $S_5$, which is not solvable. So it is not solvable by radicals. There are several similar questions, e.g. this one. It is a duplicate of this question. – Dietrich Burde Aug 26 '22 at 14:46
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    Instead of putting (Galois theory) in the subject line, why are you not using the [galois-theory] tag? – Arturo Magidin Aug 26 '22 at 14:46
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    Just one nitpick: the fact that $f'$ has two real roots is not enough to prove that $f$ has three real roots. Derivatives are insensitive to vertical translations, and with an appropriate choice of constant $c$, $f(x) + c$ may only have a single root (or two roots, one with multiplicity 2). You need to show that $f$ actually changes signs three times. – Tob Ernack Aug 26 '22 at 15:02

1 Answers1

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It’s correct. More generally, if $p\ge5$ is a prime and $f(x)\in\Bbb Q[x]$ is an irreducible degree-$p$ polynomial with precisely two non-real roots, then $f(x)$ is unsolvable since $\operatorname{Gal}(f)\cong S_p$.

This uses the result on transposition generators that is proven here (it also uses the fact that $A_n$ is simple and nonabelian for any $n\ge5$, and thus $S_n$ is unsolvable). The Galois group contains a transposition since it acts on the roots transitively, so, taking a non-real root $\alpha$, there is at least one (and exactly one!) element of the Galois group mapping $\alpha\to\overline{\alpha}$ through $i\mapsto-i$. That map is an automorphism over $\Bbb C$, so it will remain a valid element of the Galois group by fixing all the other roots. Then, the action on the roots is just to swap two, and leave all others fixed. This is a transposition.

The argument fails if there are four, six, however many more non-real roots, because we’d have a double, triple etc. transposition which might not (in combination with a $p$-cycle) generate $S_p$. There is a $p$-cycle in the Galois group since it acts transitively over $p$-distinct roots, so $p$ divides the order of this group, so it contains an element of order $p$ since $p$ is prime. The order $p$ elements of $S_p$ are precisely the $p$-cycles.

FShrike
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