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I'm having a hard time trying to prove that the polynomial

f(x) = x^p - x - 1 in Z_p[x]

is not solvable by radicals even though its Galois Group is solvable.

So far, I have shown that the polynomial is irreducible in the base field and that its Galois group is isomorphic to Z_p. Since the Galois group is cyclic, it is solvable. I have also shown that if n is a root of f then the pth root of n is not a root of f by the definition of solvable by radicals given to us, which is definition 2 in the link below:

http://www.math.brown.edu/~abrmovic/MA/f1314/251/Zijian-notes.pdf

tamefoxes
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  • See section 8.6 part C of David Cox's Galois Theory (2nd edition) for solvability in characteristic p. – KCd Dec 08 '14 at 03:33
  • Can't seem to find a pdf of it online. Do you have a link for one? – tamefoxes Dec 08 '14 at 03:36
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    No, but not everything in the world exists (yet?) online. Go to a library and look at a physical copy. If your library does not have one, they can get one for you by interlibrary loan. – KCd Dec 08 '14 at 13:35
  • You did notice that Zijian Yao begins by making an umbrella assumption that everything is in characteristic zero. Therefore I fail to see how their definition 2 would apply here. – Jyrki Lahtonen Dec 08 '14 at 18:10
  • The definition I was given is exactly how it is defined in the link with the characteristic of the field not being mentioned in mine. – tamefoxes Dec 08 '14 at 18:16

2 Answers2

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Show that the polynomial is irreducible. Show that the decomposition field of that polynomial is generated by any one of its roots; in particular, the degree of the decomposition field is then $p$. Show that thay field is not generated by the pth roots of any element of the prime field —or that if an element of that field is a pth root of the prime field, it is already in the prime field.

Mariano Suárez-Álvarez
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    Being solvable by radicals doesn't require the splitting field itself to be a radical extension, but only to lie in a radical extension. – KCd Dec 08 '14 at 03:33
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The OP specified that the characteristic zero definition from Zijian Yao's lecture notes is to be taken literally. Given that I want to put this question to rest by declaring that

The claim is false as stated.

The polynomial $x^p-x-1$ is irreducible over $\Bbb{F}_p$ (many proofs for this fact can be found by searching our site). Therefore its zeros generate the field $L=\Bbb{F}_{p^p}$ that is the unique degree $p$ extension of the prime field (and also the splitting field of $f(x)$).

But $L=\Bbb{F}_p(\zeta)$, where $\zeta$ satisfies the equation $\zeta^{p^p-1}=1$. Therefore all the zeros of $f(x)$ are "expressible by radicals" and, according to Definition 2, $f(x)$ is thus deemed solvable by radicals.

A very concrete counterexample: when $p=2$ the zeros of the polynomial $x^2-x-1=x^2+x+1$ are cubic roots of unity.

Jyrki Lahtonen
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    I think that this was an oversight by whoever gave you this exercise. They should have used $K=\Bbb{F}_p(T)$, and $f(x)=x^p-x-T$ instead. I don't see a way of proving that in the modified case the splitting field is not contained in any root tower extension, but I would be surprised if it were. – Jyrki Lahtonen Dec 10 '14 at 14:39
  • Thanks for the answer. I will post the solution to the answer when they are available. – tamefoxes Dec 11 '14 at 16:02
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    @JyrkiLahtonen : It's a bit late, but I would like to mention that your comment is exactly given as an exercise in Rotman's Galois theory, in Problem no. 94. There, the author has indeed taken the ground field to be $\mathbb{F}_p(T)$. – Hajime_Saito May 21 '23 at 08:19