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We consider $$x^5 + 5x^4 + 10x^3 + 10x^2 + 5x - 2 \in \mathbb{Q}[x]$$ and $$x^5 - x + 5 \in \mathbb{Q}[x].$$ I want to see if they are solvable by radicals or not.

I know that firstly I should determine the splitting field of our polynomials and the Galois group of it. However, I can't think of how to do this...

Could you help me, please? Thanks!!

1 Answers1

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For the second polynomial

$$p(x):=x^5-x+5$$

Theoretical side : there are results on Galois groups of such monic polynomials with $3$ terms which can be found here. If we consider its Lemma 9, as $1+1<5$, $p(x)$ is irreducible. Moreover, the discriminant of $p$ (see below) is $1952869=239 \times 8171$ without square factors. Therefore, Theorem 6 of the same paper can be applied : the Galois group of $p$ is the symmetric group $S_5$.

Practical side : here is a Sagemath program :

p=QQ['x'](x^5-x+5)
d=p.discriminant();show(d);show(factor(d))
gal=p.galois_group();
n=gal.order();show(n)
gal.is_isomorphic(SymmetricGroup(5))

giving the answers :

Discriminant : $d=1952869=239 \times 8171$

Galois group order : $n=120=5!$

With the confirmation that the boolean given by the fourth instruction is "True".

Jean Marie
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  • Intresting things also here with a paragraph devoted to polynomials of the form : $X^5-X+q$. – Jean Marie Dec 13 '23 at 15:56
  • Thank you so much!!! This is really interesting to me, I love it! I was thinking about the Abel theorem too… It says that if $f(x) \in K[x]$ is an arbitrary separable irreducible polynomial of degree $n \geq 5$ and $E$ is the splitting field of $f(x)$ then $\text{Gal}(E|K) \cong S_n$. Do you think I could use it too? –  Dec 13 '23 at 18:20
  • But what knowledge do we have of this splitting field $E$ ? – Jean Marie Dec 13 '23 at 18:25
  • Is it important? I mean, we know that $f(x)$ is solvable by radicals if and only if $\text{Gal}_K(f) = \text{Gal}(E|K)$ is solvable by radicals, right? If this Galois group is isomorphic to $S_5$… –  Dec 13 '23 at 18:30
  • Indeed. I am a little rusty, for I haven't done that much Galois theory these last years. – Jean Marie Dec 13 '23 at 18:45
  • No worries!! I love your answer, and also I had no idea there was any program that did something like that. I mean, I literally just discovered SageMath and I am really impressed. Anyway, thanks a lot! :) –  Dec 13 '23 at 18:51
  • Thanks for your thanks. It is very simple indeed to use sagemath (see my remark here). Besides, I have made a small modification : I ask the computation of the discriminant to Sage (instead of Wolfram Alpha) – Jean Marie Dec 13 '23 at 18:56