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Let $f(X)=X^5+aX+b \in \mathbb{Q}[X]$. I want to know the Galois group of $f(X)$ if all the following conditions are met:

1) $f(X)$ is irreducible in $\mathbb{Q}[X]$.

2) The discriminant $D(f)$ of polynomial $f(X)$ is a square in $\mathbb{Q}.$

3)The equation $f(X)=0$ is soluble by radicals.

So by 1) the Galois group is a transitive subgroup of $S_5$ and by 2) is contained in $A_5$, my question is the following by the 3) I know the Galois group is a solvable group? If this is the case what is the Galois group?

Edit 2: Ok, some news. I know $f(X)$ can have 1 or 5 real roots, because $f(X)$ can't have two real roots and neither four real roots because this is a contradiction with the number of possible complex roots. Futhermore $f(X)$ can't have 3 real roots because if $f(X)$ have exactly two complex roots tha Galois group will be $S_5$, a contradiction with 2). Now the question is how can i prove that $f(X)$ can't have five real roots?

julioprofe
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  • My first guess would be that the Galois group can still be either cyclic or dihedral. The holomorph $C_5\rtimes C_4$ is ruled out for not being a subgroup of $A_5$. The condition for $f(X)$ to have this special form does not really constrain the group because a Tschirnhaus transformation puts a general quintic into that form. I'm ashamed not to be conversant with the details, so it may happen that I'm wrong and a Tschirnhaus transformation involving numbers outside the spliting field of $f(X)$ is needed. – Jyrki Lahtonen Dec 16 '19 at 04:49
  • Anyway, the relevant Mathworld page does list both $C_5$ and $D_5$ as possible Galois groups of quintics in Bring form. – Jyrki Lahtonen Dec 16 '19 at 04:50
  • Locally, see this thread for more. – Jyrki Lahtonen Dec 16 '19 at 04:51
  • Ok, some news. I know $f(X)$ can have 1 or 5 real roots, because if $f(X)$ can't have two real roots and neither four real roots because this is a contradiction with the number of possible complex roots. Futhermore $f(X)$ can't have 3 real roots because if $f(X)$ have exactly two complex roots the Galois group will be $S_5$, a contradiction with 2). Now the question is how can i prove that $f(X)$ can't have five real roots? – julioprofe Dec 16 '19 at 14:04
  • @julioprofe, your polynomial cannot have five real roots, because that would mean that its derivative has four real roots, and clearly it has 0, 1 or 2. – Mees de Vries Dec 16 '19 at 14:23
  • If $f(X)$ has 5 real zeros, then $5x^4+a$ should have 4 real solutions, but that is not possible for rational $a$. – Randy Marsh Dec 16 '19 at 14:25
  • @GoranMalic Maybe the following question is stupid but why is not possible? – julioprofe Dec 16 '19 at 14:29
  • What are the real solutions to $x^4-1$? The same reasoning applies to $5x^4+a$. – Randy Marsh Dec 16 '19 at 14:40
  • Two real solutions $1$ and $-1$, so $f'(x)$ have at most two real solutions? Indeed the only "possible" case is $f'(x)$ have two real solutions but this is not possible.Therefore $f(X)$ have only one real solution. – julioprofe Dec 16 '19 at 14:45
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    Yes, $x^4=1$ has solutions $x=\xi_k$, where $\xi_k$ are the fourth roots of unity for $k=0,1,2,3$ (i.e. $\pm 1$ and $\pm i$). If $x$ is real, then $5x^4$ must be positive, so $a$ is negative, therefore $a=-c$ for some positive rational $c$. Then the solutions to $x^4=c/5$ are $x=\xi_k \sqrt[4]{c/5}$. – Randy Marsh Dec 16 '19 at 14:57

1 Answers1

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The irreducibility condition implies that the Galois group $G$ acts transitively on the 5 roots.

There are exactly 5 such groups: $S_5$, $A_5$, the Frobenius group of order $20$ (which is the normalizer of the p-Sylow and also isomorphic to the group of linear affine transformations of $\mathbf{F}_5$), the dihedral group of order 10 (a subgroup of the Frobenius group), and the cyclic group of order $5$.

The discriminant condition implies that $G$ is a subgroup of $A_5$, which rules out $S_5$ and the Frobenius group, since the linear map $x \mapsto 2x$ defines a 4-cycle which is odd.

The solvability condition rules out $A_5$, leaving the cyclic group and the dihedral group.

The form of the polynomial implies (by Descartes rule of signs, for example) that not all the roots are real. This implies that complex conjugation is non-trivial, and hence that the Galois group has even order. This rules out the cyclic group of order $5$, and thus by elimination the Galois group must be the dihedral group of order $10$.

Magister Ludi
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  • Thanks a lot I have the last question, what is the Galois group if $f(X)$ would have had five real roots? The cyclic group of order 5 right? – julioprofe Dec 16 '19 at 15:12
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    No, if the polynomial had 5 real roots there would not be enough information to distinguish the last two cases – Magister Ludi Dec 16 '19 at 19:30