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I have tried many ways.This polynomial has exactly one real and four complex roots and it is in Bring-Jerrard form, the solvability of which is about finding $\varepsilon, e$ and $c$ to equate some rational expressions with $5$ and $-10$.But this is very much calculative and I could not find any way. I know it will not be solvable if the Galois group is isomorphic to $A_5$ or $S_5$ but how to show that! Please help to find way.

Robert Shore
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Promit Mukherjee
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  • ? polgalois(x^5 + 5*x - 10)

    %1 = [120, -1, 1, "S5"]

     –  Aug 31 '20 at 18:14
  • discriminant is 32050000, not a square. so it is not A_5. –  Aug 31 '20 at 18:15
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    I have vague memories of a relationship between number of roots of $x^5 + 5x - 10$ in $\mathbb{F}_p$ for $p$ prime as $p$ varies, and the Galois group of the splitting field of $x^5 + 5x - 10$. In a quick check, it looks like I get at most three roots for each case of $7 \le p \le 61$ which suggests the Galois group is "probably" one of the smaller ones. – Daniel Schepler Aug 31 '20 at 18:18
  • Yes the polynomial is irreducible mod 3, which proves the galois group has a cycle of (5) in it (a 5-cycle). Together with the complex conjugation map having a (2)(2) type you must have S_5. –  Aug 31 '20 at 18:37
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    I understand that the Galois group has a 5 cycle but there are 4 complex roots how it can have a transposition? It has a double trabaposition but how it has a transposition bcz I know a 5 cycle and a transposition would generate S5.Please make me understand bit clearer! – Promit Mukherjee Aug 31 '20 at 18:47
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    @rain1 Doesn't the dihedral group of order 10, considered as a subgroup of $S_5$ via its action on the vertices of a regular pentagon, have a 5-cycle and an element of (2)(2) type? – Daniel Schepler Aug 31 '20 at 18:51
  • Looking at https://groupprops.subwiki.org/wiki/Subgroup_structure_of_symmetric_group:S5 the only proper subgroup there with a single orbit (necessary since the polynomial is certainly irreducible) and not contained in $A_5$ (cannot be with a non-square determinant) would be the "GA(1,5) in S5" entry $\langle (1,2,3,4,5),(2,3,5,4) \rangle$. If that's the Galois group then https://groupprops.subwiki.org/wiki/General_affine_group:GA(1,5) says it's solvable. – Daniel Schepler Aug 31 '20 at 19:20

1 Answers1

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Let's recall the following.

Theorem (Dedekind): If an irreducible polynomial $f(x)$ factors $\bmod p$ into factors with degrees $d_1, d_2, \dots$, then the Galois group of $f$ contains an element of cycle type $(d_1, d_2, \dots)$.

Theorem (Frobenius / Chebotarev): Every cycle type in the Galois group occurs this way, with density proportional to its density in the Galois group.

Furthermore, transitive subgroups of $S_5$ (and I think maybe also $S_6$?) are determined up to conjugacy by the cycle types of their elements, so factorizations mod a prime eventually identify all Galois groups of irreducible quintics. The complete list of transitive subgroups of $S_5$ (up to conjugacy) is

  • $C_5$
  • $D_5$
  • The Frobenius group $x \mapsto ax + b, a \in \mathbb{F}_5^{\times}, b \in \mathbb{F}_5$
  • $A_5$
  • $S_5$.

Using WolframAlpha we find that $f(x) = x^5 + 5x - 10$ is

  • irreducible $\bmod 3$, so the Galois group contains a $5$-cycle (but this already follows from transitivity),
  • a product of an irreducible quadratic and cubic $\bmod 7$, so the Galois group contains a permutation of cycle type $(2, 3)$

and we can stop here: already $S_5$ is the only transitive subgroup of $S_5$ containing an element of cycle type $(2, 3)$.

Qiaochu Yuan
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