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Based on this answer it appears that any irreducible polynomial $f(x)\in\mathbb{Q}[x]$ of degree $5$ is not solvable by radicals if it has only three real roots. Is this claim true?

But looking at the solvable polynomials of type $x^{5} - 2$ which have only one real root, it appears that irreducible quintics with only one real root are solvable which brings me to the counter-example $x^{5} - x - 16$ which has one real root and is not solvable by radicals (as mentioned by G. H. Hardy in his A Course of Pure Mathematics).

I guess deducing the Galois group of irreducible polynomials by noting the real/complex nature of roots is not an effective procedure and yet many textbooks provide examples of non-solvable quintics on the basis of such information.

Is my understanding correct? Or does the real/complex nature of roots play a significant role in finding Galois groups?

Update: This answer shows that real/complex nature of roots does play a role in solvability by radicals, but unfortunately the proof of the result is not provided.

Paramanand Singh
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1 Answers1

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If $p$ is an odd prime then the only solvable subgroup of $S_p$ is the linear group of all maps $$z\mapsto az+b$$ where we imagine the group acting on $\mathbb{Z}_p$ with mod $p$ arithmetic. From this it is clear to see the following theorem:

A irreducible polynomial of odd prime degree is solvable iff all roots are can be rationally expressed in terms of any two roots.

For it follows from the above expression that the only permutation which fixes two elements is the identity. Thus the splitting field is generated by any two roots. The other direction of more difficult.

In particular if there are two real roots then all roots are real. Since such an equation always has a real root, there are either one or $p$ real roots.

Rene Schipperus
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  • Ok I got it. You are talking about the famous result of Galois himself. I am yet to study its full proof but yes it does show that a solvable irreducible equation od prime degree has either one or all roots real. So that if the equation has 3 real roots then it is not solvable. I think the hard part is figuring out when the equation has only one real root and is still not solvable. +1 for the answer. – Paramanand Singh Apr 26 '17 at 05:01
  • Also this approach appears to make more sense to me than the approach outlined in the answer I linked at the start of my post. – Paramanand Singh Apr 26 '17 at 05:04