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Let $f=2x^5-5x^4+5\in \Bbb Q[x]$. Then, how to prove that it's not solvable by radicals?

Since $f$ is solvable by radicals iff $Gal(f)$ is solvable and $Gal(f)\subset S_5$ (up to isomorphism), I try to prove that $Gal(f)$ is not solvable, but how to do it?

The roots of $f$ have weird form, so I don't get any information about $Gal(f)$...

Give some hint about it and tip to solve a problem like this: polynomial is solvable by radical or not.

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If you have $Gal(f)$ as proper subgroup of $S_5$ then you can not say for sure... So you have to somehow try that this is not proper... I am also interested in this problem but i am not getting any thoughts... –  Praphulla Koushik Dec 10 '13 at 7:10… might be of some help... –  Praphulla Koushik Dec 10 '13 at 7:11

1 Answer 1

up vote 6 down vote accepted

Here are some facts that will be helpful in proving that the Galois group of $f$ is all of $S_5$ and thus not solvable.

Fact 1: The degree of an irreducible polynomial over $\mathbb{Q}$ divides the order of its Galois group.

Fact 2: If the order of a finite group is divisible by a prime $p$, then the group has an element of order $p$.

Fact 3: If a polynomial of degree $n$ has exactly two complex roots, then its Galois group viewed as a subgroup of $S_n$ contains a transposition.

Fact 4: For any prime $p$, $S_p$ is generated by any transposition together with any $p$-cycle.

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3rd fact is the most important I think... I don't know about it or may I don't see since I learn it very recently. Thank you very much! –  Arturo Dec 10 '13 at 8:07

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