Galois theory, resolvent, Frobenius group I need to prove that the Galois group of the polynomial $x^5+15x+12\in \mathbb{Q}[x]$ is the Frobenius group of order 20.
The discriminant of that polynomial is $D=2^{10}\cdot 3^4\cdot 5^5$, i.e. it is not a square and therefore not a subgroup of the alternating group $A_5$. So far so good, it shows me that it can't be $D_{10},Z_5$ or $A_5$, but how can I show, that the Galois group is not the symmetric group $S_5$?
In my paper they often mention a so-called resolvent method. But unfortunately they don't describe how this is done exactly. How is a resolvent polynomial constructed? Or how would you prove that the corresponding Galois group is not $S_5$?
Thank you already in advance.
 A: Here's some incomplete work that someone might be able to help me patch up:
Let's view $\operatorname{Gal}(f)$ as a permutation group acting on the roots of $f$.  It is a theorem that a $2$-cycle together with a $p$-cycle generates the entire symmetric group $S_p$.  Using this as motivation, we first show that $\operatorname{Gal}(f)$ contains a $p$-cycle.  Let $\alpha$ be a root of $f$ and consider the tower of fields $\mathbb{Q} \subset \mathbb{Q}[\alpha] \subset K$, where $K$ is the splitting field of $f$.  Since $f$ is irreducible (guaranteed by Eisenstein), then we know $\mathbb{Q}[\alpha]$ is a degree $5$ extension of $\mathbb{Q}$.  Therefore, $5$ divides $[K: \mathbb{Q}] = |\operatorname{Gal}(f)|$, and by Cauchy's theorem, $\operatorname{Gal}(f)$ therefore contains an element of order $5$, which is necessarily a $5$-cycle in $S_5$.
Given our work above, showing $\operatorname{Gal}(f) \ncong S_5$ is equivalent to showing that it does not contain a $2$-cycle.  Since $f$ has four complex roots, complex conjugation is out of the question (this will be a composition of $2$-cycles).  With some work that I won't include at the moment, I have also shown that $( \text{Real Root} ) \mapsto ( \text{Any of the complex roots} )$ cannot be a $2$-cycle.  
All that remains to show is that $\text{Complex root} \mapsto \text{Another complex root that is not its conjugate}$ is not a $2$-cycle.  
I will post what I have in the hopes that it'll help you or someone else finish up.  In the meanwhile, I'll give this some more thought.  Ultimately, I will delete my answer if nothing fruitful comes of it.
A: The $F_5$ resolvent of the quintic $x^5 + 15x + 12$ is $$y^{6} + 120 y^{5} + 9000 y^{4} + 540000 y^{3} + 20250000 y^{2} + 324000000 y$$ (calculated using a computer algebra system)
This has an obvious rational root $y=0$ which implies the Galois group is contained in the Frobenius group $F_5$.
This answer is probably not very helpful, since the formula for the resolvent is so long and complex it may have been just as easy to ask the computer for the Galois group directly. Anyway I hope it's of some interest.
