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 $S_5$?

• We could start looking at factorization of the polynomial modulo a prime $p$ as that is connected to the cycle structure whenever $p\nmid D$.. However, this time we want to disprove the existence of a 2-cycle, so we only expect negative information. Chebotarev's theorem tells us that if the Galois group has a 2-cycle or a 3-cycle, it will show up with a known frequency. But how high would we need to go to be convinced we never find one? – Jyrki Lahtonen Apr 3 '15 at 21:15

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.