How to compute the Galois group of $x^5+99x-1$ over $\mathbb{Q}$? I was stuck trying to compute the Galois group of $x^5 + 99x -1$. The problem asks to compute the Galois group over $\mathbb{F}_2, \mathbb{F}_3, \mathbb{F}_5, \mathbb{F}_{11}$ and $\mathbb{Q}$. I was able to handle the finite fields cases: I believe the splitting fields are $\mathbb{F}_{2^{5}}/ \mathbb{F}_2$, $\mathbb{F}_{3^4}/ \mathbb{F}_3$, $\mathbb{F}_{5^{5}}/ \mathbb{F}_5$,  and $\mathbb{F}_{11}/ \mathbb{F}_{11}$. However I am stuck on the $\mathbb{Q}$ case. 
 A: Let $E$ be the splitting field of $f = x^5 + 99x -1$ over $\Bbb{Q}$. Let $G = \operatorname{Gal}(E/\Bbb{Q})$.
There is an important result of Dedekind that tells you the following. 
Reduce the monic polynomial with integer coefficients $f$ modulo a prime $p$. The Galois group $G_p$ of the splitting field over $\Bbb{F}_{p}$ of (the reduction modulo $p$ of) $f$ will be cyclic then. Write the generator of $G_p$ as a permutation, in the form of a product of disjoint cycles. 
Then provided $p$ does not divide the discriminant of $f$, the Galois group $G$, regarded as a group of permutations of the roots of $f$, contains a permutation  with the same structure as a product of disjoint cycles. Here the discriminant would be $2434534530869$, which decomposes as $7^{2}\cdot 107 \cdot 464339983$, but it's simpler to verify that there are no multiple roots modulo the various primes we are checking.
So with $p = 2$ you find in $G$ the product of a $3$-cycle and a $2$-cycle, with $p = 3$ a $4$-cycle and with $p = 5$ a $5$-cycle. I used GAP to do these calculations (including the discriminant above), splitting $f$ into irreducible factors over each $\Bbb{F}_{p}$. The degrees of these factors tell you the cycles lengths that appear when writing the permutation as a product of disjoint cycles.
This gives you that $G$ has order at least $\operatorname{lcm}(6,4,5) = 60$, but it contains odd permutations, so it's not $A_5$, so $G = S_5$.
Barring mistakes, as usual.
