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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.

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As observed by Andreas Caranti, over $\mathbb{F}_2$ your polynomial factors into a product of irreducible polynomials as $$x^5+99x-1=x^5+x+1=(x^2+x+1)(x^3+x+1).$$ Therefore the splitting field is the compositum of $\mathbb{F}_8$ and $\mathbb{F}_4$, i.e. $\mathbb{F}_{2^6}$. The other finite field cases seem to check out. –  Jyrki Lahtonen Mar 6 '13 at 7:38
    
Except that the cubic factor should be $x^3+x^2+1$ :-) –  Jyrki Lahtonen Mar 6 '13 at 7:56
    
I am quite confused in the $\mathbb F_{11}$ case. I thought it would be like $\mathbb F_{11^4}$. Could you provide some explanations? Thanks. –  awllower Mar 6 '13 at 9:29
    
@awllower, over $\Bbb{F}_{11}$ the polynomial splits in linear factors. –  Andreas Caranti Mar 6 '13 at 11:06
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@awllower: Correct. –  Jyrki Lahtonen Mar 6 '13 at 14:11
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1 Answer

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.

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