Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
@awllower: Correct. – Jyrki Lahtonen Mar 6 '13 at 14:11

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.