Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

My question seems to be quite subtle, but let is see an example first. This example is appeared originally in the notes of J.S.Milne :

Consider $X^5-X-1$. Modulo 2, this factor as $(X^2+X+1)(X^3+X^2+1)$, and modulo 3 it is irreducible. Hence $G_f$(the Galois group of $X^5-X-1$) contains $(ik)(lmn)$ and (12345), and so also $(ik)(lmn)^3=ik$. Therefore $G_f=S_5$.

My question is :

  1. How do we know that we have to prove that $f$ is irreducible modulo 3?

  2. Can we apply the same method to compute the Galois group of other irreducible polynomial of degree 5 over $\mathbb{Q}$?

  3. Given an irreducible polynomial of degree 5, what is the general technique to compute its Galois group over $\mathbb{Q}$

Update: I also found this example in the book Abstract Algebra of Dummit and Foot on page 641. In that book, the author claims that $x^5-x-1$ is irreducible mod 3 and then irreducible over $\mathbb{Z}[x]$ which is more motivated than Milne's one. So, my question is : what is the relation between irreducibility in $\mathbb{F}_p[x]$ and irreducibility in $\mathbb{Z}[x]$ ?

share|cite|improve this question
If a quintic is known to be irreducible, then we already know that its Galois group will have a 5-cycle as an element. This is because the Galois group will act transitively on the roots, hence has order divisible by 5, hence an element ($\in S_5$) of order 5. That already narrows down the possibilities for the Galois group severely (IIRC five possibilities remain). – Jyrki Lahtonen Jan 5 '13 at 9:15
@Jyrki: No further comments. :-) – Asaf Karagila Jan 5 '13 at 11:36
  1. The same lengthy argument, if that polynomial is reducible, either it has a root in $F_3$ or factors into 2 polynomials of degree 2 and 3.

  2. He used this particular method since, if a polynomial is irreducible for some mod p (p is prime), then it is irreducible over $\mathbb Q$, but converse is not true, hence using this is not easy always.

  3. I think there is no such general procedure you need to use possible facts you know.

share|cite|improve this answer
Is there something special of Galois group of an irreducible polynomial of degree 5 ? – knot Jan 5 '13 at 9:11
I don't think so, but solvability by radicals is an important result, group $S_n$ is not solvable if $n\ge 5$, then this particular example is important because, if a Galois group contains a transposition and cycle of length n then Galois group is Sn – Ram Jan 5 '13 at 9:50

Here are a few results I find useful while computing Galois groups.

Result $1$: Let $p$ be a prime and $f(x) \in \mathbb{Q}[x]$ be an irreducible monic polynomial of degree $p$. If all but $2$ roots of $f(x)$ are in $\mathbb{R}$, then the Galois group of $f(x)$ is $S_p$.

Result $2$: Let $f(X) \in \mathbb{Z}[x]$ be a monic irreducible polynomial of degree $n$. Let $p$ be a prime number such that $p$ does not divide the discriminant of $f$. Let $f(x) \equiv q_1(x)\dots q_k(x)$ (mod $p$) be an irreducible factorization of $f(x)$ mod $p$. Let $d_i = \deg(q_i(x))$. If we view the Galois group of $f(x)$ as a subgroup of $S_n$, then this group has a permutation of type $(d_1,\dots,d_k)$.

Result $3$: Let $f(x)$ be a monic irreducible polynomial in $\mathbb{Z}[x]$ of prime degree $p$. If $q$ is a prime number which does not divide the discriminant of $f$, such that $f(x)$ mod $q$ has all but two roots in $\mathbb{F}_q$, then the Galois group of $f(x)$ over $\mathbb{Q}$ is $S_p$.

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.