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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 $\mathbb{Z}[x]$ ?

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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
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2 Answers

  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.

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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
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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 $S_p$.

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