Introduction

In the previous post we are convinced that the Galois group of a separable irreducible polynomial ff can be realised as a subgroup of the symmetric group, the elements of which permute the roots of ff. We worked on cubic polynomials over a field with characteristic not equal to 22 and 33, and this definitely works with QQ. In this post we go one step further.

Let f∈Q[X]f∈Q[X] be an irreducible polynomial of prime degree pp. Since it is also separable (see lemma 9.12.1 on the stack project), we can safely work on its Galois group GG. One immediately wants to question the position of SpSp. Indeed we have G⊂SpG⊂Sp. The question is, when does the equality hold? It is not likely to have an immediate answer. However, we have some interesting sufficient conditions, which will be discussed in this post.

Generators of the Symmetric Group

We present some handy results in finite group theory that will be used in the main result. One may skip this section until needed. I will collapse the proof in case one wants to treat it as an exercise.

Lemma 1. Let pp be a prime number. The symmetric group SpSp is generated by [12⋯p][12⋯p] and an arbitrary transposition [rs][rs].

Click to expand the proof. Proof. We prove this by presenting several sets of generators of where is a positive integer.

Now, back to the case when is prime. Put and . If then it is already proved in 5 by several conjugations. Therefore we may assume that . From now on integers may be a number in either or , depending on the context. Recall that is a field. Pick the integer such that in . By conjugation we see and generate The product of elements above is . Therefore we are still back to 5.

Computing the Galois Group

We have many good reasons to study the Galois group of something. It would be great if the group can be written down explicitly. In this section we show that the group can be revealed by the number of nonreal roots.

The Simplest Case

Proposition 1. Let f(X)∈Q[X]f(X)∈Q[X] be an irreducible polynomial of prime degree. If ff has precisely two nonreal roots, then the Galois group GG over QQ is SpSp.

Proof. Let LL be the splitting field of ff. It suffices to show that GG contains a transposition and a pp-cycle, which is [12⋯p][12⋯p]. By the Sylow's theorem, GG has a subgroup HH of order pp, which can only be cyclic. Say H=⟨σ⟩H=⟨σ⟩. Suppose σσ is of cycle type (k1,…,kr)(k1,…,kr). Then the period of σσ, which equals pp, is the least common multiple of k1,…,krk1,…,kr, where k1+⋯+kr=pk1+⋯+kr=p. This can only happen when r=1r=1 and k1=pk1=p. Therefore σσ is a pp-cycle.

In fact, σσ can be considered as [12…p][12…p]. Suppose the order of roots of ff is given, for which we have σ=[i1i2…ip]σ=[i1i2…ip]. Then If we re-order these roots, by putting the kkth root to be the original ikikth root, then we can write σ=[12…p]σ=[12…p]. (This re-ordering is, in fact, a conjugation.)

It remains to prove that GG contains a transposition. Let αα and ββ be two nonreal roots of ff. Since ¯¯¯¯αα¯ is also a root of ff (because coefficients of ff are real; if ∑pn=0anαn=0∑n=0panαn=0, then ∑pn=0an¯¯¯¯αn=∑pn=0¯¯¯¯¯¯¯¯¯¯¯anαn=¯¯¯0=0∑n=0panα¯n=∑n=0panαn¯=0¯=0) we see β=¯¯¯¯αβ=α¯. Therefore complex conjugation over Q(α)Q(α) extends to LL as an element of order 22, which is a transposition in GG. This proves our assertion. □◻

For example, consider the polynomial f(X)=X5−4X+2.f(X)=X5−4X+2. With calculus one can show that it has exactly three roots, hence it has two nonreal roots. Eisenstein's criterion shows that ff is irreducible. Therefore we are allowed to use proposition 1. The Galois group of ff is S5S5.

This also works fine when p=2p=2 or 33. The case when p=2p=2 is nothing but working around a quadratic polynomial. When f(X)f(X) is irreducible of degree 33, and it has two nonreal roots, we also know that it has an irrational root. Let the roots be a+bi,a−bi,ca+bi,a−bi,c where b≠0b≠0 and cc is irrational. We see √Δ=2bi(c−a−bi)(c−a+bi)=2bi[(c−a)2+b2]∉Q.Δ=2bi(c−a−bi)(c−a+bi)=2bi[(c−a)2+b2]∉Q. Therefore the Galois group is S3S3.

"Linear" Generalisation

It is way too ambitious to restrict ourselves in one single pair of roots. Also, it seems we have ignored the alternating group ApAp for no reason. Oz Ben-Shimol gave us a nice way to work around this (see arXiv:0709.2868). The whole paper is not easy but the result is pretty beautiful and generalised what we said above as p≥5p≥5.

Proposition 2. Let f∈Q[X]f∈Q[X] be an irreducible polynomial of prime degree p≥5p≥5. Suppose that ff has k>0k>0 pairs of nonreal roots. If p≥4k+1p≥4k+1, then the Galois group GG is isomorphic to ApAp or SpSp. If kk is odd then G≅SpG≅Sp.

The proof is done by showing that Ap⊂G⊂SpAp⊂G⊂Sp. As the index of ApAp is 22, GG can only be one of them. The solvability of GG is also concerned here.

Indeed, what we have proved in "the simplest case" is nothing but k=1k=1. When p≥5p≥5 we clearly have p≥1+4×1p≥1+4×1. This refined the result of A. Bialostocki and T. Shaska (see arXiv:math/0601397), and the inequality used to be p≥k(klogk+2logk+3).p≥k(klog⁡k+2log⁡k+3). When kk is big enough, we have k(klogk+2logk+3)≥4k+1k(klog⁡k+2log⁡k+3)≥4k+1. Oz Ben-Shimol's result is a refinement because it is saying, pp does not need to that big. He also offered a refined algorithm to compute the Galois group, which we will present below. Also, computing 4k+14k+1 is much easier than computing k2logkk2log⁡k plus something.

Input: An irreducible polynomial f(X) over Q with prime degree p >= 5
Output: The Galois group Gal(f/Q)
begin
    r:=NumberOfRealRoots(f(X))
    k:=(p-r)/2
    if k>0 and p>=4k+1 then
        if k is odd then
            Gal(f/Q)=S_p;
        else
            if ∆(f) is a complete square then
                Gal(f/Q)=A_p;
            else
                Gal(f/Q)=S_p;
            endif;
        endif;
    else
        ReductionMethod(f(X));
    endif;
end;

Here, Δ(f)Δ(f) is the discriminant of ff. We have seen that whether ΔΔ is a perfect square matters a lot. The discussion of ReductionMethod can be trailed in Oz Ben-Shimol's paper.