Introduction

In the previous post we are convinced that the Galois group of a separable irreducible polynomial \(f\) can be realised as a subgroup of the symmetric group, the elements of which permute the roots of \(f\). We worked on cubic polynomials over a field with characteristic not equal to \(2\) and \(3\), and this definitely works with \(\mathbb{Q}\). In this post we go one step further.

Let \(f \in \mathbb{Q}[X]\) be an irreducible polynomial of prime degree \(p\). Since it is also separable (see lemma 9.12.1 on the stack project), we can safely work on its Galois group \(G\). One immediately wants to question the position of \(\mathfrak{S}_p\). Indeed we have \(G \subset \mathfrak{S}_p\). 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 \(p\) be a prime number. The symmetric group \(\mathfrak{S}_p\) is generated by \([12 \cdots p]\) and an arbitrary transposition \([rs]\).

Click to expand the proof.

Proof. We prove this by presenting several sets of generators of \(\mathfrak{S}_n\) where \(n\) is a positive integer.

It is generated by cycles. This is a really, really routine verification and sometimes assumed as a fact.
It is generated by transpositions, i.e., \(2\)-cycles. It suffices to show that a cycle is a product of transpositions. Indeed, for any cycle \([i_1\dots i_k]\) in \(\mathfrak{S}_n\), we have \([i_1\cdots i_k]=[i_1i_2][i_2i_3]\cdots[i_{k-1}i_k]\). This proves our statement.
It is generated by translations of the form \([1k]\). It suffices to show that a transposition is generated as such. For any transposition \([rs]\), we have \([rs]=[1r][1s][1r]\).
It is generated by adjacent translations, i.e. the generators can be of the form \([k-1 ,k]\). This follows from the following identity:

\[[1k]=[12][23]\cdots[k-1,k][k-2,k-1]\cdots [23][12]\]

It is generated by two elements: \(\sigma=[12]\) and \(\tau=[12\cdots n]\). This follows from the following identity:

\[\tau^{k-2}\sigma\tau^{-(k-2)}=[\tau^{k-2}(1)\tau^{k-2}(2)]=[k-1,k].\]

Now, back to the case when \(n=p\) is prime. Put \(\sigma=[rs]\) and \(\tau=[12\cdots p]\). If \(s-r=1\) then it is already proved in 5 by several conjugations. Therefore we may assume that \(d=s-r>1\). From now on integers may be a number in either \(\mathbb{Z}\) or \(\mathbf{F}_p=\mathbb{Z}/p\mathbb{Z}\), depending on the context. Recall that \(\mathbf{F}_p\) is a field. Pick the integer \(w\) such that \(dw=1\) in \(\mathbf{F}_p\). By conjugation we see \(\tau\) and \(\sigma\) generate \[[1,1+d],[1+d,1+2d],\dots,[1+(w-1)d,1+wd].\] The product of elements above is \([1,1+wd]=[12]\). Therefore we are still back to 5. \(\square\)

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 \(f(X) \in \mathbb{Q}[X]\) be an irreducible polynomial of prime degree. If \(f\) has precisely two nonreal roots, then the Galois group \(G\) over \(\mathbb{Q}\) is \(\mathfrak{S}_p\).

Proof. Let \(L\) be the splitting field of \(f\). It suffices to show that \(G\) contains a transposition and a \(p\)-cycle, which is \([12\cdots p]\). By the Sylow's theorem, \(G\) has a subgroup \(H\) of order \(p\), which can only be cyclic. Say \(H=\langle \sigma \rangle\). Suppose \(\sigma\) is of cycle type \((k_1,\dots,k_r)\). Then the period of \(\sigma\), which equals \(p\), is the least common multiple of \(k_1,\dots,k_r\), where \(k_1+\dots+k_r=p\). This can only happen when \(r=1\) and \(k_1=p\). Therefore \(\sigma\) is a \(p\)-cycle.

In fact, \(\sigma\) can be considered as \([12\dots p]\). Suppose the order of roots of \(f\) is given, for which we have \(\sigma=[i_1 i_2 \dots i_p]\). Then If we re-order these roots, by putting the \(k\)th root to be the original \(i_k\)th root, then we can write \(\sigma=[12\dots p]\). (This re-ordering is, in fact, a conjugation.)

It remains to prove that \(G\) contains a transposition. Let \(\alpha\) and \(\beta\) be two nonreal roots of \(f\). Since \(\overline{\alpha}\) is also a root of \(f\) (because coefficients of \(f\) are real; if \(\sum_{n=0}^{p}a_n\alpha^n=0\), then \(\sum_{n=0}^{p}a_n\overline{\alpha}^n=\sum_{n=0}^{p}\overline{a_n\alpha^n}=\overline{0}=0\)) we see \(\beta=\overline{\alpha}\). Therefore complex conjugation over \(\mathbb{Q}(\alpha)\) extends to \(L\) as an element of order \(2\), which is a transposition in \(G\). This proves our assertion. \(\square\)

For example, consider the polynomial \[f(X)=X^5-4X+2.\] With calculus one can show that it has exactly three roots, hence it has two nonreal roots. Eisenstein's criterion shows that \(f\) is irreducible. Therefore we are allowed to use proposition 1. The Galois group of \(f\) is \(\mathfrak{S}_5\).

This also works fine when \(p=2\) or \(3\). The case when \(p=2\) is nothing but working around a quadratic polynomial. When \(f(X)\) is irreducible of degree \(3\), and it has two nonreal roots, we also know that it has an irrational root. Let the roots be \(a+bi,a-bi,c\) where \(b \ne 0\) and \(c\) is irrational. We see \[\sqrt\Delta=2bi(c-a-bi)(c-a+bi)=2bi[(c-a)^2+b^2] \not \in \mathbb{Q}.\] Therefore the Galois group is \(\mathfrak{S}_3\).

"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 \(\mathfrak{A}_p\) 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 \(p \ge 5\).

Proposition 2. Let \(f \in \mathbb{Q}[X]\) be an irreducible polynomial of prime degree \(p \ge 5\). Suppose that \(f\) has \(k>0\) pairs of nonreal roots. If \(p \ge 4k+1\), then the Galois group \(G\) is isomorphic to \(\mathfrak{A}_p\) or \(\mathfrak{S}_p\). If \(k\) is odd then \(G \cong \mathfrak{S}_p\).

The proof is done by showing that \(\mathfrak{A}_p \subset G \subset \mathfrak{S}_p\). As the index of \(\mathfrak{A}_p\) is \(2\), \(G\) can only be one of them. The solvability of \(G\) is also concerned here.

Indeed, what we have proved in "the simplest case" is nothing but \(k=1\). When \(p \ge 5\) we clearly have \(p \ge 1+4 \times 1\). This refined the result of A. Bialostocki and T. Shaska (see arXiv:math/0601397), and the inequality used to be \[p \ge k(k\log k+2\log k+3).\] When \(k\) is big enough, we have \(k(k\log{k}+2\log{k}+3) \ge 4k+1\). Oz Ben-Shimol's result is a refinement because it is saying, \(p\) does not need to that big. He also offered a refined algorithm to compute the Galois group, which we will present below. Also, computing \(4k+1\) is much easier than computing \(k^2\log{k}\) 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, \(\Delta(f)\) is the discriminant of \(f\). We have seen that whether \(\Delta\) is a perfect square matters a lot. The discussion of ReductionMethod can be trailed in Oz Ben-Shimol's paper.