Examples in Galois Theory 2 - Cubic Extensions

Let KK be a field of characteristic ≠2≠2 and 33. In this post we discuss the Galois group of a cubic polynomial ff over KK. The process is not very hard but there are some quite non-trivial observations.

Let kk be an arbitrary field and suppose f(X)∈k[X]f(X)∈k[X] is separable and, i.e., ff has no multiple roots in an algebraic closure, and of degree ≥1≥1. Let f(X)=(X−x1)⋯(X−xn)f(X)=(X−x1)⋯(X−xn) be its factorisation in a splitting field FF. Put G=G(L/k)G=G(L/k). We say that GG is the Galois group of ff over kk. Let xixi be a root of ff and pick any σ∈Gσ∈G. By definition of Galois group, we see σ(xi)σ(xi) is still a root of ff (consider the map ~σ:L[X]→L[X]σ~:L[X]→L[X] induced by σσ naturally; it is the identity when restricted to k[X]k[X]). This is to say, elements of GG permutes the roots of ff.

For example, consider L=CL=C, k=Rk=R, f(X)=X2+1f(X)=X2+1. The Galois group GG contains two elements and is generated by complex conjugation σ:a+bi↦a−biσ:a+bi↦a−bi. A root of ff is ii, and σ(i)=−iσ(i)=−i is another root.

Based on this fact, we can consider GG as a subgroup of SnSn, where nn is the degree of ff. The structure of SnSn can be extremely complicated, but for now we assume that they are well-known. The question is, what subgroup is GG inside SnSn. Let's take a look into the case when n=3n=3.

To begin with we note that we can assume that the quadratic term is 00. Let f(X)=X3+aX2+bX+cf(X)=X3+aX2+bX+c be a polynomial, then f(X−a3)=(X−a3)3+a(X−a3)2+b(X−a3)+c=X3−aX2+a23X−a327+aX2−6a33X+⋯f(X−a3)=(X−a3)3+a(X−a3)2+b(X−a3)+c=X3−aX2+a23X−a327+aX2−6a33X+⋯ and as a result aX2aX2 is cancelled. A translation does not change any property of a polynomial except the value of its roots. Therefore we can reduce our study to polynomials in the depressed form f(X)=X3+aX+b.f(X)=X3+aX+b. In fact, for all g(X)=Xn+an−1Xn−1+⋯+a0g(X)=Xn+an−1Xn−1+⋯+a0, we can cancel out an−1Xn−1an−1Xn−1 by a substitution Y=X−an−1nY=X−an−1n.

Irreducibility

Now back to our main story. First of all we study irreducibility. If ff is irreducible, then clearly it has no root in KK. On the other hand, if ff has no root in KK, does that mean ff is irreducible over KK? This does not hold in general for all polynomials. For example, the polynomial g(X)=(X2+1)2g(X)=(X2+1)2 is not irreducible yet it has no root in RR or QQ. But fortunately, 33 is a beautiful number and we can proceed. Were ff irreducible, there would be a factorisation f(X)=p1(X)p2(X)f(X)=p1(X)p2(X) with each pi(X)pi(X) being a proper factor of f(X)f(X). However, this is to say, at least one of pi(X)pi(X) has degree 11. A contradiction. We therefore have a result as follows:

Proposition 1. Let f(X)f(X) be a cubic polynomial in K[X]K[X] where charK=0,5,7,…char⁡K=0,5,7,…, then ff is irreducible over KK if and only if ff has no root in KK.

The Galois group

Notation being above, we assume that ff is irreducible. Let LL be the splitting field of ff. We claim that ff is separable. Before proving the claim, one should notice that the characteristic matters a lot. For example, X3−2X3−2 is irreducible over QQ but X3−2=(X+1)3X3−2=(X+1)3 over F3[X]F3[X] and we therefore have a triple root.

ff is separable if and only if gcd(f,f′)=0gcd(f,f′)=0. The derivative of ff, which should be simplified because ff has been, is given by f′(X)=3X2+a.f′(X)=3X2+a. It is not equal to aa because the characteristic of KK is not 33. We will show carefully that f(X)f(X) is separable by working on these two polynomials.

The first question is the value of aa and bb. If some of them is 00 then things may be easier or harder. Note first we must have b≠0b≠0 because if not then f(X)=X(X2+a)f(X)=X(X2+a) and this is not irreducible. If a=0a=0, then f(X)=X3+bf(X)=X3+b and f′(X)=3X2≠0f′(X)=3X2≠0 because charK≠3char⁡K≠3. It follows that (f,f′)=0(f,f′)=0 because either XX or X2X2 divides X3+bX3+b.

Now there only remains the most general case: a≠0a≠0 and b≠0b≠0. This is where the Euclidean Algorithm kicks in. Recall that for any three polynomials p,q,rp,q,r in K[X]K[X], we have gcd(p,q)=gcd(q,p)=gcd(q,p+rq).gcd(p,q)=gcd(q,p)=gcd(q,p+rq). This is how the Euclidean Algorithm works. Note we can write f(X)=13Xf′(X)+23aX+br0(X).f(X)=13Xf′(X)+23aX+b⏟r0(X). It follows that gcd(f,f′)=gcd(f′,r0)gcd(f,f′)=gcd(f′,r0). We next work on f′f′ and r0r0. f′(X)=92aX(23aX+b)+(−9b2aX+a)r1(X).f′(X)=92aX(23aX+b)+(−9b2aX+a)⏟r1(X). However, r0(X)r0(X) and r1(X)r1(X) has common divisor 00, which implies that ff and f′f′ has common divisor 00. Whichever the case is, we have gcd(f,f′)=0gcd(f,f′)=0 and therefore ff is separable. Note the fact that the characteristic of KK is not 22 or 33 is frequently used here, otherwise there are a lot of equations making no sense.

Where we are at? We want to ensure that ff is separable so that working with the Galois group of ff is not that troublesome. And ff is. We now back to the study of the Galois group G=G(L/K)G=G(L/K), where LL is the splitting field of ff. Let α1α1, α2α2, α3α3 be the roots of ff and pick one of them as αα. We see [K(α):K]=3[K(α):K]=3.

Since GG permutes three elements, GG has to be a subgroup of S3S3. Therefore |G|=[L:K]≥[K(α):K]=3|G|=[L:K]≥[K(α):K]=3, which implies that |G|=3|G|=3 or 66. In the first case, G=A3G=A3, the alternating group. In the second case, G=S3G=S3 and K(α)K(α) is not normal over KK because, there is an irreducible polynomial f(X)∈K[X]f(X)∈K[X] which has a root in K(α)K(α) that does not split into linear factors in K(α)K(α). This is the definition of normal extension.

The question now is, when GG is S3S3 and when it is A3A3? We get a good chance to review finite group theory. This is answered by the sign of elements in GG. To be precise, G=S3G=S3 if and only if GG has an odd element. If not then G=A3G=A3. To work with this, we recall how the sign function work. Put δ=(α1−α2)(α2−α3)(α3−α1).δ=(α1−α2)(α2−α3)(α3−α1). For any σ∈Gσ∈G, we have σ(δ)=ε(σ)δσ(δ)=ε(σ)δ, where ε(σ)ε(σ) is the sign of σσ. If we put Δ=δ2Δ=δ2, which is the discriminant, we see σ(Δ)=Δσ(Δ)=Δ. Therefore Δ∈LG=KΔ∈LG=K. But wait, since σ(δ)=±δσ(δ)=±δ, the sign is not guaranteed, we see δδ is not guaranteed to be in KK. This is where we crack the problem.

If δ∈Kδ∈K, or more precisely, √Δ∈KΔ∈K, then σ(δ)=δσ(δ)=δ and it follows that ε(σ)=1ε(σ)=1 for all σ∈Gσ∈G. This can only happen if G=A3G=A3.

If √Δ∉KΔ∉K, then δδ is not fixed by GG. There is some σ∈Gσ∈G such that σ(δ)=−δσ(δ)=−δ, which is to say that ε(σ)=−1ε(σ)=−1. This can only happen when G=S3G=S3.

We have the following conclusion.

Proposition 2. Notation being above. Assume that ff is irreducible. Then the Galois group of ff is S3S3 if and only if √δ∉Kδ∉K. The group is A3A3 if and only if √Δ∈KΔ∈K.

A dirty calculation shows that Δ=−4a3−27b2Δ=−4a3−27b2. One can show this using Vieta's formulas. You shan't feel this to be strange because in the quadratic case we have Δ=b2−4acΔ=b2−4ac and we did care if Δ>0Δ>0, which amounts to whether √Δ∈RΔ∈R.

Let's conclude this post by a handy but nontrivial example. Consider f(X)=X3−X−1f(X)=X3−X−1 The discriminant is −4⋅(−1)3−27⋅(−1)2=−23−4⋅(−1)3−27⋅(−1)2=−23, which lies in Q(√−23)Q(−23) and therefore the Galois group over it is A3A3. However, when the base field is a subfield, for example, QQ, then the Galois group is S3S3.