# Examples in Galois Theory 2 - Cubic Extensions

Let \(K\) be a field of characteristic \(\ne 2\) and \(3\). In this post we discuss the Galois group of a cubic polynomial \(f\) over \(K\). The process is not very hard but there are some quite non-trivial observations.

Let \(k\) be an arbitrary field and suppose \(f(X) \in k[X]\) is separable and, i.e., \(f\) has no multiple roots in an algebraic closure, and of degree \(\ge 1\). Let \[ f(X)=(X-x_1)\cdots(X-x_n) \] be its factorisation in a splitting field \(F\). Put \(G=G(L/k)\). We say that \(G\) is the Galois group of \(f\) over \(k\). Let \(x_i\) be a root of \(f\) and pick any \(\sigma \in G\). By definition of Galois group, we see \(\sigma(x_i)\) is still a root of \(f\) (consider the map \(\tilde\sigma:L[X] \to L[X]\) induced by \(\sigma\) naturally; it is the identity when restricted to \(k[X]\)). This is to say, elements of \(G\) permutes the roots of \(f\).

For example, consider \(L=\mathbb{C}\), \(k=\mathbb{R}\), \(f(X)=X^2+1\). The Galois group \(G\) contains two elements and is generated by complex conjugation \(\sigma:a+bi \mapsto a-bi\). A root of \(f\) is \(i\), and \(\sigma(i)=-i\) is another root.

Based on this fact, we can consider \(G\) as a subgroup of \(\mathfrak{S}_n\), where \(n\) is the degree of \(f\). The structure of \(\mathfrak{S}_n\) can be extremely complicated, but for now we assume that they are well-known. The question is, what subgroup is \(G\) inside \(\mathfrak{S}_n\). Let's take a look into the case when \(n=3\).

To begin with we note that we can assume that the quadratic term is \(0\). Let \(f(X)=X^3+aX^2+bX+c\) be a polynomial, then \[ \begin{aligned} f\left(X-\frac{a}{3}\right) &= \left( X-\frac{a}{3}\right)^3+a\left( X-\frac{a}{3}\right)^2+b\left( X-\frac{a}{3}\right)+c \\ &= X^3-aX^2+\frac{a^2}{3}X-\frac{a^3}{27} + aX^2-\frac{6a^3}{3}X+\cdots \end{aligned} \] and as a result \(aX^2\) 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 \[ f(X)=X^3+aX+b. \] In fact, for all \(g(X)=X^n+a_{n-1}X^{n-1}+\dots+a_0\), we can cancel out \(a_{n-1}X^{n-1}\) by a substitution \(Y=X-\frac{a_{n-1}}{n}\).

# Irreducibility

Now back to our main story. First of all we study irreducibility. If \(f\) is irreducible, then clearly it has no root in \(K\). On the other hand, if \(f\) has no root in \(K\), does that mean \(f\) is irreducible over \(K\)? This does not hold in general for all polynomials. For example, the polynomial \(g(X)=(X^2+1)^2\) is not irreducible yet it has no root in \(\mathbb{R}\) or \(\mathbb{Q}\). But fortunately, \(3\) is a beautiful number and we can proceed. Were \(f\) irreducible, there would be a factorisation \[ f(X)=p_1(X)p_2(X) \] with each \(p_i(X)\) being a proper factor of \(f(X)\). However, this is to say, at least one of \(p_i(X)\) has degree \(1\). A contradiction. We therefore have a result as follows:

Proposition 1.Let \(f(X)\) be a cubic polynomial in \(K[X]\) where \(\operatorname{char}K=0,5,7,\dots\), then \(f\) is irreducible over \(K\) if and only if \(f\) has no root in \(K\).

# The Galois group

Notation being above, we assume that \(f\) is irreducible. Let \(L\) be the splitting field of \(f\). We claim that \(f\) is separable. Before proving the claim, one should notice that the characteristic matters a lot. For example, \(X^3-2\) is irreducible over \(\mathbb{Q}\) but \(X^3-2=(X+1)^3\) over \(\mathbf{F}_3[X]\) and we therefore have a triple root.

\(f\) is separable if and only if \(\gcd(f,f')=0\). The derivative of \(f\), which should be simplified because \(f\) has been, is given by \[ f'(X)=3X^2+a. \] It is not equal to \(a\) because the characteristic of \(K\) is not \(3\). We will show carefully that \(f(X)\) is separable by working on these two polynomials.

The first question is the value of \(a\) and \(b\). If some of them is \(0\) then things may be easier or harder. Note first we must have \(b \ne 0\) because if not then \(f(X)=X(X^2+a)\) and this is not irreducible. If \(a=0\), then \(f(X)=X^3+b\) and \(f'(X)=3X^2 \ne 0\) because \(\operatorname{char}K \ne 3\). It follows that \((f,f')=0\) because either \(X\) or \(X^2\) divides \(X^3+b\).

Now there only remains the most general case: \(a \ne 0\) and \(b \ne 0\). This is where the Euclidean Algorithm kicks in. Recall that for any three polynomials \(p,q,r\) in \(K[X]\), we have \[ \gcd(p,q)=\gcd(q,p)=\gcd(q,p+rq). \] This is how the Euclidean Algorithm works. Note we can write \[ f(X)=\frac{1}{3}Xf'(X)+\underbrace{\frac{2}{3}aX+b}_{r_0(X)}. \] It follows that \(\gcd(f,f')=\gcd(f',r_0)\). We next work on \(f'\) and \(r_0\). \[ f'(X)=\frac{9}{2a}X\left(\frac{2}{3}aX+b\right)+\underbrace{\left(-\frac{9b}{2a}X+a\right)}_{r_1(X)}. \] However, \(r_0(X)\) and \(r_1(X)\) has common divisor \(0\), which implies that \(f\) and \(f'\) has common divisor \(0\). Whichever the case is, we have \(\gcd(f,f')=0\) and therefore \(f\) is separable. Note the fact that the characteristic of \(K\) is not \(2\) or \(3\) is frequently used here, otherwise there are a lot of equations making no sense.

Where we are at? We want to ensure that \(f\) is separable so that working with the Galois group of \(f\) is not that troublesome. And \(f\) is. We now back to the study of the Galois group \(G=G(L/K)\), where \(L\) is the splitting field of \(f\). Let \(\alpha_1\), \(\alpha_2\), \(\alpha_3\) be the roots of \(f\) and pick one of them as \(\alpha\). We see \([K(\alpha):K]=3\).

Since \(G\) permutes three elements, \(G\) has to be a subgroup of \(\mathfrak{S}_3\). Therefore \(|G|=[L:K] \ge [K(\alpha):K]=3\), which implies that \(|G|=3\) or \(6\). In the first case, \(G=\mathfrak{A}_3\), the alternating group. In the second case, \(G=\mathfrak{S}_3\) and \(K(\alpha)\) is not normal over \(K\) because, there is an irreducible polynomial \(f(X) \in K[X]\) which has a root in \(K(\alpha)\) that does not split into linear factors in \(K(\alpha)\). This is the definition of normal extension.

The question now is, when \(G\) is \(\mathfrak{S}_3\) and when it is \(\mathfrak{A}_3\)? We get a good chance to review finite group theory. This is answered by the sign of elements in \(G\). To be precise, \(G=\mathfrak{S}_3\) if and only if \(G\) has an odd element. If not then \(G=\mathfrak{A}_3\). To work with this, we recall how the sign function work. Put \[ \delta=(\alpha_1-\alpha_2)(\alpha_2-\alpha_3)(\alpha_3-\alpha_1). \] For any \(\sigma \in G\), we have \(\sigma(\delta)=\varepsilon(\sigma)\delta\), where \(\varepsilon(\sigma)\) is the sign of \(\sigma\). If we put \(\Delta=\delta^2\), which is the discriminant, we see \(\sigma(\Delta)=\Delta\). Therefore \(\Delta \in L^G=K\). But wait, since \(\sigma(\delta)=\pm\delta\), the sign is not guaranteed, we see \(\delta\) is not guaranteed to be in \(K\). This is where we crack the problem.

If \(\delta \in K\), or more precisely, \(\sqrt\Delta \in K\), then \(\sigma(\delta)=\delta\) and it follows that \(\varepsilon(\sigma)=1\) for all \(\sigma \in G\). This can only happen if \(G=\mathfrak{A}_3\).

If \(\sqrt\Delta \not\in K\), then \(\delta\) is not fixed by \(G\). There is some \(\sigma \in G\) such that \(\sigma(\delta)=-\delta\), which is to say that \(\varepsilon(\sigma)=-1\). This can only happen when \(G=\mathfrak{S}_3\).

We have the following conclusion.

Proposition 2.Notation being above. Assume that \(f\) is irreducible. Then the Galois group of \(f\) is \(\mathfrak{S}_3\) if and only if \(\sqrt\delta \not\in K\). The group is \(\mathfrak{A}_3\) if and only if \(\sqrt\Delta \in K\).

A dirty calculation shows that \(\Delta=-4a^3-27b^2\). One can show this using Vieta's formulas. You shan't feel this to be strange because in the quadratic case we have \(\Delta=b^2-4ac\) and we did care if \(\Delta>0\), which amounts to whether \(\sqrt\Delta \in \mathbb{R}\).

Let's conclude this post by a handy but nontrivial example. Consider \[ f(X)=X^3-X-1 \] The discriminant is \(-4 \cdot(-1)^3-27 \cdot (-1)^2=-23\), which lies in \(\mathbb{Q}(\sqrt{-23})\) and therefore the Galois group over it is \(\mathfrak{A}_3\). However, when the base field is a subfield, for example, \(\mathbb{Q}\), then the Galois group is \(\mathfrak{S}_3\).

Examples in Galois Theory 2 - Cubic Extensions