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

Let \(k\) be an arbitrary field andsuppose \(f(X) \in k[X]\) is separableand, i.e., \(f\) has no multiple rootsin an algebraic closure, and of degree \(\ge1\). 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 Galoisgroup, we see \(\sigma(x_i)\) is stilla root of \(f\) (consider the map \(\tilde\sigma:L[X] \to L[X]\) induced by\(\sigma\) naturally; it is theidentity 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 generatedby complex conjugation \(\sigma:a+bi \mapstoa-bi\). 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 extremelycomplicated, but for now we assume that they are well-known. Thequestion is, what subgroup is \(G\)inside \(\mathfrak{S}_n\). Let's take alook 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 \(aX^2\) iscancelled. A translation does not change any property of a polynomialexcept the value of its roots. Therefore we can reduce our study topolynomials 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 cancancel out \(a_{n-1}X^{n-1}\) by asubstitution \(Y=X-\frac{a_{n-1}}{n}\).

Irreducibility

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

\(f\) is separable if and only if\(\gcd(f,f')=0\). The derivative of\(f\), which should be simplifiedbecause \(f\) has been, is given by\[f'(X)=3X^2+a.\] It is not equal to \(a\)because the characteristic of \(K\) isnot \(3\). We will show carefully that\(f(X)\) is separable by working onthese 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 followsthat \((f,f')=0\) 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. Recallthat 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 nextwork 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 thecharacteristic of \(K\) is not \(2\) or \(3\) is frequently used here, otherwisethere are a lot of equations making no sense.

Where we are at? We want to ensure that \(f\) is separable so that working with theGalois group of \(f\) is not thattroublesome. And \(f\) is. We now backto 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 \(|G|=[L:K] \ge [K(\alpha):K]=3\), whichimplies 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 irreduciblepolynomial \(f(X) \in K[X]\) which hasa root in \(K(\alpha)\) that does notsplit into linear factors in \(K(\alpha)\). This is the definition ofnormal extension.

The question now is, when \(G\) is\(\mathfrak{S}_3\) and when it is \(\mathfrak{A}_3\)? We get a good chance toreview 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, werecall how the sign function work. Put \[\delta=(\alpha_1-\alpha_2)(\alpha_2-\alpha_3)(\alpha_3-\alpha_1).\] For any \(\sigma \in G\), wehave \(\sigma(\delta)=\varepsilon(\sigma)\delta\),where \(\varepsilon(\sigma)\) is thesign of \(\sigma\). If we put \(\Delta=\delta^2\), which is thediscriminant, we see \(\sigma(\Delta)=\Delta\). Therefore \(\Delta \in L^G=K\). But wait, since \(\sigma(\delta)=\pm\delta\), the sign is notguaranteed, we see \(\delta\) is notguaranteed to be in \(K\). This iswhere we crack the problem.

If \(\delta \in K\), or moreprecisely, \(\sqrt\Delta \in K\), then\(\sigma(\delta)=\delta\) and itfollows that \(\varepsilon(\sigma)=1\)for all \(\sigma \in G\). This can onlyhappen 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 saythat \(\varepsilon(\sigma)=-1\). Thiscan only happen when \(G=\mathfrak{S}_3\).

We have the following conclusion.

Proposition 2. Notation being above. Assume that\(f\) is irreducible. Then the Galoisgroup 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 thisusing Vieta'sformulas. You shan't feel this to be strange because in thequadratic 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 theGalois group over it is \(\mathfrak{A}_3\). However, when the basefield is a subfield, for example, \(\mathbb{Q}\), then the Galois group is\(\mathfrak{S}_3\).