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 $\mathfrak{S}_3$.