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

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 $f(X)=X^3+a{X}^2+bX+c$ be a polynomial, then

and as a result $a{X}^2$ 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

In fact, for all $g(X)=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0$, 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

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 $\mathrm{char}K=0,5,7,\dots$, 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 $gcd(f,f^{\prime })=0$. The derivative of $f$, which should be simplified because $f$ has been, is given by

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 $f^{\prime }(X)=3X^2\ne 0$ because $\mathrm{char}K\ne 3$. It follows that $(f,f^{\prime })=0$ 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

This is how the Euclidean Algorithm works. Note we can write

It follows that $gcd(f,f^{\prime })=gcd(f^{\prime },r_0)$. We next work on $f^{\prime }$ and $r_0$.

However, $r_0(X)$ and $r_1(X)$ has common divisor $0$, which implies that $f$ and $f^{\prime }$ has common divisor $0$. Whichever the case is, we have $gcd(f,f^{\prime })=0$ 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

For any $\sigma \in G$, we have $\sigma (\delta )=\epsilon (\sigma )\delta$, where $\epsilon (\sigma )$ is the sign of $\sigma$. If we put $\mathrm{\Delta }={\delta }^2$, which is the discriminant, we see $\sigma (\mathrm{\Delta })=\mathrm{\Delta }$. Therefore $\mathrm{\Delta }\in L^G=K$. 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, $\sqrt{\mathrm{\Delta }}\in K$, then $\sigma (\delta )=\delta$ and it follows that $\epsilon (\sigma )=1$ for all $\sigma \in G$. This can only happen if $G={\mathfrak{A}}_3$.

If $\sqrt{\mathrm{\Delta }}\notin K$, then $\delta$ is not fixed by $G$. There is some $\sigma \in G$ such that $\sigma (\delta )=-\delta$, which is to say that $\epsilon (\sigma )=-1$. 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 $\sqrt{\delta }\notin K$. The group is ${\mathfrak{A}}_3$ if and only if $\sqrt{\mathrm{\Delta }}\in K$.

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

Let’s conclude this post by a handy but nontrivial example. Consider

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 ${\mathfrak{S}}_3$.