Galois group of a cubic polynomial
Definition 1.
If $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in}k\mathit{}\mathrm{[}x\mathrm{]}$, the Galois group^{} of $f\mathrm{}\mathrm{(}x\mathrm{)}$ is the Galois group $\mathrm{Gal}\mathit{}\mathrm{(}K\mathrm{/}k\mathrm{)}$ of a splitting field^{} $K$ of $f\mathit{}\mathrm{(}x\mathrm{)}$.
Theorem 1.
For $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in}k\mathit{}\mathrm{[}x\mathrm{]}$, the Galois group of $f\mathit{}\mathrm{(}x\mathrm{)}$ permutes the set of roots of $f\mathit{}\mathrm{(}x\mathrm{)}$. Therefore, if the roots of $f\mathit{}\mathrm{(}x\mathrm{)}$ are ${\alpha}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{\alpha}_{n}\mathrm{\in}K$, the Galois group of $f\mathit{}\mathrm{(}x\mathrm{)}$ is isomorphic^{} to a subgroup^{} of ${S}_{n}$.
Proof. $K=k({\alpha}_{1},\mathrm{\dots},{\alpha}_{n})$, so any automorphism^{} $\sigma $ of $K$ fixing $k$ is determined by the image of each ${\alpha}_{i}$. But $\sigma $ must take each ${\alpha}_{i}$ to some ${\alpha}_{j}$ (where possibly $i=j$), since $\sigma $ is a homomorphism^{} of $K$ and thus $f(\sigma ({\alpha}_{i}))=\sigma (f({\alpha}_{i}))=0$. Thus $\sigma $ permutes the roots of $f(x)$ and is determined by the resulting permutation^{}.
We now restrict our attention to the case $k=\mathbb{Q}$. If $f(x)\in \mathbb{Q}[x]$ is a cubic, its Galois group is a subgroup of ${S}_{3}$. We can then use the knowledge of the group structure^{} of ${S}_{3}$ to anticipate the possible Galois groups of a cubic polynomial. There are six subgroups of ${S}_{3}$, and the three subgroups of order $2$ are conjugate^{}. This leaves four essentially different subgroups of ${S}_{3}$: the trivial group, the group $\u27e8(1,2)\u27e9$ that consists of a single transposition^{}, the group ${A}_{3}=\u27e8(1,2,3)\u27e9$, and the full group ${S}_{3}$. All four of these groups can in fact appear as the Galois group of a cubic.
Let $K$ be a splitting field of $f(x)$ over $\mathbb{Q}$.
If $f(x)$ splits completely in $\mathbb{Q}$, then $K=\mathbb{Q}$ and so the Galois group of $f(x)$ is trivial. So any cubic (in fact, a polynomial^{} of any degree) that factors completely into linear factors in $\mathbb{Q}$ will have trivial Galois group.
If $f(x)$ factors into a linear and an irreducible^{} quadratic term, then $K=\mathbb{Q}(\sqrt{D})$, where $D$ is the discriminant^{} of the quadratic. Hence $[K:\mathbb{Q}]=2$ and the order of $\mathrm{Gal}(K/\mathbb{Q})$ is $2$, so $\mathrm{Gal}(K/\mathbb{Q})\cong \u27e81,2\u27e9\cong \mathbb{Z}/2\mathbb{Z}$; the nontrivial element of the Galois group takes each root of the quadratic to its complex conjugate (i.e. it maps $\sqrt{D}\mapsto \sqrt{D}$). Thus any cubic that has exactly one rational root will have Galois group isomorphic to $\u27e81,2\u27e9\cong \mathbb{Z}/2\mathbb{Z}$.
The Galois groups ${A}_{3}$ and ${S}_{3}$ arise when considering irreducible cubics. Let $f(x)$ be irreducible with roots ${r}_{1},{r}_{2},{r}_{3}$. Since $f$ is irreducible, the roots are distinct. Thus $\mathrm{Gal}(K/\mathbb{Q})$ has at least $3$ elements, since the image of ${r}_{1}$ may be any of the three roots. Since $\mathrm{Gal}(K/\mathbb{Q})\subset {S}_{3}$, it follows that $\mathrm{Gal}(K/\mathbb{Q})\cong {A}_{3}$ or $\mathrm{Gal}(K/\mathbb{Q})\cong {S}_{3}$ and thus by the fundamental theorem of Galois theory^{} that $[K:\mathbb{Q}]=3$ or $6$.
Now, the discriminant of $f$ is
$$ 
This is a symmetric polynomial^{} in the ${r}_{i}$. The coefficients of $f(x)$ are the elementary symmetric polynomials in the ${r}_{i}$: if $f(x)={x}^{3}+a{x}^{2}+bx+c$, then
$$c={r}_{1}{r}_{2}{r}_{3}$$  
$$b=({r}_{1}{r}_{2}+{r}_{1}{r}_{3}+{r}_{2}r+3)$$  
$$a={r}_{1}+{r}_{2}+{r}_{3}$$ 
Thus $D$ can be written as a polynomial in the coefficients of $f$, so $D\in \mathbb{Q}$. $D\ne 0$ since $f(x)$ is irreducible and therefore has distinct roots; also clearly $\sqrt{D}\in K$ and thus $\mathbb{Q}({r}_{1},\sqrt{D})\subset K$. If $f(x)={x}^{3}+a{x}^{2}+bx+c$, then its discriminant $D$ is $18abc+{a}^{2}{b}^{2}4{b}^{3}4{a}^{3}c27{c}^{2}$ (see the article on the discriminant for a longer discussion).
If $\sqrt{D}\notin \mathbb{Q}$, it follows that $\sqrt{D}$ has degree $2$ over $\mathbb{Q}$, so that $[\mathbb{Q}({r}_{1},\sqrt{D}):\mathbb{Q}]=6$. Hence $K=\mathbb{Q}({r}_{1},\sqrt{D})$, so we can derive the splitting field for $f$ by adjoining any root of $f$ and the square root of the discriminant. This can happen for either positive or negative $D$, clearly. Note in particular that if $$, then $\sqrt{D}$ is imaginary^{} and thus $K$ is not a real field, so that $f$ has one real and two imaginary roots. So any cubic with only one real root has Galois group ${S}_{3}$.
If $\sqrt{D}\in \mathbb{Q}$, then any element of $\mathrm{Gal}(K/\mathbb{Q})$ must fix $\sqrt{D}$. But a transposition of two roots does not fix $\sqrt{D}$  for example, the map
$${r}_{1}\mapsto {r}_{2},{r}_{2}\mapsto {r}_{1},{r}_{3}\mapsto {r}_{3}$$ 
takes
$$\sqrt{D}=({r}_{1}{r}_{2})({r}_{1}{r}_{3})({r}_{2}{r}_{3})\mapsto ({r}_{2}{r}_{1})({r}_{2}{r}_{3})({r}_{1}{r}_{3})=\sqrt{D}$$ 
Then $\mathrm{Gal}(K/\mathbb{Q})$ does not include transpositions and so it must in this case be isomorphic to ${A}_{3}$. Thus $[K:\mathbb{Q}]=3$, so $K=\mathbb{Q}({r}_{1})=\mathbb{Q}({r}_{1},\sqrt{D})$ since $\sqrt{D}\in \mathbb{Q}$. This proves:
Theorem 2.
Let $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in}\mathrm{Q}\mathit{}\mathrm{[}x\mathrm{]}$ be an irreducible cubic and $K$ its splitting field. Then if $\alpha $ is any root of $f$,
$$K=\mathbb{Q}(\alpha ,\sqrt{D})$$ 
where $D$ is the discriminant of $f$. Thus if $\sqrt{D}$ is rational, $\mathrm{[}K\mathrm{:}\mathrm{Q}\mathrm{]}\mathrm{=}\mathrm{3}$ and the Galois group is isomorphic to ${A}_{\mathrm{3}}$, otherwise $\mathrm{[}K\mathrm{:}\mathrm{Q}\mathrm{]}\mathrm{=}\mathrm{6}$ and the Galois group is isomorphic to ${S}_{\mathrm{3}}$.
Thus if $f$ is an irreducible cubic with Galois group ${A}_{3}$, it must have three real roots. However, the converse^{} does not hold (see Example 4 below).
of looking at the above analysis is that for a “general” polynomial of degree $n$, the Galois group is ${S}_{n}$. If the Galois group of some polynomial is not ${S}_{n}$, there must be algebraic^{} relations^{} among the roots that restrict the available set of permutations. In the case of a cubic whose discriminant is a rational square, this relation is that $\sqrt{D}$, which is a polynomial in the roots, must be preserved.

Example 1
$f(x)={x}^{3}6{x}^{2}+11x6$. By the rational root test, this polynomial has the three rational roots $1,2,3$, so it factors as $f(x)=(x1)(x2)(x3)$ over $\mathbb{Q}$. Its Galois group is therefore trivial.

Example 2
$f(x)={x}^{3}{x}^{2}+x1$. Again by the rational root test, this polynomial factors as $(x1)({x}^{2}+1)$, so its Galois group has two elements, and a splitting field $K$ for $f$ is derived by adjoining the square root of the discriminant of the quadratic: $K=\mathbb{Q}(\sqrt{1})$. The nontrivial element of the Galois group maps $\sqrt{1}\leftrightarrow \sqrt{1}$.

Example 3
$f(x)={x}^{3}2$. This polynomial has discriminant $108=3\cdot {6}^{2}$. This is not a rational square, so the Galois group of $f$ over $\mathbb{Q}$ is ${S}_{3}$, and the splitting field for $f$ is $\mathbb{Q}(\sqrt[3]{2},\sqrt{108})=\mathbb{Q}(\sqrt[3]{2},\sqrt{3})$. This is in agreement with what we already know, namely that the cube roots of $2$ are
$$\sqrt[3]{2},\omega \sqrt[3]{2},{\omega}^{2}\sqrt[3]{2}$$ where $\omega =\frac{1+\sqrt{3}}{2}$ is a primitive cube root of unity.

Example 4
$f(x)={x}^{3}4x+2$. This is irreducible since it is Eisenstein at $2$ (or by the rational root test), and its discriminant is $202$, which is not a rational square. Thus the Galois group for this polynomial is also ${S}_{3}$; note, however, that $f$ has three real roots (since $f(0)>0$ but $$).

Example 5
$f(x)={x}^{3}3x+1$. This is also irreducible by the rational root test. Its discriminant is $81$, which is a rational square, so the Galois group for this polynomial is ${A}_{3}$. Explicitly, the roots of $f(x)$ are
$${r}_{1}=2\mathrm{cos}(2\pi /9),{r}_{2}=2\mathrm{cos}(8\pi /9),{r}_{3}=2\mathrm{cos}(14\pi /9)$$ and we see that
$$\mathrm{cos}(14\pi /9)=\mathrm{cos}(4\pi /9)=2{\mathrm{cos}}^{2}(2\pi /9)1$$ $$\mathrm{cos}(8\pi /9)=2{\mathrm{cos}}^{2}(4\pi /9)1$$ Let’s consider an automorphism of $K$ sending ${r}_{1}$ to ${r}_{3}$, i.e. sending $\mathrm{cos}(2\pi /9)\mapsto \mathrm{cos}(14\pi /9)$. Given the relations above, it is clear that this mapping uniquely determines the image of ${r}_{3}$ as well, since
$${r}_{3}=2{\mathrm{cos}}^{2}(2\pi /9)1\mapsto 2{\mathrm{cos}}^{2}(4\pi /9)1={r}_{2}$$ and thus we see how the relation imposed by the discriminant actually manifests itself in terms of restrictions^{} on the permutation group^{}.
Title  Galois group of a cubic polynomial 

Canonical name  GaloisGroupOfACubicPolynomial 
Date of creation  20130322 17:40:33 
Last modified on  20130322 17:40:33 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  10 
Author  rm50 (10146) 
Entry type  Topic 
Classification  msc 12F10 
Classification  msc 12D10 
Related topic  CardanosFormulae 