# Galois group of a cubic polynomial

###### Definition 1.

If $f(x)\in k[x]$, the Galois group  of $f(x)$ is the Galois group $\operatorname{Gal}(K/k)$ of a splitting field  $K$ of $f(x)$.

###### Theorem 1.

For $f(x)\in k[x]$, the Galois group of $f(x)$ permutes the set of roots of $f(x)$. Therefore, if the roots of $f(x)$ are $\alpha_{1},\ldots,\alpha_{n}\in K$, the Galois group of $f(x)$ is isomorphic    to a subgroup   of $S_{n}$.

Proof. $K=k(\alpha_{1},\ldots,\alpha_{n})$, 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 $\langle(1,2)\rangle$ that consists of a single transposition  , the group $A_{3}=\langle(1,2,3)\rangle$, 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 $\operatorname{Gal}(K/\mathbb{Q})$ is $2$, so $\operatorname{Gal}(K/\mathbb{Q})\cong\langle 1,2\rangle\cong\mathbb{Z}/2% \mathbb{Z}$; 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 $\langle 1,2\rangle\cong\mathbb{Z}/2\mathbb{Z}$.

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 $\operatorname{Gal}(K/\mathbb{Q})$ has at least $3$ elements, since the image of $r_{1}$ may be any of the three roots. Since $\operatorname{Gal}(K/\mathbb{Q})\subset S_{3}$, it follows that $\operatorname{Gal}(K/\mathbb{Q})\cong A_{3}$ or $\operatorname{Gal}(K/\mathbb{Q})\cong S_{3}$ and thus by the fundamental theorem of Galois theory  that $[K:\mathbb{Q}]=3$ or $6$.

Now, the discriminant of $f$ is

 $D=\prod_{i

This is a symmetric polynomial  in the $r_{i}$. The coefficients of $f(x)$ are the elementary symmetric polynomials in the $r_{i}$: if $f(x)=x^{3}+ax^{2}+bx+c$, then

 $\displaystyle c=r_{1}r_{2}r_{3}$ $\displaystyle b=-(r_{1}r_{2}+r_{1}r_{3}+r_{2}r+3)$ $\displaystyle a=r_{1}+r_{2}+r_{3}$

Thus $D$ can be written as a polynomial in the coefficients of $f$, so $D\in\mathbb{Q}$. $D\neq 0$ 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 $f(x)=x^{3}+ax^{2}+bx+c$, then its discriminant $D$ is $18abc+a^{2}b^{2}-4b^{3}-4a^{3}c-27c^{2}$ (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 $D<0$, 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 $\operatorname{Gal}(K/\mathbb{Q})$ must fix $\sqrt{D}$. But a transposition of two roots does not fix $\sqrt{D}$ - for example, the map

 $r_{1}\mapsto r_{2},\qquad r_{2}\mapsto r_{1},\qquad r_{3}\mapsto r_{3}$

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 $\operatorname{Gal}(K/\mathbb{Q})$ 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 $f(x)\in\mathbb{Q}[x]$ 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, $[K:\mathbb{Q}]=3$ and the Galois group is isomorphic to $A_{3}$, otherwise $[K:\mathbb{Q}]=6$ and the Galois group is isomorphic to $S_{3}$.

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.

1. Example 1

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

2. Example 2

$f(x)=x^{3}-x^{2}+x-1$. Again by the rational root test, this polynomial factors as $(x-1)(x^{2}+1)$, 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 $\sqrt{-1}\leftrightarrow-\sqrt{-1}$.

3. 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 $\mathbb{Q}(\sqrt{2},\sqrt{-108})=\mathbb{Q}(\sqrt{2},\sqrt{-3})$. This is in agreement with what we already know, namely that the cube roots of $2$ are

 $\sqrt{2},\omega\sqrt{2},\omega^{2}\sqrt{2}$

where $\omega=\frac{-1+\sqrt{-3}}{2}$ is a primitive cube root of unity.

4. 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 $f(1)<0$).

5. 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

 $r_{1}=2\cos(2\pi/9),r_{2}=2\cos(8\pi/9),r_{3}=2\cos(14\pi/9)$

and we see that

 $\displaystyle\cos(14\pi/9)=\cos(4\pi/9)=2\cos^{2}(2\pi/9)-1$ $\displaystyle\cos(8\pi/9)=2\cos^{2}(4\pi/9)-1$

Let’s consider an automorphism of $K$ sending $r_{1}$ to $r_{3}$, i.e. sending $\cos(2\pi/9)\mapsto\cos(14\pi/9)$. Given the relations above, it is clear that this mapping uniquely determines the image of $r_{3}$ as well, since

 $r_{3}=2\cos^{2}(2\pi/9)-1\mapsto 2\cos^{2}(4\pi/9)-1=r_{2}$
Title Galois group of a cubic polynomial GaloisGroupOfACubicPolynomial 2013-03-22 17:40:33 2013-03-22 17:40:33 rm50 (10146) rm50 (10146) 10 rm50 (10146) Topic msc 12F10 msc 12D10 CardanosFormulae