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 $⟨(1,2)⟩$ that consists of a single transposition, the group $A_3=⟨(1,2,3)⟩$, 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 $\mathrm{Gal}(K/\mathbb{Q})\cong ⟨1,2⟩\cong \mathbb{Z}/2\mathbb{Z}$; 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 $⟨1,2⟩\cong \mathbb{Z}/2\mathbb{Z}$.

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 $18abc+a^2{b}^2-4b^3-4a^{3}c-27c^2$ (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 , 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.

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. $f(x)=x^3-x^2+x-1$. 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. $f(x)=x^3-2$. 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.

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 ).

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

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

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

   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

   $$r_3=2{\cos }^2(2\pi /9)-1\mapsto 2{\cos }^2(4\pi /9)-1=r_2$$

   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	2013-03-22 17:40:33
Last modified on	2013-03-22 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