We now restrict our attention to the case $k=\Rats$ . If $f(x)\in\Rats[x]$ 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 $\Rats$ .

If $f(x)$ splits completely in $\Rats$ , then $K=\Rats$ 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 $\Rats$ will have trivial Galois group.

If $f(x)$ factors into a linear and an irreducible quadratic term, then $K=\Rats(\sqrt{D})$ , where $D$ is the discriminant of the quadratic. Hence $[K:\Rats]=2$ and the order of $\Gal(K/\Rats)$ is $2$ , so $\Gal(K/\Rats)\cong \langle1,2\rangle\cong\Ints/2\Ints$ ; 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 $\langle 1,2\rangle\cong \Ints/2\Ints$ .

The Galois groups $A_3$ and $S_3$ arise when considering irreducible cubics. Let $f(x)$ is irreducible with roots $r_1, r_2, r_3$ . Since $f$ is irreducible, the roots are distinct. Thus $\Gal(K/\Rats)$ has at least $3$ elements, since the image of $r_1$ may be any of the three roots. Since $\Gal(K/\Rats)\subset S_3$ , it follows that $\Gal(K/\Rats)\cong A_3$ or $\Gal(K/\Rats)\cong S_3$ and thus by the fundamental theorem of Galois theory that $[K:\Rats]=3$ or $6$ .

Now, the discriminant of $f$ is $$ D=\prod_{i<j} (r_i-r_j)^ $$ 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

If $\sqrt{D}\notin\Rats$ , it follows that $\sqrt{D}$ has degree $2$ over $\Rats$ , so that $[\Rats(r_1,\sqrt{D}):\Rats]=6$ . Hence $K=\Rats(r_1,\sqrt{D})$ , 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 $\sqrt{D}\in\Rats$ , then any element of $\Gal(K/\Rats)$ must fix $\sqrt{D}$ . But a transposition of two roots does not fix $\sqrt{D}$ - for example, the map $$ r_1\mapsto r_2, \qquad r_2\mapsto r_1, \qquad r_3\mapsto r_ $$ 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 $\Gal(K/\Rats)$ does not include transpositions and so it must in this case be isomorphic to $A_3$ . Thus $[K:\Rats]=3$ , so $K=\Rats(r_1)=\Rats(r_1,\sqrt{D})$ since $\sqrt{D}\in\Rats$ . This proves:

Note that one consequence of all of this is that any irreducible cubic with three real roots must have Galois group $A_3$ (and thus any irreducible cubic with Galois group $S_3$ must have two complex roots), but the converse does not hold.

One way 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.

$f(x)=x^3-6x^2+11x-6$ . 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 $\Rats$ . Its Galois group is therefore trivial.

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: $K=\Rats(\sqrt{-1})$ . 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 $\Rats$ is $S_3$ , and the splitting field for $f$ is $\Rats(\sqrt[3]{2},\sqrt{-108})=\Rats(\sqrt[3]{2},\sqrt{-3})$ . 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 $f(1)<0$ ).

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

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 $$ r_3=2\cos^2(2\pi/9)-1\mapsto 2\cos^2(4\pi/9)-1=r_ $$ and thus we see how the relation imposed by the discriminant actually manifests itself in terms of restrictions on the permutation group.