Prove that the Galois group of an irreducible cubic polynomial over $\mathbb{Q}$ is isomorphic to $S_3$ or $\mathbb{Z}_3$

Question

I know that in order to show that the Galois group of this polynomial is isomorphic to $\mathbb{Z}_3$ we must show that, if $\alpha$ is a root of the polynomial, then $[\mathbb{Q}(\alpha):\mathbb{Q}]=|G(\mathbb{Q}(\alpha):\mathbb{Q})|=3$ , but how can I show that and also how can I show that it can also be $S_3$? Really don't know where to start, but I am thinking that if $\alpha \in \mathbb{C}$ then $G(\mathbb{Q}(\alpha)/\mathbb{Q})$ must have $\{id,\sigma,\sigma^2\}$ where $\sigma$ is complex conjugation. But what about, $S_3$?

Answer 1

Here is the key fact:

The splitting field of an irreducible polynomial of degree $n$ has degree a multiple of $n$ and is at most $n!$. Its Galois group is a subgroup of $S_n$.

Therefore, the splitting field of an irreducible cubic has degree at least $3$ and at most $6$.

Thus, the Galois group of an irreducible cubic is $C_3$, $C_6$, or $S_3$.

It cannot be $C_6$ because $C_6$ is not a subgroup of $S_3$.

Answer 2

Let $f\in\mathbb{Q}[X]$ be an irreducible cubic polynomial with roots $\alpha_1,\alpha_2,\alpha_3$ in $\mathbb{C}$. We set $F = \mathbb{Q}(\omega)$, where $\omega = e^{2\pi i/3}$, and let $E = F(\alpha_1,\alpha_2, \alpha_3)$ be the splitting field over $F$ of $f$ in $\mathbb{C}$. We regard the Galois group $G = \operatorname{Gal}(E/F)$ as a subgroup of the symmetric group $S_3$ by the permuting of roots, of which there are $3!=6$ possible ways of doing this. We also have $3\mid |G|$, so the possible orders of $G$ are $3$, and $6$.

For order $3$ we note the alternating group $A_n$ is the commutator subgroup of $S_n$ with index $2$ having $n!/2$ elements. Now $A_n$ is abelian if and only if $n \le 3$, and so $|A_3|=3!/2=3$ which implies $A_3\cong \mathbb{Z}/(3)$. Hence $G\cong A_3$ if $|G|=3$.

For order $6$ we have the two groups $\mathbb{Z}/(6)$ or $S_3$ to consider. Since $\mathbb{Z}/(6)\nsubseteq S_3$, we have $G\cong S_3$ if $|G|=6$.

Note that $f$ is irreducible in $\mathbb{Q}[X]$ if and only if $G=\operatorname{Gal}(E/F)$ is a transitive subgroup of $S_3$. The only transitive subgroups of $S_3$ are $S_3$ and $A_3$. To find which it is we can look at the discriminant of $f$ in $\mathbb{Q}$. If $\operatorname{Disc} f=\square$ then $G\cong A_3$, and if $\operatorname{Disc} f\neq\square$ then $G\cong S_3$.

Hence an irreducible cubic has Galois group $G\cong A_3$ or $G\cong S_3$.