What is the Galois group of the polynomial $f(x)=x^3-3$ over $\mathbb{Q}$?

Question

What is the Galois group of the polynomial $f(x)=x^3-3$ over $\mathbb{Q}$ ?

$f(x)=0 \ $ gives $x= 3^{1/3},~ \zeta_3 3^{1/3},~ \zeta_3^2 3^{1/3}$ over $\mathbb{C}$.

Thus the permutation of these $3$ roots produces the Galois group $S_3$.

Thus the Galois group $G \cong S_3$.

Am I right ?

If I replace the polynomial by $f(x)=x^3-\frac{1}{3}$, then still we get the same Galois group $S_3$.

Am I right ?

Answer 1

the Galois group of $x^3-3$ is the group of automorphisms of the splitting field of $x^3-3$ which leave the base field $\mathbb Q$ fixed.

You have found the 3 roots: $\alpha_1 = 3^{1/3}$, $\alpha_2 = \zeta_3 3^{1/3}$, $\alpha_3 = \zeta_3^2 3^{1/3}$. The splitting field of the polynomial is then $\mathbb Q(\alpha_1, \alpha_2, \alpha_3)$.

An automorphism of $\mathbb Q(\alpha_1, \alpha_2, \alpha_3)$ is an isomorphism from that field to itself and it needs to leave the base field $\mathbb Q$ alone. If $f$ is one of these automorphisms then $\sigma(\alpha_1)$ must be $\alpha_1$, $\alpha_2$ or $\alpha_3$ because $\alpha_1^3-3 = 0$ implies $\sigma(\alpha_1)^3-3 = 0$. So every field automorphisms is a to permutations of the roots.

Notice that $\alpha_3 = \alpha_2^2/\alpha_1$ so $\sigma(\alpha_1)$ and $\sigma(\alpha_2)$ determines $\sigma(\alpha_3)$. This is not a restriction on the permutations we can choose though.

Every (all 6) permutation of the three roots is a field automorphism in this case and that means the Galois group is $S_3$, but this isn't always true.

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In the example Jyrki Lahtonen mentioned in comments: $x^3 - 3x + 1$ the Galois group is $C_3$. Let the roots of the polynomial be:

• $\beta_1 = $ -1.8793852415718167681082185546494629399
• $\beta_2 = $ 1.5320888862379560704047853011108333479
• $\beta_3 = $ 0.3472963553338606977034332535386295920

Notice that:

• $1-1/\beta_1 = \beta_2$
• $1-1/\beta_2 = \beta_3$
• $1-1/\beta_3 = \beta_1$

So if $f$ is an automorphism of the field $\mathbb Q(\beta_1, \beta_2, \beta_3)$ then $f(\beta_1)$ determines $f(\beta_2)$ and $f(\beta_3)$.

This means we can only choose 3 automorphisms and the Galois group is $C_3$.