# The Galois Group Q(∛2) over Q

Recall from the Galois Groups page that if $K$ is a field and $F$ is a field extension of $K$ then the Galois group of $F$ over $K$ is:

(1)On the Automorphisms in the Galois Group of f(x) over K Permute the Roots of f page we proved that if $K$ is a field, $F$ is a field extension of $K$, and $f(x) \in K$ then every $\phi \in \mathrm{Gal} (F/K)$ permutes the roots of $f$ in $F$.

We will now look at the Galois group $\mathrm{Gal} (\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q})$.

Observe that this Galois group is the same as the Galois group of $f(x) = x^3 - 2$ over $\mathbb{Q}$. We compute the roots of $f$. There are three roots, $r_1, r_2, r_3$ which are:

(2)By the theorem references above, every automorphism $\phi \in \mathrm{Gal} (\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q})$ must permute the roots of $f$ in $\mathbb{Q}(\sqrt[3]{2})$. But the only root of $f$ in $\mathbb{Q}(\sqrt[3]{2})$ is $r_1$.

So $\mathrm{Gal} (\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q})$ contains only one element, namely the identity automorphism, and so $\mathrm{Gal} (\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q})$ is a trivial group, that is:

(3)