The Galois Group Q 2 Over Q

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:

\begin{align} \quad \mathrm{Gal} (F/K) = \{ \phi \in \mathrm{Aut}(F) : \phi(k) = k, \: \forall k \in K \} \end{align}

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:

\begin{align} \quad r_1 = \sqrt[3]{2}, \quad r_2 = \left (\frac{-1 + \sqrt{3}i}{2} \right ) \sqrt[3]{2}, \quad r_3 = \left (\frac{-1 + \sqrt{3}i}{2} \right)^2 \sqrt[3]{2} \end{align}

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:

\begin{align} \quad \mathrm{Gal} (\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}) \cong \{ 1 \} \end{align}
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