The Galois Group Q(√2 + √3) over Q

The Galois Group Q(√2 + √3) 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)
\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 $\mathrm{Gal} (\sqrt{2} + \sqrt{3})$ over $\mathbb{Q}$. First observe that:

(2)
\begin{align} \quad Q(\sqrt{2} + \sqrt{3}) \cong Q(\sqrt{2}, \sqrt{3}) \end{align}

Observe that $\pm\sqrt{2}$ are the roots of the polynomial $x^2 - 2$ and $\pm\sqrt{3}$ are the roots of the polynomial $x^2 - 3$. So if $\phi \in \mathrm{Gal} (\mathbb{Q}(\sqrt{2} + \sqrt{3}) / \mathbb{Q})$ then we must have that:

(3)
\begin{align} \quad \phi(\sqrt{2}) &= \pm \sqrt{2} \\ \quad \phi(\sqrt{3}) &= \pm \sqrt{3} \end{align}

Now a basis for $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is:

(4)
\begin{align} \quad \{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6} \} \end{align}

Therefore, every element of $x \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ as a vector space is of the form:

(5)
\begin{align} \quad x = a + b\sqrt{2} + c\sqrt{3} + d \sqrt{6} \end{align}

Where $a, b, c, d \in \mathbb{Q}$. Therefore, if $\phi \in \mathrm{Gal} (\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q})$ then:

(6)
\begin{align} \quad \phi(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) &= a + b \phi(\sqrt{2}) + c\phi(\sqrt{3}) + c \phi(\sqrt{2} \sqrt{3}) \\ &=a + b \phi(\sqrt{2}) + c\phi(\sqrt{3}) + c \phi(\sqrt{2}) \phi(\sqrt{3}) \end{align}

There are four automorphisms $\phi_1, \phi_2, \phi_3, \phi_4 : \mathbb{Q}(\sqrt{2}, \sqrt{3}) \to \mathbb{Q}(\sqrt{2}, \sqrt{3})$ in $\mathrm{Gal} (\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q})$ depending on the values of $\phi(\sqrt{2})$ and $\phi(\sqrt{3})$. They are given by:

(7)
\begin{align} \quad \phi_1(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a + b\sqrt{2} + c \sqrt{3} + d\sqrt{6} \\ \quad \phi_2(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a - b\sqrt{2} + c \sqrt{3} - d\sqrt{6} \\ \quad \phi_3(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a + b\sqrt{2} - c \sqrt{3} - d\sqrt{6} \\ \quad \phi_4(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a - b\sqrt{2} - c \sqrt{3} + d\sqrt{6} \\ \end{align}

Now note that there are only two groups of order $4$, namely $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$. It is easy to see that for each $1 \leq i \leq 4$ we have that:

(8)
\begin{align} \quad \phi_i^2(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) = a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \end{align}

Therefore, $\phi_i^2$ gives the identity for each $1 \leq i \leq 4$. Since there are no elements in $\mathrm{Gal} (\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q})$ of order $4$ we conclude that:

(9)
\begin{align} \quad \mathrm{Gal} (\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \end{align}
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