Exercise 6.1.2

Proof. As $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the splitting field of $x^2-3$ over $\mathbb{Q}(\sqrt{2})$, and as $x^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$ (see Exercise 5.1.13), there exists by Proposition 5.1.8 a field isomorphism $\sigma :L\to L$ which is identity on $\mathbb{Q}(\sqrt{2})$ and which takes $\sqrt{3}$ on $-\sqrt{3}$. As $\sigma$ is identity on $\mathbb{Q}(\sqrt{2})$, we have also $\sigma (\sqrt{2})=\sqrt{2}$. As the restriction of $\sigma$ to $\mathbb{Q}(\sqrt{2})$ is identity, the restriction of $\sigma$ to $\mathbb{Q}$ is the identity on $\mathbb{Q}$, so $\sigma \in \mathrm{Gal}(L∕\mathbb{Q})$.

Similarly $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the splitting field of $x^2-2$ over $\mathbb{Q}(\sqrt{3})$, and $x^2-2$ is irreducible over $\mathbb{Q}(\sqrt{3})$ by the Reciprocity Theorem (see Exercise 4.3.6), so there exists by Proposition 5.1.8 a field isomorphism $\tau :L\to L$ which is identity on $\mathbb{Q}(\sqrt{3})$ and which takes $\sqrt{2}$ on $-\sqrt{2}$. As $\tau$ is identity on $\mathbb{Q}(\sqrt{3})$, we have also $\sigma (\sqrt{3})=\sqrt{3}$, and $\tau \in \mathrm{Gal}(L∕\mathbb{Q})$.

Moreover $1_L(\sqrt{2})=\sqrt{2},1_L(\sqrt{3})=\sqrt{3}$, with $1_L\in \mathrm{Gal}(L∕\mathbb{Q})$.

Finally $\mathit{\sigma \tau }=\sigma \circ \tau \in \mathrm{Gal}(L∕\mathbb{Q})$ satisfies $(\mathit{\sigma \tau })(\sqrt{2})=-\sqrt{2},(\mathit{\sigma \tau })(\sqrt{3})=-\sqrt{3}$.

All possibilities in Example 6.1.10 can occur. Consequently $|\mathrm{Gal}(L∕\mathbb{Q})|\ge 4$. As it is proved in Example 6.1.10 that $|\mathrm{Gal}(L∕\mathbb{Q})|\le 4$, then $|\mathrm{Gal}(L∕\mathbb{Q})|=4$, and

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