Exercise 6.1.6

Proof. $L=\mathbb{Q}(\sqrt{6},\sqrt{10},\sqrt{15})$.

$\sqrt{15}=\sqrt{3\cdot 5}=3\frac{\sqrt{10}}{\sqrt{6}}\in \mathbb{Q}(\sqrt{6},\sqrt{10})$, therefore

Then Exercise 1 shows that

Note : moreover, $x^2-10$ is irreducible over $\mathbb{Q}(\sqrt{6})$, otherwise the roots $\pm \sqrt{10}$ of $f$ would be in $\mathbb{Q}(\sqrt{6})$, and then

By squaring, we obtain $10=a^2+6b^2+2\mathit{ab}\sqrt{6}$. The irrationality of $\sqrt{6}$ shows that $\mathit{ab}=0,a^2+6b^2=10$. Since $\sqrt{10}$ and $\sqrt{\frac{5}{3}}$ are irrational, this system has no solution in $\mathbb{Q}\times \mathbb{Q}$.

$x^2-10$ is irreducible over $\mathbb{Q}(\sqrt{6})$, thus

Using section 6.2, as $L$ is the splitting field of the separable polynomial $(x^2-6)(x^2-10)$ over $\mathbb{Q}$, we obtain

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