Determine the minimal polynomial over $\mathbb{Q}$ for the element $\sqrt{2} + \sqrt{5}$.
$\textit{Solution}$ : We wish to find a monic, irreducible polynomial of minimal degree denoted $m_{\alpha,\mathbb{Q}}$ s.t. $m_{\alpha,\mathbb{Q}}(\alpha)=0$ where $\alpha \in{\mathbb{Q}[\sqrt{2}+\sqrt{5}]}$.
$\mathbb{Q}[\sqrt{2}+\sqrt{5}]$ will be the splitting field for $m_{\alpha,\mathbb{Q}}$.
Label $\alpha = \sqrt{2} + \sqrt{5}$. Then,
$$\alpha-\sqrt{2} = \sqrt{5} $$
Squaring both sides,
$$\alpha^2-2\sqrt{2}\alpha+2=5$$
$$\alpha^2-3=2\sqrt{2}\alpha$$
Squaring both sides again,
$$\alpha^4-6\alpha^2+9=8\alpha^2$$
$$\alpha^4-14\alpha^2+9=0$$
...Is the minimal polynomial for the element $\sqrt{2} + \sqrt{5}$ over $\mathbb{Q}$. Note that it is irreducible over $\mathbb{Q}[x]$ by the rational roots theorem.
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