Compute \[ [\QQ(\sqrt2+\sqrt3):\QQ]. \]
Solution
Set \[ \alpha=\sqrt2+\sqrt3. \] Since \[ (\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=1, \] we have \[ \alpha^{-1}=\sqrt3-\sqrt2\in\QQ(\alpha). \] Therefore \[ \sqrt3=\frac{\alpha+\alpha^{-1}}2, \qquad \sqrt2=\frac{\alpha-\alpha^{-1}}2, \] so \[ \QQ(\alpha)=\QQ(\sqrt2,\sqrt3). \]
Now \[ [\QQ(\sqrt2):\QQ]=2, \] and \(\sqrt3\notin\QQ(\sqrt2)\); otherwise writing \(\sqrt3=a+b\sqrt2\) with \(a,b\in\QQ\) and squaring would force \(2ab=0\) and then either \(3=a^2\) or \(3=2b^2\), impossible in \(\QQ\). Hence \[ [\QQ(\sqrt2,\sqrt3):\QQ(\sqrt2)]=2. \] By the tower law, \[ [\QQ(\sqrt2+\sqrt3):\QQ]=4. \]
Indeed, \(\alpha\) satisfies \[ x^4-10x^2+1=0, \] which is therefore its minimal polynomial.