Show that \([\QQ(\sqrt 2 + \sqrt 3) : \QQ] = 4\).
- Show that \(\QQ(\sqrt 2 + \sqrt 3) = \QQ(\sqrt 2 - \sqrt 3) = \QQ(\sqrt 2, \sqrt 3)\).
Solution
Set \[ \alpha=\sqrt2+\sqrt3. \]
1. One has\[ \alpha^{-1}=\sqrt3-\sqrt2. \]
Indeed, \[ (\sqrt2+\sqrt3)(\sqrt3-\sqrt2)=3-2=1. \] Thus \(\alpha^{-1}\in\QQ(\alpha)\).
2. Both \(\sqrt2\) and \(\sqrt3\) lie in \(\QQ(\alpha)\).
Using 1, \[ \alpha-\alpha^{-1}=2\sqrt2, \qquad \alpha+\alpha^{-1}=2\sqrt3. \] Hence \[ \sqrt2=\frac{\alpha-\alpha^{-1}}2, \qquad \sqrt3=\frac{\alpha+\alpha^{-1}}2 \] belong to \(\QQ(\alpha)\).
3. Therefore\[ \QQ(\sqrt2+\sqrt3)=\QQ(\sqrt2,\sqrt3). \]
The left-hand field is contained in the right because \(\alpha=\sqrt2+\sqrt3\). The reverse inclusion follows from 2.
4. Also\[ \QQ(\sqrt2-\sqrt3)=\QQ(\sqrt2,\sqrt3). \]
The element \(\sqrt2-\sqrt3=-\alpha^{-1}\) lies in \(\QQ(\alpha)\) by 1, so \[ \QQ(\sqrt2-\sqrt3)\subseteq\QQ(\alpha). \] Conversely, if \(\beta=\sqrt2-\sqrt3\), then \[ \beta^{-1}=-(\sqrt2+\sqrt3)=-\alpha, \] so \(\alpha\in\QQ(\beta)\). Hence the two simple extensions coincide, and 3 identifies them with \(\QQ(\sqrt2,\sqrt3)\).
5. The degree of this field over \(\QQ\) is \(4\).
The extension \(\QQ(\sqrt2)/\QQ\) has degree \(2\). It remains to show \(\sqrt3\notin\QQ(\sqrt2)\). If \[ \sqrt3=a+b\sqrt2 \qquad(a,b\in\QQ), \] then squaring gives \[ 3=a^2+2b^2+2ab\sqrt2. \] Thus \(2ab=0\). If \(b=0\), then \(a^2=3\), impossible in \(\QQ\); if \(a=0\), then \(2b^2=3\), also impossible in \(\QQ\). Hence \[ [\QQ(\sqrt2,\sqrt3):\QQ(\sqrt2)]=2. \] By the tower law, \[ [\QQ(\sqrt2,\sqrt3):\QQ]=2\cdot2=4. \] Using 3 proves \[ [\QQ(\sqrt2+\sqrt3):\QQ]=4. \]