- Prove that \[ \mathbb Q(\sqrt2+\sqrt3)=\mathbb Q(\sqrt2-\sqrt3)=\mathbb Q(\sqrt2,\sqrt3). \]
- Prove that this extension has degree \(4\) over \(\mathbb Q\).
Solution
Set \[ \alpha=\sqrt2+\sqrt3. \] Since \[ \alpha^{-1}=\sqrt3-\sqrt2, \] we have \[ \sqrt2-\sqrt3=-\alpha^{-1}\in\mathbb Q(\alpha). \] Therefore \[ \sqrt2=\frac{\alpha-\alpha^{-1}}2, \qquad \sqrt3=\frac{\alpha+\alpha^{-1}}2 \] belong to \(\mathbb Q(\alpha)\). The reverse inclusion is immediate, so \[ \mathbb Q(\alpha)=\mathbb Q(\sqrt2,\sqrt3)=\mathbb Q(\sqrt2-\sqrt3). \]
Now \([\mathbb Q(\sqrt2):\mathbb Q]=2\). Also \(\sqrt3\notin\mathbb Q(\sqrt2)\): if \[ \sqrt3=a+b\sqrt2\qquad(a,b\in\mathbb Q), \] then squaring gives \[ 3=a^2+2b^2+2ab\sqrt2. \] If \(ab\ne0\), this would make \(\sqrt2\) rational; if \(a=0\) or \(b=0\), it would force respectively \(b^2=3/2\) or \(a^2=3\), impossible in \(\mathbb Q\). Hence \[ [\mathbb Q(\sqrt2,\sqrt3):\mathbb Q(\sqrt2)]=2. \] By the tower law, \[ [\mathbb Q(\alpha):\mathbb Q]=4. \] Equivalently, \(\alpha\) has minimal polynomial \[ x^4-10x^2+1. \]