Find the minimal polynomial of \(\sqrt2+\sqrt3\) over \(\mathbb Q\). Justify your answer.
Solution
1. Let\[ \alpha=\sqrt2+\sqrt3. \]Then \(\alpha\) is a root of\[ f(x)=x^4-10x^2+1. \]
We have \[ \alpha^2=5+2\sqrt6, \] so \[ \alpha^2-5=2\sqrt6. \] Squaring gives \[ (\alpha^2-5)^2=24, \] hence \[ \alpha^4-10\alpha^2+1=0. \]
2. The field \(\mathbb Q(\alpha)\) contains both \(\sqrt2\) and \(\sqrt3\).
Since \[ (\sqrt2+\sqrt3)(\sqrt3-\sqrt2)=1, \] we have \[ \alpha^{-1}=\sqrt3-\sqrt2\in\mathbb Q(\alpha). \] Therefore \[ \sqrt3=\frac{\alpha+\alpha^{-1}}2, \qquad \sqrt2=\frac{\alpha-\alpha^{-1}}2. \] Thus \(\mathbb Q(\sqrt2,\sqrt3)\subseteq\mathbb Q(\alpha)\), while the reverse inclusion is immediate from the definition of \(\alpha\).
3. We have\[ [\mathbb Q(\alpha):\mathbb Q]=4. \]
By 2, \[ \mathbb Q(\alpha)=\mathbb Q(\sqrt2,\sqrt3). \] Now \([\mathbb Q(\sqrt2):\mathbb Q]=2\), and \(\sqrt3\notin\mathbb Q(\sqrt2)\). Indeed, if \(\sqrt3=a+b\sqrt2\) with \(a,b\in\mathbb Q\), squaring gives \[ 3=a^2+2b^2+2ab\sqrt2, \] so \(ab=0\); either case is impossible. Hence adjoining \(\sqrt3\) gives another quadratic extension, and the total degree is \(4\).
4. Therefore the minimal polynomial of \(\sqrt2+\sqrt3\) over \(\mathbb Q\) is\[ \boxed{x^4-10x^2+1}. \]
The polynomial in 1 is monic of degree \(4\) and annihilates \(\alpha\). By 3, the minimal polynomial of \(\alpha\) has degree \(4\), so it must equal this polynomial.