Let \(F\) be a field and let \(K\) be a field containing \(F\). Suppose that \(\alpha, \beta \in K\) are both algebraic over \(F\). Prove that \(\alpha + \beta\) is algebraic over \(F\).
What is the minimal polynomial of \(\sqrt{5} + 2\sqrt{7}\) over \(\mathbb{Q}\)?
Solution
1. In part (a), \(\alpha+\beta\) is algebraic over \(F\).
Since \(\alpha\) is algebraic over \(F\), \([F(\alpha):F]<\infty\). Since \(\beta\) is algebraic over \(F\subseteq F(\alpha)\), \([F(\alpha,\beta):F(\alpha)]<\infty\). By the tower law, \([F(\alpha,\beta):F]=[F(\alpha,\beta):F(\alpha)]\,[F(\alpha):F]<\infty\). Every element of a finite extension is algebraic [@DF04], and \(\alpha+\beta\in F(\alpha,\beta)\).
Let \(\gamma=\sqrt5+2\sqrt7\).
2. \(\gamma\) is a root of \(f(x)=x^4-66x^2+529\).
\(\gamma^2=5+4\sqrt{35}+28=33+4\sqrt{35}\), so \((\gamma^2-33)^2=16\cdot35=560\), which expands to \(\gamma^4-66\gamma^2+529=0\).
3. \(\QQ(\gamma)=\QQ(\sqrt5,\sqrt7)\), a field of degree \(4\) over \(\QQ\).
The field \(\QQ(\gamma)\) contains \(\sqrt{35}=(\gamma^2-33)/4\) and hence \(\sqrt{35}\,\gamma=14\sqrt5+5\sqrt7\). The linear system \(\sqrt5+2\sqrt7=\gamma\), \(14\sqrt5+5\sqrt7=\sqrt{35}\,\gamma\) has determinant \(5-28=-23\ne0\), so \(\sqrt5,\sqrt7\in\QQ(\gamma)\); the reverse inclusion is clear. Finally \([\QQ(\sqrt5):\QQ]=2\), and \(\sqrt7\notin\QQ(\sqrt5)\): if \((a+b\sqrt5)^2=7\) with \(a,b\in\QQ\), then \(ab=0\), and neither \(a^2=7\) nor \(5b^2=7\) has a rational solution. Thus \([\QQ(\sqrt5,\sqrt7):\QQ]=4\).
4. The minimal polynomial of \(\gamma\) over \(\QQ\) is \(\boxed{x^4-66x^2+529}\).
By step 3 the minimal polynomial has degree \(4\), and by step 2 it divides the monic quartic \(f\), so it equals \(f\).
5. Q.E.D.
Step 1 answers part (a) and step 4 answers part (b).