Let \(E/F\) be a field extension with only finitely many intermediate fields. Prove that \(E/F\) is simple.
Solution
1. The extension \(E/F\) is finite algebraic.
If \(\alpha\in E\) were transcendental over \(F\), then \[ F(\alpha)\supsetneq F(\alpha^2)\supsetneq F(\alpha^4)\supsetneq\cdots \] would give infinitely many intermediate fields, contradiction. Hence \(E/F\) is algebraic.
If \(E/F\) were not finitely generated, start with \(F_0=F\). Whenever \(F_i\ne E\), choose \(\alpha_i\in E\setminus F_i\) and put \[ F_{i+1}=F_i(\alpha_i). \] This produces infinitely many distinct intermediate fields, again a contradiction. Therefore \(E/F\) is finitely generated and algebraic, hence finite.
2. The finite extension is simple.
If \(F\) is finite, then \(E\) is a finite field, so \(E^\times\) is cyclic. A generator of \(E^\times\) generates \(E\) as a field over \(F\).
Now suppose \(F\) is infinite. The proper intermediate fields are proper \(F\)-vector subspaces of the finite-dimensional vector space \(E\). By hypothesis there are only finitely many of them, and a finite-dimensional vector space over an infinite field cannot be the union of finitely many proper subspaces. Choose \[ \alpha\in E \] outside the union of all proper intermediate fields. Then \(F(\alpha)\) is an intermediate field. It cannot be proper, so \[ E=F(\alpha). \]