Let \(K\subseteq E\subseteq F\). If \(F/K\) is algebraic, normal, or separable, respectively, must \(F/E\) have the same property? Answer this question in each of the three cases.
Solution
All three answers are yes.
1. Algebraicity passes from \(F/K\) to \(F/E\).
Let \(\alpha\in F\). Since \(F/K\) is algebraic, some nonzero \(f\in K[x]\) satisfies \(f(\alpha)=0\). Because \(K\subseteq E\), the same polynomial lies in \(E[x]\). Hence \(\alpha\) is algebraic over \(E\).
2. Separability passes from \(F/K\) to \(F/E\).
Let \(p_K\) and \(p_E\) be the minimal polynomials of \(\alpha\in F\) over \(K\) and \(E\). Since \(p_K(\alpha)=0\), minimality over \(E\) gives \[ p_E\mid p_K \] in \(E[x]\). If \(F/K\) is separable, then \(p_K\) has no repeated roots, so neither does its divisor \(p_E\). Thus \(F/E\) is separable.
3. Normality passes from \(F/K\) to \(F/E\).
With the same notation, \(p_E\mid p_K\) in \(E[x]\). If \(F/K\) is normal, then \(p_K\) splits completely over \(F\). Hence \(p_E\) also splits completely over \(F\). Therefore every minimal polynomial over \(E\) of an element of \(F\) splits in \(F\), so \(F/E\) is normal.