necessarily over \(E ?\) Answer this question for each of the words “algebraic,” “normal,” and “separable” in the blanks.
Solution
Let \(F\subseteq E\subseteq K\) be fields. Each of the three properties passes from \(K/F\) to \(K/E\).
1. If \(K/F\) is algebraic, then \(K/E\) is algebraic.
For \(\alpha\in K\), a nonzero \(p\in F[x]\) with \(p(\alpha)=0\) also lies in \(E[x]\).
2. If \(K/F\) is normal, then \(K/E\) is normal.
\(K\) is the splitting field over \(F\) of a family \(\mathcal F\subseteq F[x]\). Then \(\mathcal F\subseteq E[x]\), its roots lie in \(K\), and they generate \(K\) over \(F\), hence over \(E\supseteq F\); so \(K\) is the splitting field of \(\mathcal F\) over \(E\).
3. If \(K/F\) is separable, then \(K/E\) is separable.
For \(\alpha\in K\), the minimal polynomial \(m_{\alpha,E}\) divides \(m_{\alpha,F}\) in \(E[x]\), since \(m_{\alpha,F}\in E[x]\) kills \(\alpha\). As \(m_{\alpha,F}\) has distinct roots, so does its divisor \(m_{\alpha,E}\).