- If \(F\) is over \(K\), and \(E\) is an intermediate extension of \(F\) over \(K\), is \(F\) necessarily over \(E ?\) Answer this question for each of the words “algebraic,” “normal,” and “separable” in the blanks.
Solution
The answer is yes in all three cases.
1. If \(F/K\) is algebraic, then \(F/E\) is algebraic.
Let \(\alpha\in F\). Since \(F/K\) is algebraic, there is a nonzero polynomial \(f(x)\in K[x]\) with \(f(\alpha)=0\). Because \(K\subseteq E\), the same polynomial belongs to \(E[x]\). Thus \(\alpha\) is algebraic over \(E\). Since \(\alpha\) was arbitrary, \(F/E\) is algebraic.
2. If \(F/K\) is separable, then \(F/E\) is separable.
Let \(\alpha\in F\). Since \(F/K\) is separable, the minimal polynomial \(m_{\alpha,K}(x)\) has distinct roots in a splitting field. The minimal polynomial \(m_{\alpha,E}(x)\) divides \(m_{\alpha,K}(x)\) in \(E[x]\): indeed, \(m_{\alpha,K}(\alpha)=0\) and \(m_{\alpha,E}\) is the monic polynomial of least positive degree in \(E[x]\) annihilating \(\alpha\). A divisor of a separable polynomial is separable, so \(m_{\alpha,E}\) is separable. Hence every element of \(F\) is separable over \(E\), and therefore \(F/E\) is separable.
3. If \(F/K\) is normal, then \(F/E\) is normal.
Because \(F/K\) is normal and algebraic, there is a family \(\mathcal S\subseteq K[x]\) such that \(F\) is the splitting field of \(\mathcal S\) over \(K\). Since \(K\subseteq E\subseteq F\), we may also regard \(\mathcal S\) as a family in \(E[x]\). All of its roots already lie in \(F\), and \(F\) is generated over \(K\) by those roots; hence it is also generated over the larger base field \(E\) by the same roots. Thus \(F\) is the splitting field of \(\mathcal S\) over \(E\), so \(F/E\) is normal.
4. Therefore algebraicity, separability, and normality each pass from \(F/K\) to \(F/E\) for every intermediate field \(K\subseteq E\subseteq F\).
Combine 1, 2, and 3.