- If \(F\) is over \(E\), and \(E\) is \(\quad\) over \(K\) is \(F\) necessarily over \(K\) ? Answer this question for each of the words “algebraic,” “normal,” and “separable” in the blanks.
Solution
1. Algebraicity is transitive: if \(F/E\) and \(E/K\) are algebraic, then \(F/K\) is algebraic.
Fix \(\alpha\in F\). Since \(\alpha\) is algebraic over \(E\), there is a relation \[ a_0+a_1\alpha+\cdots+a_d\alpha^d=0 \] with \(a_i\in E\), not all zero. Put \(E_0=K(a_0,\ldots,a_d)\). Because \(E/K\) is algebraic and only finitely many coefficients are adjoined, \(E_0/K\) is finite. The displayed relation shows that \(\alpha\) is algebraic over \(E_0\), hence \(E_0(\alpha)/E_0\) is finite. Therefore \[ [K(\alpha):K]\le [E_0(\alpha):K]=[E_0(\alpha):E_0][E_0:K]<\infty. \] Thus \(\alpha\) is algebraic over \(K\). Since \(\alpha\) was arbitrary, \(F/K\) is algebraic.
2. Separability is transitive: if \(F/E\) and \(E/K\) are separable algebraic extensions, then \(F/K\) is separable.
Fix \(\alpha\in F\), and let \(m_{\alpha,E}(x)\in E[x]\) be its minimal polynomial over \(E\). Let \(E_0/K\) be the finite extension generated by the finitely many coefficients of \(m_{\alpha,E}\). Since \(E/K\) is separable, \(E_0/K\) is finite separable.
Because \(F/E\) is separable, \(m_{\alpha,E}\) has no repeated root in an algebraic closure. The same polynomial lies in \(E_0[x]\), and squarefreeness is unchanged after extending scalars. The minimal polynomial \(m_{\alpha,E_0}\) divides \(m_{\alpha,E}\) in \(E_0[x]\), so it too has no repeated root. Hence \(\alpha\) is separable over \(E_0\), and \(E_0(\alpha)/E_0\) is finite separable.
A tower of finite separable extensions is separable: if \(L/M\) and \(M/K\) are finite separable, then the number of \(K\)-embeddings of \(L\) into an algebraic closure equals \[ [L:M][M:K]=[L:K], \] because each of the \([M:K]\) embeddings of \(M\) extends to exactly \([L:M]\) embeddings of \(L\). Hence \(L/K\) is separable. Applying this to \[ K\subseteq E_0\subseteq E_0(\alpha) \] shows that \(\alpha\) is separable over \(K\). Therefore \(F/K\) is separable.
3. Normality is not transitive.
Take \[ K=\mathbb Q,\qquad E=\mathbb Q(\sqrt2),\qquad F=\mathbb Q(2^{1/4}). \] The extension \(E/K\) is quadratic, hence normal. Also \(F/E\) is quadratic, because \(2^{1/4}\) satisfies \[ x^2-\sqrt2\in E[x], \] and this polynomial splits in \(F\) as \[ (x-2^{1/4})(x+2^{1/4}); \] therefore \(F/E\) is normal.
However \(F/K\) is not normal. The minimal polynomial of \(2^{1/4}\) over \(\mathbb Q\) is \(x^4-2\), irreducible by Eisenstein at \(2\). Its roots are \[ \pm 2^{1/4},\qquad \pm i2^{1/4}. \] The field \(F\subset\mathbb R\) contains the first two roots but not the nonreal roots. Hence \(x^4-2\) does not split over \(F\), so \(F/\mathbb Q\) is not normal.
4. Therefore the answers are: algebraic — yes; normal — no; separable — yes.
Combine 1, 2, and 3.