Give an example of each of the following:
Fields \(F \subset K \subset L\) such that \(L/K\) and \(K/F\) are Galois, \(L/F\) is algebraic, but \(L/F\) is not Galois.
A pair of fields \(F \subset K\) such that \(K\) is finite dimensional over \(F\) but \(K\) is not separable over \(F\).
Solution
1. For part (a), take\[ F=\mathbb Q,\qquad K=\mathbb Q(\sqrt2), \qquad L=\mathbb Q(\sqrt[4]{2}). \]
Let \(a=\sqrt[4]{2}>0\), so \(a^2=\sqrt2\). Eisenstein’s criterion at two makes \(T^2-2\) and \(T^4-2\) irreducible over \(\mathbb Q\) [@DF04]. Hence \([K:F]=2\), \([L:F]=4\), and the tower law gives \([L:K]=2\). In particular both inclusions are strict and \(L/F\) is algebraic.
The field \(K\) is the splitting field of \(T^2-2\) over \(F\), with distinct roots \(\sqrt2,-\sqrt2\). The field \(L=K(a)\) is the splitting field of \(T^2-\sqrt2\) over \(K\), with distinct roots \(a,-a\). Thus both \(K/F\) and \(L/K\) are Galois, being splitting fields of separable polynomials [@DF04].
However, \(L\subseteq\mathbb R\) contains \(a\) but not the nonreal root \(ia\) of the irreducible polynomial \(T^4-2\). That polynomial therefore does not split over \(L\), so \(L/F\) is not normal and hence is not Galois.
2. For part (b), take\[ F=\mathbb F_2(t),\qquad K=F(u),\qquad u^2=t, \]where \(t\) is transcendental over \(\mathbb F_2\).
The polynomial \(T^2-t\) has no root in \(F\). Indeed, a root written as \(A(t)/B(t)\), with nonzero \(A,B\in\mathbb F_2[t]\), would give \(A(t)^2=tB(t)^2\). The left side has even degree and the right side odd degree, an impossibility. Since a reducible quadratic over a field has a root, \(T^2-t\) is irreducible. Thus the quotient \(F[T]/(T^2-t)\) is a field, and its element \(u=[T]\) gives the asserted extension \(K\) with \([K:F]=2\).
In characteristic two one has \(T^2-t=(T-u)^2\) over \(K\). The minimal polynomial of \(u\) over \(F\) therefore has a repeated root and is not separable. Hence \(K/F\) is a finite-dimensional, nonseparable extension, as required.