Give an example of each of the following or explain why no such examples exist. One or two sentence answers will suffice; you don’t need to give formal proofs.
A Galois extension \(K/F\) and an intermediate field \(E\) (\(F \subset E \subset K\)) such that \(E/F\) is not Galois.
A Galois extension \(K/F\) and an intermediate field \(E\) (\(F \subset E \subset K\)) such that \(K/E\) is not Galois.
A finite field of size 8.
A subfield \(E\) of the splitting field of \(x^5 + x - 1\) with \([E:\mathbb{Q}] = 7\).
A field extension \(K/F\) which is finite dimensional but not separable.
An irreducible fifth degree polynomial over \(\mathbb{Q}\) which is solvable by radicals.
The source writes the extensions in parts (a) and (b) as \(F/E\) and \(E/K\). With \(F\subset E\subset K\) these are the extensions \(E/F\) and \(K/E\), as stated here.
Solution
1. In part (a), take \(F=\mathbb Q\),\(E=\mathbb Q(a)\), and \(K=\mathbb Q(a,\zeta)\), where\(a=\sqrt[3]{2}>0\) and \(\zeta=e^{2\pi i/3}\).
The polynomial \(T^3-2\) is Eisenstein at two and therefore irreducible over \(\mathbb Q\) [@DF04]. Thus \([E:F]=3>1\). The field \(E\) is real, whereas \(\zeta\) is nonreal, so \(E\subset K\) is also strict. The roots \(a,\zeta a,\zeta^2a\) generate \(K\), because their ratio recovers \(\zeta\). Hence \(K\) is a splitting field in characteristic zero and \(K/F\) is Galois [@DF04]. But \(E\) contains only the real root of the irreducible polynomial \(T^3-2\), so \(E/F\) is not normal and is not Galois.
2. No example exists in part (b): \(K/E\) is always Galois.
For \(a\in K\), let \(p_a(T)\in F[T]\) and \(q_a(T)\in E[T]\) be its monic minimal polynomials over \(F\) and \(E\). The extension \(K/F\) is algebraic, so both exist. Since \(p_a(a)=0\), division by the minimal polynomial over \(E\) gives \(q_a\mid p_a\) in \(E[T]\). Normality and separability of \(K/F\) imply that \(p_a\) splits into distinct linear factors in \(K[T]\). Its divisor \(q_a\) therefore does the same. Thus every element of \(K\) is separable over \(E\), and every irreducible polynomial over \(E\) with a root in \(K\) splits over \(K\). This proves that \(K/E\) is algebraic, normal, and separable, hence Galois.
3. In part (c), take \(\mathbb F_2[T]/(T^3+T+1)\).
The polynomial is nonzero at both \(0\) and \(1\). A reducible cubic over a field has a linear factor, so this polynomial is irreducible over \(\mathbb F_2\). The quotient is consequently a field [@DF04]. Polynomial division gives the basis \(1,[T],[T^2]\) over \(\mathbb F_2\), so it has exactly \(2^3=8\) elements.
4. No example exists in part (d).
Let \(L\) be the splitting field of \(T^5+T-1\) over \(\mathbb Q\). This is a finite Galois extension, since the base field has characteristic zero [@DF04]. Its automorphisms act faithfully on the distinct roots, because those roots generate \(L\). There are at most five such roots, so this action embeds \(\operatorname{Gal}(L/\mathbb Q)\) into \(S_5\), adding fixed letters if necessary. Lagrange’s theorem gives \([L:\mathbb Q]=|\operatorname{Gal}(L/\mathbb Q)|\mid120\). For an intermediate field \(E\), the tower law makes \([E:\mathbb Q]\) a divisor of \([L:\mathbb Q]\) and hence of \(120\). Since \(7\nmid120\), degree seven is impossible.
5. In part (e), take \(F=\mathbb F_2(t)\) and \(K=F(u)\)with \(u^2=t\), where \(t\) is transcendental.
The polynomial \(T^2-t\) has no root in \(F\): an equality \((A(t)/B(t))^2=t\), with nonzero polynomials \(A,B\), would give \(2\deg A=1+2\deg B\), impossible by parity. It is therefore irreducible, giving \([K:F]=2\). In \(K[T]\) it equals \((T-u)^2\), so the minimal polynomial of \(u\) has a repeated root. Hence \(K/F\) is not separable.
6. In part (f), take \(f(T)=T^5-2\).
Eisenstein’s criterion at two proves irreducibility over \(\mathbb Q\) [@DF04]. Let \(b=\sqrt[5]{2}\) and choose a primitive fifth root of unity \(\xi\). The roots of \(f\) are precisely \(\xi^j b\), \(0\leq j<5\). They all lie in the radical tower \[ \mathbb Q\subset\mathbb Q(\xi) \subset\mathbb Q(\xi,b): \qquad \xi^5=1\in\mathbb Q,\quad b^5=2\in\mathbb Q(\xi). \] Each step adjoins an element having a positive integral power in the preceding field. Thus every root of the irreducible fifth-degree polynomial lies in a radical extension, which is exactly solvability by radicals.