Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 17F (Galois Theory)

Here one must distinguish two conventions for a radical extension. The solvability theorem says that a finite extension with solvable normal-closure Galois group is contained in a tower of radical adjunctions. Under the convention that this containment defines a radical extension, this is the requested assertion.

Let $N/K$ be the finite normal closure and let $m$ be divisible by the orders of its Galois group elements. Adjoin a primitive $m$th root of unity, obtaining $K^{\prime }=K({\zeta }_m)$, which itself is a radical adjunction since ${\zeta }_m^m=1$. The Galois group of $N{K}^{\prime }/K^{\prime }$ embeds in the solvable group of $N/K$, so it is solvable. A composition series with cyclic prime-order quotients gives a tower of cyclic prime-degree extensions above $K^{\prime }$. Each is a radical adjunction, as the following argument shows.

For a cyclic extension $E/F$ of prime degree $p$ containing ${\zeta }_p$, let $\sigma$ generate its Galois group. The $F$-linear operator $\sigma$ satisfies $X^p-1$, a product of distinct linear factors in characteristic zero. Thus it is diagonalizable. Because $\sigma \ne 1$, it has an eigenvector $a\ne 0$ with eigenvalue ${\zeta }_p^j\ne 1$. Then $a^p\in F$ and $a\notin F$, so $E=F(a)$ by primeness of the degree. Applying this at every step produces a radical tower containing $L$.

If “radical extension” instead means that $L/K$ itself must be such a tower, the printed statement is false without this qualification. For example the real cubic subfield of $\mathbb{Q}({\zeta }_7)$ is cyclic and solvable, but contains no nonrational $a$ with $a^n\in \mathbb{Q}$: all three conjugates are real roots of the same equation $X^n=a^n$, which has at most two real roots, while any nonrational element of the cubic field has three distinct conjugates. Consequently no nontrivial radical adjunction can start a tower inside it. Containment in a radical tower is the valid general conclusion.

Take the Vandermonde determinant $\Delta ={\prod }_{i<j}(x_i-x_j)$. Its square is a symmetric polynomial, hence a polynomial in the elementary symmetric polynomials and belongs to $K$. It is nonzero because the variables are algebraically independent. A transposition of two variables fixes every $e_i$ and therefore fixes $K$, but sends $\Delta$ to $-\Delta \ne \Delta$ in characteristic zero. Thus $\boxed{\alpha =\Delta \notin K,{\alpha }^2\in K}$.

In characteristic two, let $K={\mathbb{F}}_2(t)$ and $L=K(u)$ with $u^2+u=t$. The Artin–Schreier polynomial is irreducible: if $r\in K$ solved it, a finite pole of $r$ would create a pole of even order in $r^2+r$, impossible for $t$; at infinity a pole of $r$ likewise gives even order, while $t$ has order one. A constant $r$ cannot solve it either. Its derivative is one, so this degree-two extension is separable field extension. Adjoining $\sqrt{\alpha }$ in characteristic two is either trivial or purely inseparable field extension, since the derivative of $X^2-\alpha$ vanishes. Hence $\boxed{L\ne K(\sqrt{\alpha })\text{ for every }\alpha \in K}$.

For the second example take $\boxed{K={\mathbb{F}}_{11}}$. Every extension still has characteristic eleven, and $X^{11}-1=(X-1{)}^{11}$ in it. Thus the only eleventh root of unity is one, whose order is one, and there is no primitive eleventh root in any extension.