Figure 1.
A normalised root tower for an extension $K/F$ is a root tower with prime indices. In this entry I will prove that the splitting field of every solvable polynomial can be imbedded in a field admitting a normalised root tower, and then I compute them for some specific examples.
I use $z_n$ to denote the primitive $n$th root of unity $e^{2\pi i/n}$. I make heavy use of the following lemma from Jacobson which gives a criteria for a radical extension.
Lemma (Jacobson). Let $p$ be a prime and suppose $F$ contains $p$ distinct $p$th roots of unity. Let $E/F$ be cyclic and $p$-dimensional. Then $E=F(d)$ where $d^p\in F$.
Before proving the result for general solvable extensions, I focus on a simpler case.
Lemma. For every prime $p$, every cyclotomic field $F(z_p)/F$ can be imbedded in a larger extension $K/F$ admitting a normalised root tower.
Proof. The Galois group $\Gal(F(z_p)/F)$ is abelian and hence admits a series of normal subgroups
$$1 = G_1\subset G_2\subset \cdots \subset G_k = G,$$
where each $p_1=[G_{i+1}:G_i]$ is prime. This corresponds to a tower of simple extensions
$$F\subset F(x_1)\subset F(x_1,x_2)\subset \cdots\subset F(x_1,\ldots,x_k) = F(z_p).$$
Although the degrees are all prime, the base fields of each extension $F(x_1,\ldots,x_{l+1})/F(x_1,\ldots,x_l)$ do not necessarily contain the requisite roots of unity, and thus it may not be a radical extension. Thus we need to augment each base fields to construct our desired root tower.
Firstly, we have $p_1=[F(x_1):F]<p$, and so by induction there exists a normalised root tower from $F$ to $F_1$ where $F_1\supset F(z_{p_1})$. Since $F_1$ contains the required $p_1$th roots of unity, we can take it to be the new base field and 'lift up' the extension $F(x_1)/F$ to obtain $F_1(x_1)/F_1$. Since $F(x_1)$ is normal, it is the splitting field of $x_1$'s minpoly $p(x)$ over $F$, and thus $F_1(x_1)$ is also the splitting field of $p(x)$ over $F_1$. Thus we can conclude by Jacobson that $F_1(x_1)/F_1$ is a radical extension, and consequently it has degree either 1 or $p$. Therefore, $F_1(x_1)/F$ admits a normalised root tower.
Now we take $F_1(x_1)$ as the new base field in place of $F$, and likewise we have a root tower from $F_1(x_1)$ to $F_2$ with $F_2\supset F(z_{p_2})$. Going through the same motions we conclude that $F_2(x_2)/F_2$ is a radical extension, so $F_2(x_2)/F_1(x_1)$ has a normalised root tower. We continue in this manner until all the $x_i$'s have been exhausted, and our desired extension has the form $F_n(x_n)$. Diagramatically, this process is represented by the following diagram:
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Figure 1. |
We are done. $\qed$
Now on to the main result. Let $f(x)$ have solvable Galois group $G$. Then its composition series
$$G = G_1\supset G_2\supset \cdots \supset G_n = 1$$
has factors with prime indices. It corresponds to a tower of simple extensions with prime degrees, much like in the lemma. If $n>1$ then the argument is completely the same as in the lemma.
If $n=1$, then $G$ has prime order $p$, and the splitting field is a simple extension $F(x)/F$. By the lemma we have that $F(z_p)/F$ is contained in some $K/F$ admitting a normalised root tower, and $K(x)/K$ is a radical extension, so we are done. $\qed$
One might wonder why we couldn't have proved the main result straight away without the lemma, via induction on the degree. The reason is that in the case $n=1$, the extension $F(x)/F$ has no proper nontrivial subextensions to perform the induction on.
Now I will proceed to compute normalised root towers for $\Q(z_5)/\Q$ and $\Q(z_7)/\Q$.
Let $z=z_5$. Then $\Q(z)/\Q$ can be decomposed as a tower
$$\Q \subset \Q(z+\ol z) \subset \Q(z).$$
We have that $z+\ol z$ is fixed precisely by the subgroup $\\{1,\eta:z\mapsto z^4\\}$ of $\Gal(\Q(z)/\Q)\cong\Z_4$, so it has degree 2 over $\Q$. Its minpoly is easily seen to be $x^2+x-1$, whose roots are $(\pm\sqrt5-1)/2$. Therefore $\Q(z+\ol z)=\Q(\sqrt5)$. Since $z$ lies to the right of the $y$-axis, its $x$ component $z+\ol z$ is positive so $z+\ol z=(\sqrt5-1)/2$. The $y$-component is given by
$$+\sqrt{1-\left({\sqrt5-1 \over 2}\right)^2} = {\sqrt{2\sqrt{5}-5}\over 2}.$$
Thus
$$z = {\sqrt{5}-1+i(\sqrt{2\sqrt5-5}) \over 2} = {\sqrt{5}-1+\sqrt{5-2\sqrt5} \over 2},$$
and we have $\Q(z)=\Q(\sqrt5,\sqrt{5-2\sqrt5})$. The normalised root tower is thus
$$\Q \subset \Q(\sqrt5) \subset \Q(\sqrt5, \sqrt{5-2\sqrt5}).$$
Let $z=z_7$. The Galois group $\Gal(\Q(z)/\Q)$ is generated by the automorphism $\eta:z\mapsto z^3$, so the automorphism $z\mapsto z^{3^2}=z^2$ generates a subgroup of order 3. The corresponding subfield is $\Q(z+z^2+z^4)$, where $z+z^2+z^4\nin\Q$ because it is not fixed by every automorphism. Its minpoly is $x^2+x+2$, whose roots are $(\pm\sqrt{-7}+1)/2$. Therefore $\Q(z+z^2+z^4)=\Q(\sqrt{-7})$. In fact, $z+z^2+z^4=(\sqrt{-7}+1)/2$, because its $y$-component
$$\sin(2\pi/7)+\sin(4\pi/7)+\sin(8\pi/7)$$
is positive, owing to the fact that
$$\sin(2\pi/7)+\sin(4\pi/7) > \sin(2\pi/7) > \sin(6\pi/7) = -\sin(8\pi/7).$$
Now we augment $\Q(\sqrt{-7})$ with a primitive 3rd root of unity $\omega=e^{2\pi i/3}$ and apply the procedure in the proof of Jacobson's lemma to express $\Q(\omega,z)/\Q(\omega,\sqrt{-7})$ as a radical extension. Since $\Q(z)/\Q(\sqrt{-7})$ has degree 3, the extension $\Q(\omega,z)/\Q(\omega,\sqrt{-7})$ has degree either 1 or 3; but $z\nin\Q(\omega,\sqrt{-7})$ since otherwise it has degree $\le 4$ over $\Q$ which is not the case. Thus the degree is 3 and its Galois group is forced to be generated by the automorphism $\eta:z\mapsto z^2$. The Lagrange resolvents are
$$z+z^2+z^4,\quad z+z^2\omega+z^4\omega^2 \quad\text{and}\quad z+z^2\omega^2+z^4\omega.$$
The first resolvent is in $\Q(\omega,\sqrt{-7})$, but the latter two are not because they are not fixed by $\eta$. Let's choose the second resolvent. In the proof of Jacobson's lemma it is shown that the cube of the resolvent is an element of $\Q(\omega,\sqrt{-7})$, and that the resolvent generates $\Q(\omega,z)$. Thus we just have to compute the cube to express $\Q(\omega,z)/\Q(\omega,\sqrt{-7})$ as a radical extension. Annoyingly, I don't know of an easier way to calculate the cube without manually expanding everything out, so I enlisted the help of Mathematica. Luckily, it simplified down to a relatively tame expression:
$$(z+z^2\omega+z^4\omega^2)^3 = 2(4+z+z^2+z^4) + \omega(3+6z+6z^2+6z^4).$$
Subbing in $\omega=(\sqrt{-3}-1)/2$ (which I assume is known) and $z+z^2+z^4=(\sqrt{-7}+1)/2$ we have the element
$${12-3\sqrt{21}+6\sqrt{-3}-\sqrt{-7} \over 2}.$$
Thus the normalised root tower is
$$\Q \subset \Q(\sqrt{-7}) \subset \Q(\sqrt{-3},\sqrt{-7})=\Q(\omega,\sqrt{-7}) \subset \Q(\sqrt{-3},\sqrt{-7},\sqrt[3]{12-3\sqrt{21}+6\sqrt{-3}-\sqrt{-7} \over 2}).$$