I study the book "Introduction To Field Theory" by Iain Adamson (https://archive.org/details/IntroductionToFieldTheory), and struggle with the Theorem 26.5. on page 166:
Let $F$ be a field of characteristic zero. If the polynomial $f$ in $F[x]$ is solvable by radicals, then the Galois group of any splitting field of $f$ over $F$ is solvable.
Adamson defines "solvable by radicals" on page 160/161:
Let $F$ be a field of characteristic zero; a field $E$ containing $F$ is said to be an extension of $F$ by radicals if there exists a sequence of subfields $F = E_0, E_1, ..., E_{r-1}, E_r = E$ such that for $i = 0, ..., r- 1, E_{i+1} = E_i(\alpha_i)$, where $\alpha_i$, is a root of an irreducible polynomial in $E_i[x]$ of the form $X^{n_i} - a_i$. A polynomial $f$ in $F[x]$ is said to be solvable by radicals if there exists a splitting field of $f$ over $F$ which is contained in an extension of F by radicals.
I don't see where in the proof of Theorem 26.5 the irreducibility of $X^{n_i} - a_i$ is used. Why does Adamson require the polynomial to be irreducible?
Update, July 8th.: I would like to add some definitions. Let us call a radical extension with irreducible polynominals $X^{n_i} - a_i$, an irreducible radical extension. And a polynominal with splitting field contained in an irreducible radical extension, solvable by irreducible radicals. When Adamson says "solvable by radicals", he actually means "solvable by irreducible radicals". In the cited Theorem Adamson then states: "If the polynominal is solvable by irreducible radicals, then the Galois group is solvable". But I think what he actually proves is the stonger statement: "If the polynominal is solvable by radicals, then the Galois group is solvable". Don't be confused when you read the theorems in the book: Adamson does not use the term "solvable by irreducible radicals".
It's probably so the Galois group of the extension $K_n(E_{i+1})/K_n(E_i)$ is easier to handle. Consider, for instance, $\mathbb Q(\mathrm i,\sqrt2)/\mathbb Q(\mathrm i)$ (the $\mathrm i$ is the root of unity he adjoins at the beginning of the proof). This is the splitting field of $X^4-4$ over $\mathbb Q(\mathrm i)$, but also of the minimal polynomial $X^2-2$ of $\sqrt2$. The order of the Galois group is the degree of the minimal polynomial. If we choose the minimal polynomial to construct our extension, we can read off the order of the cyclic factors directly. If reducible polynomials were allowed, the orders of the groups would not be readily available. It would also make the degree $n$ of the polynomial $X^n-e$ whose roots he adjoins at the start unneccessarily large. I don't think any of these matter for the proof, but maybe he considered it more elegant?