PlanetPhysics/Algebraically Solvable Equations Definition

An equation $$\begin{matrix} x^n+a_1x^{n-1}+\ldots+a_n = 0, \end{matrix}$$ with coefficients $$a_j$$ in a field $$K$$, is algebraically solvable, if some of its roots may be expressed with the elements of $$K$$ by using rational operations (addition, subtraction, multiplication, division) and root extractions. I.e., a root of (1) is in a field \,$$K(\xi_1,\,\xi_2,\,\ldots,\,\xi_m)$$\, which is obtained of $$K$$ by adjoining to it in succession certain suitable radicals $$\xi_1,\,\xi_2,\,\ldots,\,\xi_m$$.\, Each radical may be contain under the root sign one or more of the previous radicals, $$\begin{matrix} \begin{cases} \xi_1 = \sqrt[p_1]{r_1},\\ \xi_2 = \sqrt[p_2]{r_2(\xi_1)},\\ \xi_3 = \sqrt[p_3]{r_3(\xi_1,\,\xi_2)},\\ \cdots\qquad\cdots\\ \xi_m = \sqrt[p_m]{r_m(\xi_1,\,\xi_2,\,\ldots,\,\xi_{m-1})}, \end{cases} \end{matrix}$$ where generally\, $$r_k(\xi_1,\,\xi_2,\,\ldots,\,\xi_{k-1})$$\, is an element of the field $$K(\xi_1,\,\xi_2,\,\ldots,\,\xi_{k-1})$$\, but no $$p_k$$'th power of an element of this field.\, Because of the formula $$\sqrt[jk]{r} = \sqrt[j]{\sqrt[k]{r}}$$ one can, without hurting the generality, suppose that the indices $$p_1,\,p_2,\,\ldots,\,p_m$$ are prime numbers.\\

Example. \, Cardano's formulae show that all roots of the cubic equation\; $$y^3+py+q = 0$$\; are in the algebraic number field which is obtained by adjoining to the field\, $$\mathbb{Q}(p,\,q)$$\, successively the radicals $$\xi_1 = \sqrt{\left(\frac{q}{2}\right)^2\!+\!\left(\frac{p}{3}\right)^3}, \quad \xi_2 = \sqrt[3]{-\frac{q}{2}\!+\!\xi_1}, \quad \xi_3 = \sqrt{-3}.$$ In fact, as we consider also the equation (4), the roots may be expressed as $$\begin{matrix} \begin{cases} y_1 = \xi_2-\frac{p}{3\xi_2}\\ y_2 = \frac{-1\!+\!\xi_3}{2}\cdot\xi_2-\frac{-1\!-\!\xi_3}{2}\cdot\!\frac{p}{3\xi_2}\\ y_3 = \frac{-1\!-\!\xi_3}{2}\cdot\xi_2-\frac{-1\!+\!\xi_3}{2}\cdot\!\frac{p}{3\xi_2} \end{cases} \end{matrix}$$