algebraically solvable

An equation

$x^n+a_1{x}^{n-1}+\mathrm{\dots }+a_n=0,$(1)

with coefficients $a_j$ in a field $K$, is algebraicallysolvable, if some of its roots (http://planetmath.org/Equation) maybe expressed with the elements of $K$ by using rationaloperations (addition, subtraction, multiplication, division)and root extractions.  I.e., a root of (1) is in a field $K({\xi }_1,{\xi }_2,\mathrm{\dots },{\xi }_m)$  which is obtained of$K$ by adjoining (http://planetmath.org/FieldAdjunction) to it insuccession certain suitable radicals${\xi }_1,{\xi }_2,\mathrm{\dots },{\xi }_m$.  Each radical may under the root sign one or more ofthe previous radicals,

$\{\begin{array}{cc}{\xi }_1=\sqrt[p_1]{r_1},\hfill & \\ {\xi }_2=\sqrt[p_2]{r_2({\xi }_1)},\hfill & \\ {\xi }_3=\sqrt[p_3]{r_3({\xi }_1,{\xi }_2)},\hfill & \\ \mathrm{\cdots }\hspace{1em}\hspace{1em}\mathrm{\cdots }\hfill & \\ {\xi }_m=\sqrt[p_m]{r_m({\xi }_1,{\xi }_2,\mathrm{\dots },{\xi }_{m-1})},\hfill & \end{array}$

where generally  $r_k({\xi }_1,{\xi }_2,\mathrm{\dots },{\xi }_{k-1})$  is an element of the field$K({\xi }_1,{\xi }_2,\mathrm{\dots },{\xi }_{k-1})$  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 (http://planetmath.org/Root) $p_1,p_2,\mathrm{\dots },p_m$ 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

$${\xi }_1=\sqrt{{\left(\frac{q}{2}\right)}^2+{\left(\frac{p}{3}\right)}^3},{\xi }_2=\sqrt[3]{-\frac{q}{2}+{\xi }_1},{\xi }_3=\sqrt{-3}.$$

In fact, as we consider also the equation (4), the roots may be expressed as

$\{\begin{array}{cc}{y}_1={\xi }_2-{\displaystyle \frac{p}{3{\xi }_2}}\hfill & \\ y_2={\displaystyle \frac{-1+{\xi }_3}{2}}\cdot {\xi }_2-{\displaystyle \frac{-1-{\xi }_3}{2}}\cdot {\displaystyle \frac{p}{3{\xi }_2}}\hfill & \\ y_3={\displaystyle \frac{-1-{\xi }_3}{2}}\cdot {\xi }_2-{\displaystyle \frac{-1+{\xi }_3}{2}}\cdot {\displaystyle \frac{p}{3{\xi }_2}}\hfill & \end{array}$