PlanetPhysics/Algebraically Solvable Equations Definition

An equation

$\begin{array}{c}{x}^n+a_1{x}^{n-1}+\dots +a_n=0,\end{array}$

with coefficients $a_j$ in a [[../CosmologicalConstant2/|field]] $K$, is algebraically solvable , if some of its roots may be expressed with the elements of $K$ by using rational [[../Cod/|operations]] (addition, subtraction, multiplication, division) and root extractions. I.e., a root of (1) is in a field \,$K({\xi }_1,{\xi }_2,\dots ,{\xi }_m)$\, which is obtained of $K$ by adjoining to it in succession certain suitable radicals ${\xi }_1,{\xi }_2,\dots ,{\xi }_m$.\, Each radical may be contain under the root sign one or more of the previous radicals,

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

where generally\, $r_k({\xi }_1,{\xi }_2,\dots ,{\xi }_{k-1})$\, is an element of the field $K({\xi }_1,{\xi }_2,\dots ,{\xi }_{k-1})$\, but no $p_k$'th [[../Power/|power]] of an element of this field.\, Because of the [[../Formula/|formula]] $\sqrt[jk]{r}=\sqrt[j]{\sqrt[k]{r}}$ one can, without hurting the generality, suppose that the indices $p_1,p_2,\dots ,p_m$ are prime numbers.\\

Example. \, Cardano's formulae show that all roots of the cubic equation\; $y^3+py+q=0$\; are in the [[../CoIntersections/|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

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