Talk:PlanetPhysics/Algebraically Solvable Equations Definition

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An equation

\begin{align} x^n+a_1x^{n-1}+\ldots+a_n = 0, \end{align} with coefficients $a_j$ in a \htmladdnormallink{field}{http://planetphysics.us/encyclopedia/CosmologicalConstant2.html} $K$, is {\em algebraically solvable}, if some of its roots may be expressed with the elements of $K$ by using rational \htmladdnormallink{operations}{http://planetphysics.us/encyclopedia/Cod.html} (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{align*} \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{align*} 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 \htmladdnormallink{power}{http://planetphysics.us/encyclopedia/Power.html} of an element of this field.\, Because of the \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html} $$\sqrt[jk]{r} = \sqrt[j]{\sqrt[k]{r}}$$ one can, without hurting the generality, suppose that the indices $p_1,\,p_2,\,\ldots,\,p_m$ are prime numbers.\\

\textbf{Example.}\, \htmladdnormallink{Cardano's formulae}{http://planetmath.org/encyclopedia/CardanosFormulae.html} show that all roots of the cubic equation\; $y^3+py+q = 0$\; are in the \htmladdnormallink{algebraic}{http://planetphysics.us/encyclopedia/CoIntersections.html} 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{align*} \begin{cases} \displaystyle y_1 = \xi_2-\frac{p}{3\xi_2}\\ \displaystyle y_2 = \frac{-1\!+\!\xi_3}{2}\cdot\xi_2-\frac{-1\!-\!\xi_3}{2}\cdot\!\frac{p}{3\xi_2}\\ \displaystyle y_3 = \frac{-1\!-\!\xi_3}{2}\cdot\xi_2-\frac{-1\!+\!\xi_3}{2}\cdot\!\frac{p}{3\xi_2} \end{cases} \end{align*}

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