AlgebraicallySolvableEquationsDefinition 2009-04-18 09:23:37 2009-04-18 09:23:37 Definition new title % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here An equation \begin{align} x^n+a_1x^{n-1}+\ldots+a_n = 0, \end{align} with coefficients $a_j$ in a field $K$, is {\em algebraically solvable}, if some of its \PMlinkname{roots}{Equation} 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 \PMlinkname{adjoining}{FieldAdjunction} to it in succession certain suitable radicals $\xi_1,\,\xi_2,\,\ldots,\,\xi_m$.\, Each radical may be \PMlinkescapetext{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 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 \PMlinkname{indices}{Root} $p_1,\,p_2,\,\ldots,\,p_m$ are prime numbers.\\ \textbf{Example.}\, \PMlinkexternal{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 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{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*}