algebraically solvable equation

An equation

$\displaystyle x^n+a_1x^{n-1}+\ldots+a_n = 0,$ (1)

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{align*}\begin{cases}\xi_1 = \sqrt[p_1]{r_1},\\ \xi_2 = \sqrt[p_2]{r_2(\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

$\displaystyle \sqrt[jk]{r} = \sqrt[j]{\sqrt[k]{r}}$

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

$\displaystyle \xi_1 = \sqrt{\left(\frac{q}{2}\right)^2\!+\!\left(\frac{p}{3}\right)^3}, \quad
\xi_2 = \sqrt[3]{-\frac{q}{2}\!+\!\xi_1}, \quad \xi_3 = \sqrt{-3}.$

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}\\ \display...
...\cdot\xi_2-\frac{-1\!+\!\xi_3}{2}\cdot\!\frac{p}{3\xi_2} \end{cases}\end{align*}    



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