# algebraically solvable

An equation

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

with coefficients $a_{j}$ in a field $K$, is algebraically solvable, if some of its roots (http://planetmath.org/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 adjoining (http://planetmath.org/FieldAdjunction) to it in succession certain suitable radicals   $\xi_{1},\,\xi_{2},\,\ldots,\,\xi_{m}$.  Each radical may under the root sign one or more of the previous radicals,

 $\displaystyle\begin{cases}\xi_{1}=\sqrt[p_{1}]{r_{1}},\\ \xi_{2}=\sqrt[p_{2}]{r_{2}(\xi_{1})},\\ \xi_{3}=\sqrt[p_{3}]{r_{3}(\xi_{1},\,\xi_{2})},\\ \cdots\qquad\cdots\\ \xi_{m}=\sqrt[p_{m}]{r_{m}(\xi_{1},\,\xi_{2},\,\ldots,\,\xi_{m-1})},\end{cases}$

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 indices (http://planetmath.org/Root) $p_{1},\,p_{2},\,\ldots,\,p_{m}$ 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

 $\xi_{1}=\sqrt{\left(\frac{q}{2}\right)^{2}\!+\!\left(\frac{p}{3}\right)^{3}},% \qquad\xi_{2}=\sqrt{-\frac{q}{2}\!+\!\xi_{1}},\qquad\xi_{3}=\sqrt{-3}.$

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

 $\displaystyle\begin{cases}\displaystyle y_{1}=\xi_{2}-\frac{p}{3\xi_{2}}\\ \displaystyle y_{2}=\frac{-1\!+\!\xi_{3}}{2}\cdot\xi_{2}-\frac{-1\!-\!\xi_{3}}% {2}\cdot\!\frac{p}{3\xi_{2}}\\ \displaystyle y_{3}=\frac{-1\!-\!\xi_{3}}{2}\cdot\xi_{2}-\frac{-1\!+\!\xi_{3}}% {2}\cdot\!\frac{p}{3\xi_{2}}\end{cases}$

## References

• 1 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet.  Tiedekirjasto No. 17.   Kustannusosakeyhtiö Otava, Helsinki (1950).
Title algebraically solvable AlgebraicallySolvable 2015-04-15 13:48:08 2015-04-15 13:48:08 pahio (2872) pahio (2872) 9 pahio (2872) Definition msc 12F10 algebraic solvability solvable algebraically RadicalExtension KalleVaisala