A polynomial
<I>f</i>(<i>x</I>)  is  solvable by radicals over a field <i>F</i> if <I>f</i>(<i>x</I>)
  splits in some extension
<I>F</i>(<i>a<sub>1</sub>,a<sub>2</sub>, &#8230,a<sub>n</sub></i>) 
of <I>F</i> and there exist positive integers
<i>k<sub>1</sub>, k<sub>2</sub>,&#8230,k<sub>n</sub></i>
such that
<i>a<sub>1</sub><sup>k<sub>1</sub></sup> &#8712  F</i> and
<i>a<sub>i</sub><sup>k<sub>i</sub></sup> &#8712  F</i>(<i>a<sub>1</sub>, a<sub>2</sub>,&#8230,a<sub><i>i -1</sub></i>) for <i> i = 2,&#8230,n</i>.