I am currently studying the third edition of Ian Stewart's book "Galois Theory". In the book, solvability of a polynomial by radicals is defined as follows:
Let $f$ be a polynomial over a subfield $K$ of $\mathbb{C}$, and let $\Sigma$ be the splitting field for $f$ over $K$. We say that $f$ is solvable by radicals if there exists a field $M$ containing $\Sigma$ such that $M:K$ is a radical extension.
My question is this: why do we require $\Sigma$ to be contained in a field $M$ that is a radical extension of $K$? Why do we not simply require $\Sigma$ to be a radical extension of $K$? In Stewart's book, he addresses my question by stating that "we want everything in the splitting field $\Sigma$ to be expressible by radicals, but it is pointless to expect everything expressible by the same radicals to be inside the splitting field". However, I do not fully understand his explanation. If anyone can elaborate on Stewart's explanation by offering further motivation for the definition it would be much appreciated. Thanks.
The problem is that there may exist a 'solution', i.e. an expression in radicals for a given equation, but in order to write down this solution, you must allow for intermediary results that are not in the field generated by the roots.
One instance where this happens is the so called casus irreducibilis. This is a cubic equations with 3 real roots. However, if you want to write down the roots of the equation, you will find (experimentally) that you need to use complex numbers somewhere along the way. But the field generated by the roots of the equation is necessarily a real field so you can only write down those solutions by considering a larger field. (The wikipedia article contains a more formal statement.)