In case it is relevant, the following definition comes from Galois Theory by Ian Stewart.
"Let $f$ be a polynomial over a subfield $K$ of $\mathbb {C}$, and let $\sum$ be the splitting field of $f$ over $K$. We say that $f$ is soluble by radicals if there exists a field $M$ containing $\sum$ such that $M:K$ is a radical extension."
The problem with the above definition is that $f$ is a polynomial over different fields.
Take $f = t^5-6t+3$. It is a usual example of a polynomial that has roots not expressible in a finite number of terms consisting of rationals, arithmetic operators and $n$th roots.
If we consider $f$ to be a polynomial over $\mathbb {Q}$, then by definition, $f$ is not soluble by radicals. Yet if we are to consider $f$ a polynomial over $\mathbb {C}$, then we could claim that $f$ is soluble by radicals.
It seems that a more proper definition would be:
Let $f$ be a polynomial over a subfield $K$ of $\mathbb {C}$, and let $\sum$ be the splitting field of $f$ over $K$. We say that $f$ is soluble by radicals in $K$ (or maybe over $K$) if there exists a field $M$ containing $\sum$ such that $M:K$ is a radical extension.
Meaning that the roots of $f$ can be expressed in a finite number of terms consisting of elements in $K$, arithmetic operators and $n$th roots.
Since it is different from what the book states, would such an interpretation of the meaning of solubility by radicals be correct?
I would really appreciate any help/thoughts.