Im reading about solvability by radicals in different books, more concretely in Fields and Galois Theory by Patrick Morandi and Introduction to abstract algebra by Benjamin Fine, etc. In the second book, it is stated the following theorem:
Theorem 15.6.1. Suppose $K$ is a Galois extension of $F$ with $char(F) = 0$. Then $K$ is an extension of $F$ by radicals if and only if $Gal(K/F)$ is a solvable group.
But this is said not to be the case in the Morandi's book (and many others), as there are extensions with solvable Galois groups that arent radical.
My reasoning of why they were able to prove theorem 15.6.1. is that they assume that a convenient $n$'th primitive root of the unity is already in $K$ so that they can indeed construct a simple radical chain from $F$ to $K$. Without that key assumption (which isnt mentioned in the theorem), the theorem is false. Am I correct? If not, which are the cases that the theorem holds?
Edit: In the book Galois Theory by David Cox, he provides the following counterexample:
Example 8.2.3. Let $\mathbb{Q} \subset L$ be the splitting field of $f=x^3+x^2-2x+1 \in \mathbb{Q}[x]$...
Im not going to write the whole text, but he basically proves that the extension is Galois, is contained in a radical extension but is not radical itself.
To tackle this "problem", he defines what being a solvable extension means:
Definition 8.2.4. A field extension $F\subset L$ is solvable (sometimes called solvable by radicals) if there is a field extension $L\subset M$ such that $F\subset M$ is radical.
After this and some other results, he provides this theorem (assuming characteristic 0):
Theorem 8.3.3. Let $F\subset L$ be a Galois extension. Then the following are equivalent: (a) $F\subset L$ is a solvable extension. (b) $Gal(L/F)$ is a solvable group.
As you can see, this is where my confusion starts, since being a solvable Galois group is apparently equivalent to 2 clearly non equivalent propositions.