degree 3 Galois extension of $\mathbb{Q}$ not radical

I have the following question.

I have the following result in Dummit and Foote abstract algebra (Theorem 39 p. 628) that says that over a field of $char=0$ then a polynomial $f(x)$ is soluble by radicals if and only if its Galois group is soluble group.

But I also have a result in one of my exams that says that a Galois extension of degree 3 over $\mathbb{Q}$ is not a radical extension.

How is it that the two statements aren't contradicting one another? since I thought that if the Galois group was of size 3 then its cyclic so soluble, so if $G$ is the Galois group of some irreducible cubic then wouldn't the theorem say that it is soluble by radicals?

Or is there a difference between a polynomial being soluble by radicals and its splitting field being a radical extension?

Thank you

Solution 1:

Yes, there are some differences : an extension $L/K$ is solvable by radicals if there is an extension $L_1/L$ and a sequences of extensions $K \subset K_1 \subset \ldots \subset K_n \subset L_1$ where each step is a radical extension.

If $L/K$ is cyclic of order $n$ and $K$ contains an $n$th root of unity, then we do have that $L/K$ is radical (there exists some $x \in L$ such that $x^n \in K$ and $L = K(x)$).

However, if $K$ doesn't contain those roots of unity, then no radical extension of order $n$ of $K$ is Galois. So we have to add them first, and then we obtain an extension $K \subset K(\zeta_n) \subset L_1 = L(\zeta_n)$ where the first step is soluble by radicals, and the second is radical. So $L$ is solvable by radicals, but can be very very far from being a radical extension.

The simplest example is $K = \Bbb Q(2\cos \frac {2\pi}7) = \Bbb Q[c]/(c^3+c^2-2c-1)$. It is a normal extension of $\Bbb Q$ whose Galois group is cyclic of order $3$ (generated by $c \mapsto c^2-2$). And in order to solve it by radicals and express $c$ with radicals, you will need to use $\sqrt{-3}$ somewhere.