Counter example of a radical extension that is not Galois/normal over $\mathbb{Q}(\omega)$? Most proofs of Galois theorem stating that "an equation is solvable in radicals if and only if its Galois group is solvable," show the left to right direction by induction on the height of the radical extension. Starting with a cyclotomic field, adjoining "enough" roots of the unity (e.g. adjoining the product of all the ) and continuing with successive simple radical extensions, which are all Galois and Abelian over the previous one (because we ensured that there were enough roots of unity to start with). The key element in proving that all these extensions are solvable over the base field is then to define a solvable extension as an extension which normal closure has solvable Galois group (equivalently such that there exist an extension which Galois group is solvable) (def (a)), this makes "being a solvable extension" transitive (it is distinguished) as proved in Lang Algebra. Doing this makes it possible to complete the proof by induction, while "being Galois and solvable" (def (b)) does not go through (since Galois extensions are not distinguished).
Many texts also state that the (radical) extensions arising in the above mentioned induction are not Galois in general. While this is indeed not the case in general (as with the traditional example $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt[4]{2})$ since it's missing $i\sqrt[4]{2}$), one might wonder if this is true or not for radical extensions where the base field contains enough roots of unity.
More precisely, the question is to determine whether the following statement is true.

Let $\omega_n$ be a primitive $n^\textrm{th}$ root of the unity.
Let $x = (x_1, \ldots, x_h)$ be sequence of elements in the algebraic closure of $\mathbb{Q}$, such that $x_{i+1}^{m_{i+1}} \in \mathbb{Q}(x_1, \ldots, x_{i})$ for some minimal $m_i > 1$ for all $i < h$.
Is there a natural number $N$, such that in the radical extension series
$$
 \mathbb{Q}(\omega_{N}) = K \subset K(x_1) \subset \ldots \subset K(x_1, \ldots, x_n)
$$
each $K(x_1, \ldots, x_{i+1})/K$ is Galois ?

Definition
Let's call $N_x$ the smallest of such natural numbers if it exists, and set $N_x = \infty$ if there is no such natural number.
The above question amounts to «Is $N_x$ always finite?»
In a typical proof of Galois theorem above, since the primitive root of the unity that is chosen is typically a divisor of $d_x = \prod_i m_i$ (one can also pick the lcm afaik).

*

*If it were true that $N_x | d_x$ for all $x$ then the proof above could be carried out with def (b) instead of (a). (The answers seem to indicate this is not true though.)

*However if $N_x$ is finite for all $x$, then for each $x$ one can use def (b) by starting the radical series with $\mathbb{Q}(\omega_{N_x})$ (instead of $\mathbb{Q}(\omega_{d_x})$ as in the proof above).

I conjectured the above statement is false, i.e. that there are $x$s such that $N_x > d_x$ and even such that $N_x = \infty$.
Hence I am searching for a (minimal?) counter example, i.e. a radical extension
$$
 \mathbb{Q}(\omega_N) = K \subset K(x_1) \subset \ldots \subset K(x_1, \ldots, x_h)
$$
such that each for all $N$ such that each $K(x_1, \ldots, x_{i+1})/K(x_1, \ldots, x_i)$ is Galois, $K(x_1, \ldots, x_n)/K$ is not Galois.
(Another way to state it is to start with $K$ being an extension of $\mathbb{Q}$ containing all roots of unit.)
I think the following is a radical extensions of total degree $6$ over $\mathbb{Q}(\omega)$, where $\omega = \frac{\sqrt{3}}{2} + \frac{i}{2}$ is a primitive 12th root of unity and where each intermediate extension is Galois over the previous one:
$$K = \mathbb{Q}(\omega) \subset F = K\left(\sqrt[3]{2}\right) \subset K\left(\sqrt[3]{2}, \sqrt{1+ \sqrt[3]{2}}\right) = E.$$
However, I would like to claim $E$ is not Galois over $K$, even though all 12th roots of unity are in $K$.
Indeed a conjugate of $\sqrt{1+ \sqrt[3]{2}}$ over $K$ is a square root $\alpha$ of $1 + j\sqrt[3]{2}$ -- their minimal polynomial is $X^6 - 3 X^4 + 3 X^2 - 3$ -- but $\alpha$ would not be in $E = \textrm{span}_F\left(1, \sqrt{1+ \sqrt[3]{2}}\right).$
I reduced the problem to checking that $\frac{1 + j\sqrt[3]{2}}{1+ \sqrt[3]{2}}$ is not a square in $F$ (unless I made a mistake) and some famous online computer algebra system tells that if I compare it to $(x+y\sqrt[3]{2}+z\sqrt[3]{2}^2)^2$ then $z$ needs to be expressed using a square root of an element of $F$ that does not look like it's in $F$, but I am still not 100% sure this is correct...
Does someone has indications on better ways to find a counter example, check this one or prove the theorem that I believe is wrong in the first place? (i.e. that all radical extension over a field with enough roots of unity are Galois.)

EDIT 25/01/2021
I made my question more precise by showing the statement for which I am searching for a counter example.
Unless I am mistaken, the answers and their comments below show that both

*

*$x_1 ^ 2 = 3, \quad x_2 ^ 2= 1 + x_1$

*$y_1 ^ 3 = 2, \quad y_2 ^ 2= 1 + y_1$
are indeed counter examples of $N | d$ since

*

*$d_x = 4$ and $N_x = 8$

*$d_y = 6$ and $N_y > 12$.

They also show that $x$ is not a counter example that $N_x < \infty$, and I think the finiteness status of $N_y$ is unknown.
 A: The polynomial $f:=(x^2-1)^3-2$ has Galois group $G \simeq S_4 \times C_2$ over $\mathbb{Q}$, so the commutator subgroup $[G,G] \simeq A_4$. Let $L$ be the splitting field (over $\mathbb{Q}$) of $f$. This contains the $12$-th roots of unity, i.e. contains your $K$. Since $[K:\mathbb{Q}]=4=[G:[G,G]]$, $L$ contains no other roots of unity and $\operatorname{Gal}(L/K)\simeq A_4$. Since $A_4$ has no normal subgroups of order $2$ the extension $E/K$ is not Galois.
Certifying that $G \simeq S_4 \times C_2$ could e.g. be done using (an exact version of a small part of) Algorithm 6.3.10 in Cohen's "A Course in Computational Algebraic Number Theory" or using more basic Galois resolvents, or...
A: I think I have a better solution: $\sqrt{1  + \sqrt{3}}$. Since the "accident" in the comment above does not happen. $\sqrt{3} -1= \frac{ (\sqrt{3}- 1)^2 }{2}  (1+ \sqrt{3})$.
Edit: it does not work, cf comment below.
A: The smallest counterexample (in terms of degree) should be to take the fixed field of a non-central involution in any realization of the dihedral group $D_4$.
EDIT: Here are some details, since the answer apparently was too cryptic: Take for example $K=\mathbb{Q}(i)$. It is known that there exists a Galois extension $M/K$ with Galois group the dihedral group $D_4$ of order $8$ (this follows for example from Shafarevich's theorem on solvable groups over number fields, but it is a much more easy special case, probably explicit polynomials can be found in the relevant databases). The fixed field $L$ of say the involution $(1 3)\in D_4\subseteq S_4$ is not Galois over $K$ (since the subgroup generated by the involution is not normal), but it is a degree four extension of $K$ decomposing into two steps of degree two, both which are therefore radical.
EDIT2: Here is how to get such an example over the maximal cyclotomic extension $\mathbb{Q}^{\rm cycl}=\mathbb{Q}^{\rm ab}$ of $\mathbb{Q}$:
By a result of Kuyk (see also Theorem 16.11.3 in the book Field Arithmetic by Fried-Jarden), Hilbert's irreducibility theorem holds over $\mathbb{Q}^{\rm ab}$, i.e. the field is Hilbertian in the sense of Lang. According to the tables in the book Inverse Galois Theory by Malle-Matzat, $f=x^4-2x^2+t$ gives a regular realization of the group $D_4$ over $\mathbb{Q}$, i.e. if $F$ is the splitting field of $f$ over $\mathbb{Q}(t)$, then $F/\mathbb{Q}(t)$ has group $D_4$ and $\mathbb{Q}$ is algebraically closed in $F$.
In particular, $f$ still has group $D_4$ over $\mathbb{Q}^{\rm ab}(t)$,
and so the Hilbertian property gives infinitely many $a\in\mathbb{Q}^{\rm ab}$ for which $f_a=x^4-2x^2+a$ has Galois group $D_4$ over $\mathbb{Q}^{\rm ab}$ (see e.g. Proposition 16.1.5 in the book by Fried-Jarden). The above construction then gives a non-Galois radical degree $4$ extension of $\mathbb{Q}^{\rm ab}$.
