Let $\mathbb{Q}_{\mathrm{Br}}$ be the field of bring-jerrard rational numbers, this is, the field defined by the infinite algebraic extension $\mathbb{Q}_{\mathrm{Br}}=\mathbb{Q}(S)$, were $$S=\{\tau\in\mathbb{C}|\tau^5+p\tau+q=0,\{p,q\}\subseteq\mathbb{Q}\}$$ That $\mathbb{Q}_{\mathrm{Br}}$ is a field comes from the fact it is a normal algebraic extension. I want to explore the algebraical limits of this extension. A good place to start is by defining the polynomial ring $\mathbb{Q}_{\mathrm{Br}}[x]$ together with the equivalence relation: $$f\sim g\Leftrightarrow\mathrm{Deg}(f)=\mathrm{Deg}(g)$$ Which gives us the partition set $$\overset{\Large\sim}{\mathbb{Q}_{\mathrm{Br}}}[x]=\{[n]\in\mathbb{N}\times\mathcal{P}(\mathbb{Q}_{\mathrm{Br}}[x])|[n]=(n,S);f\in S\Leftrightarrow\mathrm{Deg}(f)=n\}$$ Where $\mathcal{P}(\cdot)$ denotes the power set. $[n]$ is then the subset of all polynomials $f\in\mathbb{Q}_{\mathrm{Br}}[x]$ with degree $n$. Given this setting, my question is: What is the least $n\in \mathbb{N}$ such that $[n]$ contains a polynomial whose galois group is not solvable in $\mathbb{Q}_{\mathrm{Br}}$?
Trivially, we should have $n>5$. We can prove this informally by noting that every polynomial can be reduced to its Bring-Jerrard form, which, by definition, has a solution in radicals within $\mathbb{Q}_{\mathrm{Br}}$; so its Galois group is solvable... however, i can't prove if any greater $n$ will satisfy this condition. My intuition tells me that this is a mirror situation of that of plynomials in $\mathbb{Q}$, in which case $n=10$. Any insight on the matter will be very appreciated!
There are polynomials of degree as low as 6 that remain irreducible in $\mathbb{Q}_\mathrm{Br}$.
The reason being that each composition factor of a composition series for any Galois group of a finite normal subfield of $\mathbb{Q}_\mathrm{Br}$ must be a subgroup of the symmetric group $S_5$: This follows straight from the definition - any finite subextension $K\subseteq \mathbb{Q}_\mathrm{Br}$ is contained in $\mathbb{Q}(S')$ for some finite subset $S'=\{\alpha_1,...,\alpha_n\}\subseteq S$. If we write $L_i$ for the splitting field of $\alpha_i$, we therefore obtain a tower of normal extensions $$\mathbb{Q}\subseteq L_1\subseteq L_1L_2\subseteq ... \subseteq L_1...L_n.$$ Here, the final extension contains $K$ (as it contains $S'$) and clearly every successive extension has some subgroup of $S_5$ as its Galois group.
In particular, every composition factor that can ever arise in any composition series of a Galois group of a finite normal subextension must have this property (by Jordan-Hölder). But not every group admits such a series - the smallest extension would be one of degree 6 that has either $A_6$ or $S_6$ as Galois group of its splitting field (although the smallest nonsolvable Galois group not of this form is rather $\mathrm{PSL}_3(\mathbb{F}_2)$ - but this only arises from degree 7 extensions at a minimum). In fact "most" polynomials of degree 6 will have one of these as their Galois group, one concrete instance is $x^6-x+1$ (with Galois group being $S_6$).