Given that $F$ is a field, let $F\_n$ be the completion of $F$ with respect to roots of degree $n$ polynomials. For example this would make $\mathbb{Q}\_2$ the field of (ruler and compass) constructible numbers. Clearly $F\_n \subseteq F\_m$ whenever $n \leq m$ as any degree $n$ polynomial can be multiplied by $x^{m-n}$ to make a degree $m$ polynomial with the same non-trivial roots.


I'm interested in what can be said about the strictness of the chain:


$$F\_1 \subseteq F\_2 \subseteq F\_3 \subseteq F\_4 \subseteq \dots$$




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I'm curious because (assuming the characteristic of $F$ is not $2$ or $3$) the existence of a general quartic formula involving only square and cube roots tells us that $F\_3 = F\_4$. In the case of $\mathbb{Q}$ it appears as though this is the *only* equality, but I'm not sure if this is true.


On the other hand if we let $R$ denote the field of radical numbers (complex numbers which lie in radical extensions of the rationals), then $R\_1 = R\_2 = R\_3 = R\_4$ but $R\_4 \subsetneq R\_5$. So different fields can have interesting patterns.




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I have two main questions. One in the special case of $\mathbb{Q}$ and the other for general fields:



> 
> Does $\mathbb{Q}\_n = \mathbb{Q}\_{n+1}$ happen only when $n = 3$?
> 
> 
> 



> 
> Given any sequence $\phi : \mathbb{N} \rightarrow \{=,\subsetneq\}$, does there exist a field $F$ such that:
> $$F\_1 \space \phi(1) \space F\_2 \space \phi(2) \space F\_3 \space \phi(3) \dots$$
> 
> 
> 


Obviously any sequence where $\phi(3)$ is $\subsetneq$ requires the field to have characterstic $2$ or $3$ as mentioned above. I'm not sure if this is a special location in the chain, and maybe it's better to ask the second question where $\phi(3)$ is restricted to being $=$.



If degree $N$ polynomials always have a root, does there still exist an irreducible polynomial of degree $N+1$?
The answer to the first question is yes: For $n\ge5$ let $L$ be a Galois extension of $\mathbb Q$ with Galois group the alternating group $A\_n$. Suppose that $L$ is contained in the composition $E$ of (w.l.o.g. finitely many) splitting fields of polynomials of degree $\le n-1$.


This yields a group $G$ which is a subgroup of a direct product of groups $S\_k$, $k\le n-1$ (possibly with several copies for the same $k$), such that $G/N=A\_n$ for a normal subgroup $N$. In view of the simplicity of $A\_n$, this is not possible.


