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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:

$$F_1 \subseteq F_2 \subseteq F_3 \subseteq F_4 \subseteq \dots$$

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.

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 $=$.

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  • $\begingroup$ @PeterTaylor I'm not sure I follow your argument. By my current definition we have $(F_n)_m = F_{\max(n,m)}$ for any field, unless you're pointing out something I've overlooked? $\endgroup$
    – Sam Forster
    Commented Aug 1, 2023 at 11:14
  • $\begingroup$ My error: I was reading completion and interpreting extension. $\endgroup$
    – Peter Taylor
    Commented Aug 1, 2023 at 11:30
  • $\begingroup$ See link.springer.com/article/10.1007/BF02986170 for a characterization of implications between the statements expressing the existence of roots for all polynomials of given degrees. $\endgroup$
    – Emil Jeřábek
    Commented Aug 1, 2023 at 16:46

1 Answer 1

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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.

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    $\begingroup$ I think the notion of "completion" in the question is intended to include iteration, so that the embedding would be into an iterated wreath product, not a direct product, but the argument still works in the same way. $\endgroup$
    – Joachim König
    Commented Aug 1, 2023 at 13:58
  • $\begingroup$ Thank you very much! In hindsight I should have split my questions into two posts, as my post is technically still open despite your answer. But this settles my first question, thank you. $\endgroup$
    – Sam Forster
    Commented Aug 1, 2023 at 15:34
  • $\begingroup$ @SamForster You can still accept this answer and re-post your second question separately. $\endgroup$
    – Timothy Chow
    Commented Aug 2, 2023 at 12:05
  • $\begingroup$ @TimothyChow Thanks, I didn't know if that would be considered spam. $\endgroup$
    – Sam Forster
    Commented Aug 2, 2023 at 19:48

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