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$\mathbb{F}/\mathbb{Q}$ is a finite extension. Denote $\dim_{\mathbb{Q}}\mathbb{F}=n$. How (possibly) many subfields does $\mathbb{F}$ has that extend $\mathbb{Q}$?

Thoughts:

if $A=\{v_1,...,v_n\}$ is a basis of $\mathbb{F}$ over $\mathbb{Q}$, then for any $i \in \{1...n\}$, $\mathbb{Q}(v_i)$ is a different extension of Q, and a subfield of $\mathbb{F}$... But there can exist two different subsets of $A$, denoted $B,C\subseteq A$ such that $\mathbb{Q}(B)=\mathbb{Q}(C)$ (for example: $\mathbb{Q}(\sqrt[4]{2})=\mathbb{Q}(\sqrt{2},\sqrt[4]{2})$), so there at least n subfields. but that number cannot exceed $2^n$ (all subsets of A)...

How should I approach this issue?

user26857
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Daniel
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  • Galois theory does give you some kind of answers to this question, but it is not clear to me what the answer will look like. But your approach won't give much information. For example, what about the subfields generated by linear combinations such as $v_1+kv_2$ for varying $k\in\Bbb{Q}$? – Jyrki Lahtonen Jun 19 '15 at 07:48
  • you are right, but I sense that F should then be a simple extension, F=Q(x) for some x... – Daniel Jun 19 '15 at 07:52
  • What Galois theory says is the following: Let $N$ be a normal closure, so $N/\Bbb{Q}$ is Galois, and $\Bbb{F}\subseteq N$. Let $G=Gal(N/\Bbb{Q})$ and $H=Gal(N/\Bbb{F})$. Then the number of intermediate fields between $\Bbb{Q}$ and $\Bbb{F}$ is equal to the number of intermediate subgroups $K$, $H\subset K\subset G$. – Jyrki Lahtonen Jun 19 '15 at 07:52
  • Does this statement has a name? =] – Daniel Jun 19 '15 at 07:53
  • As evidence that the question is a bit trickier than one might initially think let me point out the following. If we replace $\Bbb{Q}$ with a non-perfect field it is possible to have infinitely many intermediate fields. See e.g. here. – Jyrki Lahtonen Jun 19 '15 at 07:55
  • But there is a (natural) supremum on that number when dealing with Q, am I wrong? – Daniel Jun 19 '15 at 07:58
  • Daniel, that statement is just part of the Galois correspondence. Also see Pete L. Clark's explanation. – Jyrki Lahtonen Jun 19 '15 at 08:00
  • Yes, over the rationals we do get an upper bound $2^{n-1}-2$ as follows. The above subgroup $H$ is of index $n$ in $G$. There are $2^n$ collections of cosets of $H$ in $G$, $2^{n-1}$ of those contain $H$. Any intermediate subgroup $K$ is a non-trivial such collection, so there are at most $2^{n-1}-2$ such intermediate subgroups. This bound is obviously not tight. To get a more realistic upper bound we need to summon a group theorist (like Derek Holt) to the scene. – Jyrki Lahtonen Jun 19 '15 at 08:03
  • At the other end it is known (IIRC?) that for all $n$ we can find an extension $\Bbb{F}$ of degree $n$ such that there are no intermediate fields between $\Bbb{F}$ and $\Bbb{Q}$. This happens for example when $G=S_n$ and $H=S_{n-1}$. There are no subgroups in between $H$ and $G$, so no intermediate fields either. The question mark here is about me being ignorant about the state of the art in inverse Galois theory. Does there exists an extension $N$ such that $Gal(N/\Bbb{Q})=S_n$? When $n$ is a prime this is easy, but I don't recall a general result. – Jyrki Lahtonen Jun 19 '15 at 08:14
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    See this related question. – Jyrki Lahtonen Jun 19 '15 at 10:03
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    Note that OP already asked this question once before, http://math.stackexchange.com/questions/1324251/what-is-the-possible-number-supremum-of-subfields-of-mathbbf – Gerry Myerson Jun 19 '15 at 10:23
  • Experts in group theory helped me in coming up with an upper bound $2^{(\log_2n)^2}$. See this question for more details. – Jyrki Lahtonen Jun 26 '15 at 16:36

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