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Is there a field extension $K/F$ of degree $n>1$ with $n$ not prime, such that every element $x \in K \setminus F$ has degree $n$ ? What happens if $F$ has characteristic $0$?


The motivation of this question is to notice that if $[K:F]=p$ is prime (or $=1$), then any element $x \in K\setminus F$ has degree $p$, and in particular $K/F$ has no non trivial subextension (actually there is a proper subextension $F \subsetneq L \subsetneq K$ iff there is some $x \in K$ of degree $\neq 1, \neq n$). My question is about the converse of this property.

Obviously if $n$ is not prime, we can find some non trivial $F$-vector subspaces, but they might not be subfields of $K$. I tried to work with subfields fixed by subgroups of $\mathrm{Aut}_F(K)$, but this group may be trivial!

If $K/F$ is separable (e.g. $\mathrm{char}(F)=0$), then we can write $K=F(a)$. But $a^2$ or some polynomials in $a$ might also have degree $n$ over $F$, and I don't see how to get a polynomial $P(a) \in K \setminus F$ in $a$ with degree $<n$ (which would solve my problem).

I know that the degree of any $x \in K$ over $F$ must divides $n$, but the converse doesn't hold (even if $K/F$ is Galois ; related to the fact that $A_4$ has no subgroup of order $6$).

I know that $F(X)/F$ has no non-trivial finite subextension (as well as any purely transcendental extension), even if $F(X^2)/F$ is a proper (infinite) subextension.

[Notice that a Galois extension $K/F$ has no non-trivial Galois subextension $L/F$ iff $\mathrm{Gal}(K/F)$ simple group].

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    $\begingroup$ Yes. This is possible. For a simple example consider the following. Assume that $E$ is the splitting field of a quartic over $F$ such that $Gal(E/F)\simeq S_4$. If we let $K$ be the fixed field of a subgroup $S_3\le S_4$, then $[K:F]=4$, but there are no intermediate fields between $K$ and $F$ because $S_3$ is a maximal subgroup of $S_4$. You can do the same with other composites in place of $4$. I won't post an answer, because the question has been handled earlier. $\endgroup$ – Jyrki Lahtonen Feb 2 '17 at 10:54
  • $\begingroup$ @JyrkiLahtonen : Thank you very much! It wasn't difficult actually but I just didn't think about maximal subgroups... $\endgroup$ – Watson Feb 2 '17 at 10:56
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    $\begingroup$ See for example here. Ofir (+1) had the same idea. In case you wonder about whether $S_{n-1}$ (= a point stabilizer of $S_n$ acting on $\{1,2,\ldots,n\}$) is maximal, I sketched an argument in a comment under this answer. $\endgroup$ – Jyrki Lahtonen Feb 2 '17 at 10:57
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    $\begingroup$ Related (particular case): math.stackexchange.com/questions/1018535. $\endgroup$ – Watson Feb 2 '17 at 13:25
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    $\begingroup$ Also useful: math.stackexchange.com/questions/1865538 $\endgroup$ – Watson Feb 2 '17 at 21:00
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If you know about Galois theory, then start with some $G$ Galois extension $K/F$ and let $H\leq G$ be a maximal subgroup. Then $K^H/F$ doesn't have any nontrivial subextension, so that every element in it which is not in $F$ has degree $[K^H:F]=[G:H]$. Now you have the problem of finding a group which has a maximal subgroup of index which is not prime. For example, over $\mathbb{Q}$ you have Galois extensions with the symmetric group, and these always have maximal subgroups of index which is not prime. More specifically, $S_{n-1}$ is maximal in $S_n$ of index $n$ which you can choose such that $n$ is not prime.

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