Let $F$ be an arbitrary field such that $[\overline{F}: F]=\infty$. Here $\overline{F}$ denotes the algebraic closure. The question in the title of this post can be rephrased as:

Question 1. For each $n\in\mathbb{N}$, can we find a field extension $L$ of $F$ such that $[L:F]=n$?

Let me remark that the condition $[\overline{F}: F]=\infty$ is necessary. Indeed, by the Artin-Schreier theorem, the condition $[\overline{F}:F]<\infty$ in fact forces $[\overline{F}:F]=2$, in which case $F$ can only have extensions of degree at most $2$. I suspect that Question 1 may have a negative answer for small values of $n$, so I am happy to consider the weaker problem:

Question 2. Does there exist a number $n_0$ (that only depends on $F$) such that for each $n\geq n_0$, there exists a field extension $L$ of $F$ such that $[L:F]=n$?

In case there are any separability issues, feel free to assume that $F$ is perfect.

Added later: Looks like this question has already been asked. See this MSE thread. Before we close my question as a duplicate (which is the right thing to do), I was wondering if anyone has any additional details or more in the case when $F$ is a perfect field. Having the details of the construction is especially appreciated!

  • $\begingroup$ Reusing the argument in Starfall's answer (in the other thread) shows that you can choose a set of prime numbers $S$ and then a field $F$ such that if $[K:F]=n$ then all the prime factors of $n$ are in $S$. Furthermore, that is the only constraint on $n$ in that setting. $\endgroup$ Aug 26, 2018 at 7:29
  • $\begingroup$ Starfall's argument was for a bit more general case. What I had in mind was to use as $F$ the subfield of the algebraic closure $\overline{k}$ of a finite field $k$. Then $Gal(\overline{k}/k)=\prod_p\Bbb{Z}_p$, and we can let $F$ be the fixed field of the closed subgroup $\prod_{p\notin S}\Bbb{Z}_p$. $\endgroup$ Aug 26, 2018 at 7:37
  • $\begingroup$ @JyrkiLahtonen: Thank you! Feel free to post your thoughts in the answer box below. $\endgroup$
    – Prism
    Aug 27, 2018 at 0:33
  • $\begingroup$ Maybe you can take $F$ to be a maximal subfield of $\mathbb C$ not containing $\sqrt{2}$. As I recall every finite extension of $F$ is cyclic of degree $2^n$ for some $n$. $\endgroup$
    – SMM
    Sep 1, 2018 at 11:49
  • $\begingroup$ @SMM: Thanks! That would be a really nice answer for this question! So I would be very interested in seeing the proof for this assertion. $\endgroup$
    – Prism
    Sep 1, 2018 at 15:59

1 Answer 1


Let $\bar{\mathbb{Q}}$ be the algebraic closure of $\mathbb Q$. Consider a maximal subfield $F$ of $\bar{\mathbb{Q}}$ not containing $\sqrt{2}$; this is straight-forward by Zorn's lemma. Clearly, $F$ is not algebraically closed and its algebraic closure $\bar F$ equals $\bar{\mathbb{Q}}$.

First, let us note that $[\bar{\mathbb{Q}}:F]=\infty$. If not, by Artin-Schreier we have $\bar{\mathbb{Q}}=F(i)$; in particular $i\notin F$. So $\sqrt 2\in F(i)$ and we may write $\sqrt 2=a+bi$ for some $a,b\in F$. By squaring we get $2= a^2-b^2+2abi$, and since $i\notin F$ we have $2ab=0$. It is not possible that $b=0$ because then $\sqrt 2=a\in F$, so $a=0$, i.e. $\sqrt 2=bi$. So $i\sqrt 2=-b\in F$. But now we can write $0$ as the sum of squares $(i\sqrt 2)^2+1^2+1^2$ in $F$ contradicting Artin-Schreier.

Now, if $F\lneq E\leq\bar{\mathbb{Q}}$ is a finite extension, then by the choice of $F$ (the maximality condition) we have $\sqrt 2\in E$ so $F\lneq F(\sqrt 2)\leq E$, hence $[E:F]$ is even by the chain rule. This already says that $F$ is an example of a perfect field (even of characteristic $0$) for which assertions from Q1 and Q2 don't hold. But one can prove more: $[E:F]$ must be of degree $2^n$ for some $n$. Let $E'$ be the normal closure of $E$ over $F$, so $E'/F$ is Galois. Write $[E':F]=2^mk$, where $m\geq 1$ (we saw that all extensions are of even degree) and $k$ is odd. By Galois correspondence there exists a extension $L/F$ of degree $k$, corresponding to a Sylow 2-subgroup of $Gal(E'/F)$. Since $k$ is odd and proper extensions of $F$ are of even degree, it must be $L=F$ and $k=1$. So $[E':F]=2^m$ and by $F\leq E\leq E'$ we have $[E:F]=2^n$ for some $n$.

  • $\begingroup$ Thank you so much! This is a wonderful argument, and you explained it clearly. I'll wait a couple of days before accepting and awarding the bounty (in case other answers show up). $\endgroup$
    – Prism
    Sep 2, 2018 at 11:24
  • $\begingroup$ By the way, do you know if $F$ actually has a field extension of degree $2^n$ for every $n\geq 1$? I am just curious! $\endgroup$
    – Prism
    Sep 5, 2018 at 14:41
  • $\begingroup$ @Prism Yes. Note that $[\bar{\mathbb Q}:F]=\infty$ says that for every $N$ you have some extension of degree greater than $N$. Such an extension $E/F$ as we have seen is of degree $2^n$ for some $n$. By considering the normal closure, we may assume that $E/F$ is Galois. Since $F$ has a unique extension of order $2$ ($F(\sqrt 2)$), by correspondence $Gal(E/F)$ has a unique subgroup of index $2$. It is known that $2$-groups with a unigue subgroup of index $2$ are cyclic. So $Gal(E/F)$ has a subgroup of each index $2^m$ for $m\leq n$, and by correspondence we have extension of degree $2^m$. $\endgroup$
    – SMM
    Sep 5, 2018 at 14:52
  • $\begingroup$ Beautiful! Thank you! $\endgroup$
    – Prism
    Sep 5, 2018 at 16:16

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