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It is well-known that a finite field has a unique extension of degree $n$, in a given algebraic closure, for every $n \geq 1$.

Is there a field $F$ such that, in some given algebraic closure of $F$, there are exactly two extensions of $F$ of degree $n$, for every $n \geq 3$ (or $n$ big enough)?

(Here, "exactly two" means that $F \subset K, L \subset \overline F$ simply satisfy $K \neq L$. In particular, I'm not identifying isomorphic fields or $F$-algebras.)

A more general question would be to replace "exactly two" by "exactly $N$" (for some fixed integer $N > 1$). Actually, the most general setting is the study of the maps $d_F$ (resp. $d_{F, \text{iso}}$, resp. $d_{F, F\text{-iso}}$) which assign, to every integer $n \geq 2$, the cardinal of the set of (resp. field isomorphism classes, resp. $F$-algebras isomorphism classes of) extensions of $F$ of degree $n$. For instance:

  • Can $d_F$ be a constant map $n \mapsto c$ for some $1<c<\aleph_0$ ?

  • Can $d_F$ be a bounded function (different from the constant function $1$) ?

  • Can $d_F$ take both finite and infinite values?

It can be shown that a local field of characteristic $0$ has finitely many extensions of some given degree, but the number might grow with the degree (instead of being bounded by $N$ as I want).

Thank you!

Watson
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    Could a field really have exactly two quadratic extensions?? – Angina Seng Jun 24 '18 at 15:46
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    If $F^{\times} / F^{\times, 2}$ is finite, it must have an odd number of non-trivial elements. I could have asked maybe "exactly three extensions of degree $n \geq 2$" ? I hope that choosing "exactly two extensions of degree $n \geq 3$" might be interesting. – Watson Jun 24 '18 at 16:05
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    Reference for the case of quadratic extensions : https://math.stackexchange.com/questions/1725761/ – Watson Jun 24 '18 at 18:46
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    The general question might include https://mathoverflow.net/questions/16778/what-are-the-possible-sets-of-degrees-of-irreducible-polynomials-over-a-field (which focuses on $d_F(n)=0$ or $\neq 0$), at least if $F$ is a perfect field. – Watson Jun 24 '18 at 19:14
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    Using the answer here, or the MO question 16778 cited above, one can construct, for every fixed prime $p$, a field $F = F(p)$ with $d_F(n) > 0 \iff n$ is a power of $p$. But this doesn't give any upper bound on $d_F(n)$. – Watson Jun 25 '18 at 15:40
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    The local field $\Bbb{F}_p((x))$ has infinitely many non-isomorphic extensions of degree $p$. See here. – Jyrki Lahtonen Jun 26 '18 at 06:44
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    Have you tried to exploit Kummer theory or something similar? The idea being that if $K^/(K^)^n$ is not cyclic, it has "many" (generically, more than 2) subgroups of order $n$, and so if $K$ contains the appropriate roots of unity it already has "many" abelian extensions of degree $n$. This might at least exclude some cases. – Torsten Schoeneberg Jun 27 '18 at 00:03

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