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Let $K$ be an algebraic number field, i.e. a finite extension of $\mathbb{Q}$. Let $L/K$ be an infinite dimensional algebraic extension of $K$. Are there infinitely many infinite dimensional intermediate subfields of $L/K$?

It's easy to see that there are infinitely many finite dimensional intermediate subfileds of $L/K$, but I have no idea about the question.

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Not necessarily. Clearly it is possible for there to be infinitely many subfields of infinite degree, take $\overline{\mathbf{Q}}/\mathbf{Q}$ for example. On the other hand, there exist infinite extensions of any number field $K$ with no intermediate fields of infinite degree over $K$.

For a concrete example, if $L = K_\infty/K$ is a $\mathbf{Z}_p$-extension (i.e. its Galois group is isomorphic to $\mathbf{Z}_p$, the $p$-adic integers). Such an extension always exists, e.g. as a finite index subfield of $K(\mu_{p^\infty})$, the field given by adjoining all $p$-power roots of unity. Since $\mathbf{Z}_p$ has no nontrivial closed subgroups of infinite index, the fundamental theorem of Galois theory implies that there are no intermediate fields of infinite degree over $K$.

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Consider the set $P$ of prime numbers. The number of infinite subsets of $P$ it self is infinite, in fact uncountable. For each such infinite subset $A\subset P$ consider the extension of $K_A$ obtained by adjoining all square roots $\surd p,\ p\in A$. It can be shown that they are all distinct subfields. %K% itself may contain atmost a finite number of square roots of $p$, but it will not affect. (For example, to start with we can redefine $P$ to the infinite set of primes such that $\surd p$ is not in $K$).

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  • $\begingroup$ How do you prove that $K_A \subset L$? $\endgroup$ – Makoto Kato Mar 12 '15 at 2:57
  • $\begingroup$ I misunderstood the question: the construction I gave merely gives one example of infinite extension containing infinitely many subextensions. $\endgroup$ – P Vanchinathan Mar 12 '15 at 3:00

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