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Assume that $L/K$ is an extension of fields and $[L:K]=n$, with $n$ composite. Assume that $p\mid n$, can we always produce a subextension of degree $p$ and if not under what conditions can it be done? I would guess this is very false, but I couldn't come up with any trivial counterexamples.

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If $L/K$ is Galois with a Galois group which is non-abelian and simple, then $2|n$ but there is no extension $K'/K$ contained in $L$ of degree $2.$ – jspecter Dec 3 '11 at 7:51
up vote 4 down vote accepted

We know that equations of degree 2 and 3 have solutions in radicals, but most equations of degree 6 (or any degree 5 or greater) don't. So let $f$ be a polynomial over $K$ of degree 6 not solvable in radicals, let $\alpha$ be a root of $f$ in some extension, let $L=K(\alpha)$. If there were an intermediate field $E$ of degree 2 or 3, then you could express $\alpha$ in radicals over $E$, and those radicals in radicals over $K$, contradicting the assumption that you couldn't solve $f$ in radicals over $K$.

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You are very right when you write "I would guess this is very false": here is a precise statement.

Proposition 1
For any $n\gt 1$ there exists a field extension $\mathbb Q\subset K$ of degree $[K:\mathbb Q]=n \:$ with no intermediate extension $\mathbb Q \subsetneq k\subsetneq K$.
Let $\mathbb Q \subset L$ be a Galois extension with Galois group $S_n$, the permutation group of $n$ elements .
The Galois correspondence associates to the subgroup $S_{n-1}\subset S_n$ an extension $\mathbb Q \subset K$. The intermediate fields $\mathbb Q \subsetneq k\subsetneq K$ bijectively correspond (by Galois theory again)
to intermediate subgroups $S_{n-1}\subsetneq G\subsetneq S_n$. However an easy exercise in group theory shows that no such group exists and thus no intermediate field $k$ exists either.

An analogous problem
It is well known that constructible numbers $c\in \mathbb C$ (those that can be constructed by straightedge and compass) have degree a power of two: $[\mathbb Q(c):\mathbb Q]=2^r$ . However the converse is false :
Proposition 2
For every $n=2^r\geq 4$ there exists a non-constructible number $c\in \mathbb C$ which is not constructible even though it has degree $n=2^r$.
Take an irreducible polynomial $P(X)\in \mathbb Q[X]$ whose Galois group is $S_n$. I claim that one of its roots $c $ is the required nonconstructible number of degree $n$.
Else all its roots $c_1, ...,c_n$ would be constructible and all the numbers of its splitting field $K=\mathbb Q(c_1,...,c_n)$ would be constructible too (the constructible numbers form a field).
But this is contradictory because a primitive element $c_0$ of $K$ (i.e. $K=\mathbb Q(c_0$)) should then also have degree some power of $2$, whereas its degree is actually $n!$, a number divisible by $3$ since $n\geq 4$.

The existence of polynomials over $\mathbb Q$ with Galois group $S_n$ is proved for example in Dummit-Foote's Abstract algebra, Chapter 14, Proposition 42.

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Dear Georges: +1. A detail: Our friend pki (who asked the question) didn't require that the ground field be $\mathbb Q$. This makes it much easier to build a Galois extension with Galois group $S_n$. --- By the way, I noticed that Theorem 2.2 p. 4 in these notes of our friend Keith Conrad is very similar to your answer; but I like your wording very much! – Pierre-Yves Gaillard Dec 3 '11 at 10:06
Dear @Pierre-Yves, yes it is true, pki didn't require anything of the ground field . I have added an illustration showing the interest of having polynomials over $\mathbb Q$ with Galois group $S_n$. Anyway, thanks a lot for your interest in this answer and the reference to Keith's handout. – Georges Elencwajg Dec 3 '11 at 23:50

Suppose $L/K$ is Galois with group $G$. Then subextensions of degree $m$ over $K$ correspond to normal subgroups of $G$ of index $m$. Given an abelian group you can always find subgroups of order $p$, where $p$ is a prime divisor of $\vert G \vert =n$. So to find a counterexample, you should (with $L/K$ Galois) start with non-abelian Galois groups. I'm pretty sure there exists a group of order $n$ which doesn't have a normal subgroup of index $m$ for all $m\vert n$. (Note that if $m$ is prime, such a (not necesarily normal) subgroup exists).

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Just subgroups. They don't have to be normal. – jspecter Dec 3 '11 at 8:10

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