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Let $K$ be a number field and let $p$ be a prime number.

Let $L$ be a degree $p$ field extension of $K$.

Does there exist a degree $p$ field extension $M$ of $\mathbf{Q}$ such that $$M\otimes_{\mathbf{Q}} K = L?$$

If not, when does this happen?

Does it happen almost always?

I would be willing to hear about any positive thing that can be done in this direction. I'm thinking about Weil restrictions of varieties. That's where this question came up.

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1 Answer 1

up vote 4 down vote accepted

The question can be reformulated as follows.

Let $L$ be a number field and $p$ a prime. Let $K$ be a subfield of $L$ of index $p$. Does there exist a subfield $M$ of $L$ of order $p$ such that $MK = L$? (One direction of the equivalence is clear; in the other direction, if $MK = L$ then the natural map $M \otimes_{\mathbb{Q}} K \to MK$ is an isomorphism by comparing dimensions.)

The answer is not always. If $L$ is Galois over $\mathbb{Q}$ with Galois group $C_4$, then $L$ has a unique quadratic subfield, which we'll select as $K$, so it cannot admit a second quadratic subfield $M$. As an explicit example take $L = \mathbb{Q}(\zeta_5)$ and $K = \mathbb{Q}(\sqrt{5})$.

More generally, suppose that $L$ is Galois over $\mathbb{Q}$ with Galois group $G$ and write $|G| = pn$. Then by Cauchy's theorem $G$ has an element of order $p$, so the fixed field of this element is a suitable choice for $K$. In order for $M$ to exist, $G$ needs to have subgroups of index $p$, but it may not have any. For example, suppose $L$ has full Galois group $S_n$ (this is the generic case in the sense that the splitting field of a generic irreducible polynomial of degree $n$ has this property) and $n \ge 5$. By the simplicity of $A_n$ it follows that the minimal index of a subgroup of $S_n$ which is not $A_n$ is $n$, so if $p < n$ is an odd prime then $S_n$ has no subgroups of index $p$ and consequently $L$ has no subfields of degree $p$.

On the other hand, we have the following corollary of the fundamental theorem of Galois theory.

Proposition: Let $M, K$ be subextensions of a Galois extension $L$ (we do not name the base field as it is irrelevant here) with Galois group $G$. Then $MK = L$ if and only if $\text{Gal}(L/M) \cap \text{Gal}(L/K) = \{ e \}$ (where the intersections take place in $G$).

Thus if $|G| = pn$ where $p \nmid n$, then to find $M$ it is necessary and sufficient to find a subgroup of $G$ of index $p$. For example, this is possible if $n = q^k$ for some prime $q \neq p$ by the first Sylow theorem. More generally, it is necessary and sufficient to find a subgroup of index $p$ with trivial intersection with $\text{Gal}(L/K)$ (hence to write $G$ as an internal Zappa-Szép product of a cyclic group of order $p$ and some other subgroup).

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Simply amazing... –  Harry Aug 5 '12 at 17:52

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