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Let $K/F$ be a field extension with $char(F)=0$ and degree $n>1$ - a composite number.

Can we always write $K = F_1 F_2$ when $F_1 \cap F_2 = F$ ?

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Can you do this for $\mathbb Q(\sqrt[4]2)$? No intermediate field of degree 2 other than $\mathbb Q(\sqrt 2)$ comes to my mind. – Hagen von Eitzen Jan 31 '13 at 16:52
up vote 4 down vote accepted

Well, a trivial solution would be taking $F_1 = F$ and $F_2 = K$. But there are examples where this is really the only possibility. For example, if $K/F$ is a Galois extension with Galois group $\mathbb Z/p^m \mathbb Z$ for a prime $p$, then the intermediate fields are linearly ordered by inclusion. If $F_1,F_2$ are two intermediate fields, w.l.o.g. $F_1 \subseteq F_2$, then $F_1 \cap F_2 = F_1$ and $F_1F_2 = F_2$. Therefore, the only possibility to have $F_1 F_2 = K$ and $F_1 \cap F_2 = F$ is choosing $F_1 = F$ and $F_2 = K$.

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