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Concerning finite field extensions, with $E_1E_2$ the compositum:

I proved $$[E_1E_2:K]=[E_1E_2:E_2] \cdot [E_2:K] \leq [E_1:K] \cdot [E_2:K]$$ Is it true that $[E_1E_2:K]$ has to divide $ [E_1:K] \cdot [E_2:K]$? Or is there a counterexample?

I proved that $E_1E_2$ is normal over $K$ if $E_1$ and $E_2$ are. I assume the other way around does not necessarily have to be true, does anyone know a counterexample?

I proved that Gal $(E_1E_2/K)$ is abelian if Gal $(E_1/K)$ and Gal $(E_2/K)$ are abelian by defining the following injective homomorhpism.

$$\Phi : \textrm{Gal}(E_1E_2/K) \to \textrm{Gal}(E_1/K) \times \textrm{Gal}(E_2/K)$$ $$\sigma \mapsto (\sigma|_{E_1},\sigma|_{E_2})$$

I assume the other way around does not necessarily have to be true, does anyone know a counterexample?

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2 Answers 2

up vote 1 down vote accepted
  1. $[E_1E_2:K]$ divides $[E_1:K][E_2:K]$
  2. Converse of «$E_1/K$, $E_2/K$ normal $\Rightarrow E_1E_2/K$ normal»
  3. Converse of «$G(E_1/K)$, $G(E_2/K)$ abelian $\Rightarrow G(E_1E_2/K)$ abelian»

Rene's answer provides a counterexample to the first two, with $E_1=\Bbb Q(\sqrt[3]{2})$, $E_2=\Bbb Q(\zeta_3\sqrt[3]{2})$.

The third is true, because $G(E_1/K)$ and $G(E_2/K)$ are quotients of $G(E_1E_2/K)$.

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You mean to say $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\zeta_3 \sqrt[3]{2})$. –  Rene Schipperus May 26 at 2:47

The following considerations may be helpful. Let $f(x)$ be an irreducible polynomial of degree 3, whose splitting field has degree 6. Let $\alpha_1$ and $\alpha_2$ be two distinct roots and $E_i=\mathbb{Q}(\alpha_i)$. Then each $E_i$ has degree 3 but the composite is the splitting field of degree 6.

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This seems correct, thanks! –  user132531 May 26 at 1:53

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