Let $K_1/K$ and $K_2/K$ two galois extension. I have a theorem that says that $K_1K_2/K$ is a galois extension and that if $K_1\cap K_2=K$ then $$\Gal(K_1K_2/K)\cong \Gal(K_1/K)\times \Gal(K_2/K).$$

I would like to generalize it for 3 extension field. i.e.

Let $K_1/K,K_2/K$ and $K_3/K$ three galois extension. Then, $K_1K_2K_3/K$ is galois and if $K_i\cap K_j=K$ for all $i\neq j$ then $$\Gal(K_1K_2K_3/K)\cong \Gal(K_1/K)\times \Gal(K_2/K)\times \Gal(K_3/K).$$

Now I tried to prove it using the first theorem below, but I can't conclude. The fact that $K_1K_2K_3/K$ is a galois extension is obvious. But for the rest, I really have difficulties.

My idea was to set $L=K_1K_2$, and thus show that $$\Gal(LK_3/K)\cong \Gal(L/K)\times \Gal(K_3/K),$$ using the 1st theorem below, but I can't prove the $L\cap K_3=K$.

Any idea ?

  • 3
    $\begingroup$ You have to assume $K_i \cap (K_j K_r) = K$ for all $i \neq j \neq r$. For instance, consider $\mathbb Q(\sqrt 2), \mathbb Q(\sqrt 3)$ and $\mathbb Q(\sqrt 6)$. $\endgroup$ – Watson Dec 29 '15 at 22:25

In order to have the isomorphism

$$\Gal(K_1K_2K_3/K)\cong \Gal(K_1/K)\times \Gal(K_2/K)\times \Gal(K_3/K),$$

you have to assume that each $K_i$ is linearly independant not only to each $K_j$, but to each product $K_j K_r$ with $i \neq j \neq r$.

For instance, let consider $K_1 = \mathbb Q(\sqrt 2), K_2 = \mathbb Q(\sqrt 3)$ and $K_3 = \mathbb Q(\sqrt 6)$ : if you take any two of them, they have trivial intersection. But $K_1K_2K_3 = \mathbb Q(\sqrt 2, \sqrt 3)$ ; therefore, $\Gal(K_1K_2K_3 / \mathbb Q)$ has cardinality 4, whereas $\Gal(K_1/\mathbb Q)\times \Gal(K_2/\mathbb Q)\times \Gal(K_3/\mathbb Q)$ has cardinality 8.

Let suppose $K_i \cap K_j K_r = K$ for all indices $i \neq j \neq r$. Then, as you already showed the case $n=2$ :

$$\Gal(K_1(K_2K_3) / K) \cong \Gal(K_1/K) \times \Gal(K_2K_3/K) \\ \cong \Gal(K_1/K)\times \Gal(K_2/K)\times \Gal(K_3/K)$$

  • $\begingroup$ Thank you for your response. As you present the thing, it looks like only $K_1\cap (K_2K_3)=K$ (and not $K_iK_j\cap K_r$ for all $i\neq j\neq r$. What do you think ? $\endgroup$ – user301068 Dec 29 '15 at 22:39
  • $\begingroup$ I also used $K_2 \cap K_3 = K$ in the last isomorphism. $\endgroup$ – Watson Dec 29 '15 at 22:51

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