Let $F/K$ be a Galois extension and let $E_1, E_2$ be intermediate subfields of $F/K$ with corresponding subgroups $H_1,H_2$ of $Aut(F/K)$. Prove that the intersection of $E_1$ and $E_2$ corresponds to $<H_1H_2>$,and $E_1E_2$ corresponds to the intersection of $H_1$ and $H_2$.
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Let $H = \langle H_1H_2\rangle$, then for any $\sigma \in H_1H_2$, and $x\in E_1\cap E_2$, $\sigma(x) = x$. So $$ E_1\cap E_1\subset K^H $$ Also, since $H_1 < H$, $K^H \subset K^{H_1} = E_1$. Similarly, $K^H \subset E_2$. Hence $$ E_1\cap E_2 = K^H $$
Note that $E_1E_2$ is, be definition, the smallest field containing both $E_1$ and $E_2$. Hence, if $H = H_1\cap H_2$, then $$ E_1E_2 \subset K^H \Rightarrow H \subset Aut(F/E_1E_2) \qquad (\ast) $$ Also, if $\sigma \in Aut(F/E_1E_2)$, then $$ \sigma\lvert_{E_1E_2} = id \Rightarrow \sigma\lvert_{E_1} = id \Rightarrow \sigma \in H_1 $$ and similarly $\sigma \in H_2$. Hence, $$ Aut(F/E_1E_2) \subset H \qquad (\ast\ast) $$ By $(\ast)$ and $(\ast\ast)$, we see that $$ E_1E_2 = K^H $$
Prahlad Vaidyanathan
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Thank you .. you helped alot – user291864 Dec 01 '15 at 06:06
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@Prahlad Vaidyanathan: Is it true that the Galois group of $E_1 \cap E_2$ is the subgroup $<H_1,H_2>$ ? It is clear that $<H_1,H_2> \subset \ Gal (F/(E_1 \cap E_2 ).$ I don't know if the converse is true ! – user371231 Mar 01 '18 at 10:44
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It follows from the Galois correspondence and what is proved in (1) above. – Prahlad Vaidyanathan Mar 01 '18 at 16:25