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Let $k$ be a field, $\bar{k}/k$ be a Galois extension with $G=Gal(\bar{k}/k)$ be an Abelian group(may be infinite). If $K,L$ are intermediate fields, denote $G_K=Gal(\bar{k}/K), G_L=Gal(\bar{k}/L)$. Is it true $G_K G_L=G_{L\cap K}$? Generally, is it true $\langle G_{K_i}\mid i\in I\rangle=G_{\cap K_i}$?(I guess it is true when thinking of $G_{\cap K_i}$ being topologically generated by$G_{K_i}$, which I don't know what does that exactly mean.)

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You have to be very careful: when you write $\prod G_{K_i}$, this is interpreted as the cartesian (exterior) product of the $G_{K_i}$; it's as if you had written "$G_K\times G_L$" instead of $G_KG_L$. For an infinite family of groups, you are better off writing $\langle G_{K_i}\mid i\in I\rangle$ and not using the $\prod$ sign. –  Arturo Magidin Oct 1 '11 at 21:04
    
Thanks, I've edited as you suggested. –  Li Zhan Oct 2 '11 at 0:57

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$G_K G_L=G_{L\cap K}$ is a consequence of Galois correspondence: $G_K G_L$ is the smallest group containing both $G_K$ and $G_L$, so it fixes the largest field contained in both $K$ and $L$, i.e. $K\cap L$. In the infinite degree case, Galois correspondence is between fields and closed subgroups; you are thus right, you need to take the closure of $\prod G_{K_i}$.

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Since $G$ is profinite, closed subgroups of $G$ are compact, so $G_K G_L$ is a continuous image of the compact set $G_K \times G_L$. So it's compact and thus it's closed -- you don't need to take the closure. –  David Loeffler Oct 1 '11 at 21:04
    
Thank you, you are absolutely right! –  Li Zhan Oct 2 '11 at 1:00

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