Suppose $k\subset E$ is Galois and let $F$ and $F'$ be two intermediate fields. Let $FF'$ be the smallest intermediate field containing $F$ and $F'$. Also let $G$ denote $\text{Gal}(E/k)$.

Let $H=\text{Gal}(E/F)$ and $H'=\text{Gal}(E/F')$, prove that $\text{Gal}(E/F\cap F')=\langle H, H'\rangle \subset G$.

Well, so far I've been able to prove one of the inclusions:

First remember that $$ \langle H, H'\rangle = \bigcap_{\substack{L \leq G, \\ H \cup H' \subseteq L. }}L $$ Having the fundamental theorem of Galois theory, is obvious that $H \subseteq \text{Gal}(E/F\cap F')$ and $H' \subseteq \text{Gal}(E/F\cap F')$, then $H\cup H' \subseteq \text{Gal}(E/F\cap F')$; and since $\text{Gal}(E/F\cap F') \leq G$, we have $\langle H, H'\rangle \subseteq \text{Gal}(E/F\cap F')$.

Maybe proving that $\langle H, H'\rangle \supseteq \text{Gal}(E/F\cap F')$ is just as straightforward, but I have not been able to do it.

  • $\begingroup$ I guess you mean by $H$ galois group of $E$ over $F$ and by $H'$ galois group of $E$ over $F'$.. $\endgroup$
    – user87543
    Dec 8, 2013 at 5:28
  • $\begingroup$ Yes, I'm sorry, I forgot to wrote it. And I've just edited the question adding that. $\endgroup$
    – pjox
    Dec 8, 2013 at 5:33
  • $\begingroup$ You define in the first line $\;FF'\;$...and then it doesn't appear anymore! Shall we understand that $\;F\cap F'\;$ really means $\;FF'\;$ ? $\endgroup$
    – DonAntonio
    Dec 8, 2013 at 6:45
  • $\begingroup$ No, they are different sets, $FF'$ is a field containing both $F$ and $F'$; while $F\cap F'$ is in fact a subfield of both $F$ and $F'$. To summarize $FF' \supseteq F \supseteq F\cap F'$ and $FF' \supseteq F' \supseteq F\cap F'$. I just mentioned $FF'$ because I thought it might be useful, I don't know, perhaps for completing the diamond diagram you get when you actually make the diagram of extensions... but so far I haven't found it useful myself. :( $\endgroup$
    – pjox
    Dec 8, 2013 at 7:08

1 Answer 1


If you already know the Galois correspondence (subfields of $E$ correspond 1-1 with subgroups of $G$, and the correspondence is inclusion reversing), then since $F \cap F'$ is clearly the largest subfield contained in both $F$ and $F'$, the corresponding subgroup must be the smallest subgroup containing both $H$ and $H'$, which is $\langle H, H' \rangle$.

  • $\begingroup$ Ohhh,of course! Since $F\cap F'$ is the largest subfield contained in both $F$ and $F'$, and $E^{\langle H, H'\rangle}$ is clearly a subfield of both $F$ and $F'$, then $E^{\langle H, H'\rangle} \subseteq F\cap F'$; and therefore, by the Galois correspondence we have $\langle H, H'\rangle \supseteq \text{Gal}(E/F\cap F')$. $\endgroup$
    – pjox
    Dec 8, 2013 at 7:31

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