Let $L_1/K$ and $L_2/K$ be Galois extensions contained in a (bigger) field $M$ and let $F$ be the smallest subfield of $M$ containing both $L_1$ and $L_2$ (in other words, $F$ is the field generated by products of elements in $L_1$ with $L_2$). Prove that if $L_1 \cap L_2 = K$, then $|F:K| = |L_1:K||L_2:K|$.

If you wish, you can assume that $F$ is Galois and that "$\leq$" holds (I know how to prove both). In general I tried to use the Fundamental Theorem of Galois Theory in the setting that $L_1 \cap L_2 = K$ gives that the corresponding groups of $L_1, L_2$ generate $\Gamma(F:K)$, but this does not give anything about cardinalities.

Any help appreciated!

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    $\begingroup$ From the point of view of the Galois group, you have normal subgroups $N_1$ and $N_2$, corresponding to $L_1$ and $L_2$, respectively, whose intersection corresponds to the trivial subgroup (because the compositum of $L_1$ and $L_2$ is all of $F$), and that generate the entire Galois group (because their intersection is $K$). And then you are trying to show that under these conditions, $|G|=|G/N_1||G/N_2|$. $\endgroup$ – Arturo Magidin May 17 at 19:13
  • $\begingroup$ Good point. So any idea how to show |N_1||N_2| = |G| for non-intersecting normal subgroups which generate $G$? $\endgroup$ – DesmondMiles May 17 at 19:23
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    $\begingroup$ If you know $L_1 = K(a)$ (the primitive element theorem) then there is an elementary argument : let $L_2/K,K(a)/K$ be two finite extensions and $f \in K[x]$ the minimal polynomial of $a$. Then $[L_2(a):L_2] = [K(a):K]$ iff $f $ is irreducible in $L_2[x]$. If it is not the case let $f(x) = g(x)h(x) \in L_2[x]$ and $b$ be one of the coefficients of $g,h$ which is not in $K$. If also $K(a)/K$ is a normal extension then $f$ splits completely in $K(a)[x]$ thus $b \in K(a)$ and $K(a) \cap L_2 \supset K(b)$. $\endgroup$ – reuns May 17 at 19:31
  • $\begingroup$ Yep, that works, thank you! @ArturoMagidin I also figured out how to finish your idea, thank you! $\endgroup$ – DesmondMiles May 17 at 19:33

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