# Cardinalities of Galois extensions and compositum

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!

• 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, $|G|=|G/N_1||G/N_2|$. – Arturo Magidin May 17 at 19:13
• Good point. So any idea how to show |N_1||N_2| = |G| for non-intersecting normal subgroups which generate $G$? – DesmondMiles May 17 at 19:23
• 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 $K(a) \cap L_2 \supset K(b)$. – reuns May 17 at 19:31
• Yep, that works, thank you! @ArturoMagidin I also figured out how to finish your idea, thank you! – DesmondMiles May 17 at 19:33