# Compositum of fields with trivial intersection

Let $E/F$ be a finite extension. Let $L,K$ be two intermediate fields with $L\cap K = F$, and also $$[L : F] [K:F] = [E:F].$$ Must it hold that the compositum $LK$ equals $E$? If we assume that $E/F$ and $K/F$ are Galois, then this follows from basic Galois theory. Namely, since $K/F$ is Galois we have $$\mathrm{Gal}(E/F) = \mathrm{Gal}(E/K\cap L) = \mathrm{Gal}(E/K)\mathrm{Gal}(E/L)$$ and $$\mathrm{Gal}(E/KL) = \mathrm{Gal}(E/K) \cap \mathrm{Gal}(E/L).$$ We have $$|\mathrm{Gal}(E/K)\mathrm{Gal}(E/L)| = \frac{|\mathrm{Gal}(E/K)||\mathrm{Gal}(E/L)|}{| \mathrm{Gal}(E/K) \cap \mathrm{Gal}(E/L)|}.$$ Therefore we must have

$$|\mathrm{Gal}(E/KL)| = \frac{|\mathrm{Gal}(E/K)||\mathrm{Gal}(E/L)|}{|\mathrm{Gal}(E/K)\mathrm{Gal}(E/L)| }= \frac{[E : K][E : L]}{[E: F]} = 1.$$

However, I am not sure whether this is still true if either $E$ or $K$ is not assumed Galois over $F$. Does anyone have any ideas?

• One thing I do know is that if there are any counterexamples, we must have that $[L:F]$ and $[K:F]$ are not relatively prime. Commented Mar 8, 2013 at 5:01

This is false in general. Let me begin by offering an example from the side of finite groups (the Galois correspondence is inclusion reversing, so...). Let $G=D_8$ be the group of symmetries of a regular octagon. We know that $G$ is generated by a rotation $r$ of order $8$, and a reflection $s$ of order $2$ such that $srs=r^{-1}$, and it follows that $sr^i=r^{8-i}s$ for all integers $i$. It follows that $|G|=16$ and that $r^4$ is the only non-trivial element in the center of $G$. Geometrically this is obvious, because $r^4$ is the rotation by the angle $\pi$, i.e. scalar multiplication by $-1$.

Inside $G$ we have two subgroups $H_1=\{1,s,r^4, sr^4\}$ and $H_2=\{1,sr,r^4,sr^5\}$. Both of these are generated by reflections w.r.t. to a pair of orthogonal lines - one pair gotten by rotating the other by the angle $\pi/8$. Geometrically: one pair of the the orthogonal lines connects two pairs of antipodal vertices of the octagon, the other pair connects antipodal midpoints of edges. Anyway, the key observations are the following:

1. $\langle H_1,H_2\rangle=G$. This is because $r=s(sr)$ is in the subgroup generated by both $H_1$ and $H_2$. In other words no proper subgroup of $G$ contains both $H_1$ and $H_2$.
2. $H_1\cap H_2=\{1,r^4\}$ is non-trivial.

Translating this to field extensions forming a counterexample is straightforward. Let $E/F$ be a Galois extension with Galois group $G$. Let $K=\operatorname{Inv}(H_1)$ and $L=\operatorname{Inv}(H_2)$. The intersection $L\cap K$ is then invariant under $H_1$ and $H_2$, hence under all of $G$, so $L\cap K=F$. OTOH the compositum $LK$ is the smallest field containing both $L$ and $K$, and thus the field of invariants of $H_1\cap H_2$. Here $[K:F]=[G:H_1]=4$, $[L:F]=[G:H_2]=4$, $[E:F]=|G|=16=4\cdot4$, but $[LK:F]=[G:(H_1\cap H_2)]=8<16$.

Translating everything to the group side by an application of the Galois correspondence was the key. After that it was just tinkering, trial and error.

Observe that it suffices to find a finite group $G$ together with two subgroups $H_1$ and $H_2$ such that i) $G=\langle H_1, H_2\rangle$, ii) $|G|=[G:H_1]\cdot[G:H_2]$ and ii) $|H_1\cap H_2|>1$.

• Can you add concrete examples for $E,K,L$ with $F=\mathbb{Q}$? Commented Mar 10, 2013 at 17:10
• @Martin: I'm afraid I'm not very good with reverse Galois theory. I would cop out by selecting $E=k(x_1,x_2,x_3,\ldots,x_8)$, realize $D_8$ as a group of permutations of the 8 transcendental elements, and then let $F=Inv(D_8)$. Thinking... Commented Mar 10, 2013 at 17:16
• @Martin: This question and the links provided there may help answer your question. Commented Mar 10, 2013 at 18:10

Try to show the following, $[LK:F]\leq [L:F][K:F]$, the equality holds in the case of $([L:F],[K:F])=1$, in your case would imply $[LK:F]=[E:F]$, and $E=LK$.

We need to find fields satisfy $L\cap K=F$ and $[LK:F]<[L:F][K:F]$.

Let $M$ be the splitting field of $x^4-x^3+1$, then one can check $\mathrm{Gal}(M/\mathbb{Q})\cong S_4$. Let $K=M^{S_3}, L=M^{\{(1),(14)\}}$. Then $KL=M$, $K\cap L=\mathbb{Q}$ and $$[KL:\mathbb{Q}]=24 < 48=[K:\mathbb{Q}][L:\mathbb{Q}].$$

Remark: The idea comes to my mind is considering those number fields (finite extension of $\mathbb{Q}$) with composite degree but has no nontrivial intermediate fields. In the above example, $K$ is a field with degree $4$ but has no quadratic subfields. For existence of such fields, see Field extension of composite degree has a non-trivial sub-extension . The polynomial $x^4-x^3+1$ is quoted in Andreas Caranti's answer of the above link. The method to construct such fields need Galois theory and group-theoretical arguments as I know, so this idea essentially has no difference with Jyrki Lahtonen's nice answer.