# $E_1/F$ and $E_2/F$ finite field extensions, does degree of compositum $E_1E_2$ over $F$ divide the product $[E_1:F] [E_2:F]$?

Suppose $$E_1/F$$ and $$E_2/F$$ are finite field extensions. The degree of the composite field $$E_1E_2$$ over $$F$$ is less or equal to the product of the degree of $$E_1$$ over $$F$$ times the degree of $$E_2$$ over $$F$$, i.e. $$[E_1E_2:F] \leq [E_1:F] [E_2:F].$$ See for example: The degree of a field extension is smaller than the product of the degrees of field extensions of intermediate fields that generate the field.

I suspect that $$[E_1E_2:F]$$ divides the product $$[E_1:F] [E_2:F]$$. Do you have a suggestion on how to prove this? Note we have the towers $$F\subset E_1\subset E_1E_2$$ and $$F\subset E_2\subset E_1E_2$$ that might be useful.

Consider $$E_1=\mathbb{Q}(\sqrt{2})$$, and $$E_2=\mathbb{Q}(\zeta\sqrt{2})$$, where $$\zeta$$ is a complex primitive cubic root of unity. Each of those has degree $$3$$ over $$\mathbb{Q}$$, because they are given by extending with a root of the irreducible polynomial $$x^3-2$$.
The compositum $$E_1E_2=\mathbb{Q}(\sqrt{2},\zeta)$$ is the splitting field of $$x^3-2$$, which has degree $$6$$ over $$\mathbb{Q}$$. But $$6$$ does not divide $$[E_1:\mathbb{Q}][E_2:\mathbb{Q}]=9$$.
Of course, if $$\gcd([E_1:F],[E_2:F])=1$$, then the degree of the compositum will be equal to the product, since it will be a multiple of each and less than or equal to their product.