For a field $F$ with finite extensions $K$ and $K'$, is it true that $[KK' : F] [K\cap K':F] = [K:F][K':F]$?

Question

I'm assuming that all these fields are contained in some larger field $\Omega$. Assume (if it helps) that $K'$ and $K$ are separable over $F$. For the examples of $\mathbb{Q}(i)$ and $\mathbb{Q}(e^{2\pi i/3})$ over $\mathbb{Q}$, or if $F\subset K\subset K'$, it appears to be true. But is it true in general?

Answer 1

If at least one of the extensions $ K/F $ or $ K'/F $ is Galois, then this follows from natural irrationality. Indeed, wlog assume $ K/F $ is Galois; then the map $ \textrm{Gal}(KK'/K') \to \textrm{Gal}(K/K \cap K') $ given by restriction to $ K $ is an isomorphism by Galois theory, thus we have the equality $ [KK' : K'] = [K : K \cap K'] $; substituting this immediately gives the desired result. However, if neither of them are Galois, then the equality can fail. For instance, let $ M/F $ be any Galois extesion with Galois group $ S_5 $, and find subfields $ K, K' $ of degrees $ 60 $ and $ 24 $ over $ F $ respectively; for example, by taking the fixed fields of the subgroups generated by a $ 2 $-cycle and a $ 5 $-cycle. These intersect trivially, since any $ 2 $-cycle and $ 5 $-cycle generate the group $ S_5 $; so plugging into the identity gives $ 60 \cdot 24 = 120 $, which is absurd.

For a concrete example, take $ F = \mathbf Q $ and $ M $ the splitting field of $ X^5 - 4X + 2 $ over $ \mathbf Q $.