Let $k$ be an arbitrary base field and $K, L, M$ some fields over $k$ contained in a fixed overfield $\Omega$.
Question: Are there some "reasonable" assumptions (ie beyond a bunch of really "boring" assumptions) on $K, L, M$ which imply that
$$ K(M\cap L)=MK\cap LK $$
holds? Maybe something with linear disjointness? Concrete conjecture( see for motivation below why this guess): If $L$ and $K$ linearly disjoint to $M$? If yes, how to see in that case the inclusion $K(M\cap L)\supset MK\cap LK$?
Original motivation leading me to this problem: Even thought this looks elementary I still haven't found a satisfying argument why following equality holds: assume $k \subset K \subset L$ an arbitrary tower of not neccessary finite algebraic extensions of $k$ and $M=k(t)$ the field of Laurent series. Then I expect that it should hold
$$ (k(t) \cdot K) \cap L = K $$
independently how the involved fields are embedded in overfield $\Omega$, but I still not see how to deduce the inclusion $ (k(t) \cdot K) \cap L \subset K $, even thought that it looks "plausible". Does somebody see a quick trick how to seee it?
Since in this special situation $L=LK$ and $M$ linearly disjoint to $L$, we clearly have $L \cap M =k$, this can be translated to the problem above.
I noticed that here was posted identical question, with primary focus on number fields, but unfortunarely not received any answer.
#UPDATE: As @Wojowu pointed out posing certain assumtions on linear disjointness relations between the involved fields not give the statement. What about if we pose the "very" restrictive assumtion that $K \subset L$ algebraic over $k$ and $M$ purely transcendental over $k$, like in "original motivation" part? (Not sure where it can be really weakened)
