Degree of composition extension field, given disjoint subfields

Question

Let $L/K$ be a field extension with $K \subset N,M \subset L$, such that $N \cap M = K$, does it hold that $[NM :K] = [N : K][M : K]?$

Where to compositon $NM$ is defined as $K(N \cup M)$.

My intuition says yes, but I don't know how to prove the statement.

Answer 1

Consider $K=\mathbb Q$, $M=\mathbb Q(\sqrt[3]2)$ and $N=\mathbb Q(\varepsilon_3\sqrt[3]2)$, where $\varepsilon_3$ is the third primitive root of unity.

Clearly, $MN=\mathbb Q(\sqrt[3]2,\varepsilon_3)$. By calculating degrees of chains $\mathbb Q\subset\mathbb Q(\sqrt[3]2)\subset \mathbb(\sqrt[3]2,\varepsilon_3)$ and $\mathbb Q\subset\mathbb Q(\varepsilon_3)\subset \mathbb(\sqrt[3]2,\varepsilon_3)$ we get $[MN:\mathbb Q]=6$.

Since $M$ is a real field and $N$ is not, $M\neq N$, so $\mathbb Q\subseteq M\cap N\subsetneq M$. Since $[M:\mathbb Q]=3$ (minimal polynomial is $x^3-2$) is prime, by the chain rule $[M\cap N:\mathbb Q]=1$, i.e. $M\cap N=\mathbb Q$. Also $[N:\mathbb Q]=3$ (minimal polynomial is $x^3-2$).