Tower law

Equivalently, if $[M:K]<\infty$ then $[M:L]$ and $[L:K]$ are finite and the same formula holds.

A standard proof uses bases: if $\{\ell_1,\dots,\ell_r\}$ is a $K$-basis of $L$ and $\{m_1,\dots,m_s\}$ is an $L$-basis of $M$, then $\{m_i\ell_j : 1\le i\le s,\ 1\le j\le r\}$ is a $K$-basis of $M$.

Examples

1. $\mathbb{Q}\subset \mathbb{Q}(\sqrt2)\subset \mathbb{Q}(\sqrt2,\sqrt3)$. Here $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$ and $[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}(\sqrt2)]=2$, so the tower law gives
   $$[\mathbb{Q}(\sqrt2,\sqrt3):\mathbb{Q}]=2\cdot 2=4.$$

1. Finite fields: $\mathbb{F}_p \subset \mathbb{F}_{p^2} \subset \mathbb{F}_{p^6}$. Then $[\mathbb{F}_{p^2}:\mathbb{F}_p]=2$ and $[\mathbb{F}_{p^6}:\mathbb{F}_{p^2}]=3$, hence
   $$[\mathbb{F}_{p^6}:\mathbb{F}_p]=3\cdot 2=6.$$

1. Cyclotomic example: $\mathbb{Q}\subset \mathbb{Q}(\zeta_3)\subset \mathbb{Q}(\zeta_9)$, where $\zeta_n$ is a primitive n-th root of unity. One has $[\mathbb{Q}(\zeta_3):\mathbb{Q}]=2$ and $[\mathbb{Q}(\zeta_9):\mathbb{Q}(\zeta_3)]=3$, so $[\mathbb{Q}(\zeta_9):\mathbb{Q}]=6$.