Transcendence degree is additive in finite towers

Choose a transcendence basis $S=\{s_1,\dots ,s_r\}$ of $K$ over $k$ and a transcendence basis $T=\{t_1,\dots ,t_m\}$ of $L$ over $K$. Then $r$ and $m$ are the two given transcendence degrees.

The union $S\cup T$ is algebraically independent over $k$. Indeed, a polynomial relation over $k$ among $S\cup T$ would also be a relation over $K$, contradicting algebraic independence of $T$ over $K$. Moreover $L$ is algebraic over $K(T)$, and $K$ is algebraic over $k(S)$, so [L2] shows that $L$ is algebraic over $k(S,T)$. Therefore [L1] makes $S\cup T$ a transcendence basis of $L$ over $k$.

Since $S\cup T$ is a transcendence basis of $L$ over $k$ with $r+m$ elements, ${\mathrm{trdeg}}_{k}L=r+m={\mathrm{trdeg}}_{k}K+{\mathrm{trdeg}}_{K}L$.