Theorem 4

— Let $K, E$ and $F$ be three fields such that $K \subset E \subset F$. If $S$ is a transcendence basis of $E$ over $K$ and $T$ a transcendence basis of $F$ over $E$, then $S \cap T$ is empty and $S \cup T$ is a transcendence basis of $F$ over $K$.

For $F$ is algebraic over $E(T)$; moreover, $E(T)$ is algebraic over the field $K(S \cup T) = K(S)(T)$, because $E$ is algebraic over $K(S)$ ($V$, p. 18, Cor. 2); therefore ($V$, p. 19, Prop. 3) $F$ is algebraic over $K(S \cup T)$. On the other hand, $T$ being algebraically free over $E$ is so $a fortiori$ over $K(S)$, hence ($V$, p. 107, Prop. 4) $S \cup T$ is algebraically free over $K$ and $S \cap T = \varnothing$.