Proposition 16

Let $\Omega$ be an extension of K and E, F two subextensions of $\Omega$, of finite degree over K.

a) If $E \subset F$, then $[F : K]_s = [F : E]_s \cdot [E : K]_s$, and $[F : K]_i = [F : E]_i \cdot [E : K]_i$.

b) Let $K'$ be a subextension of $\Omega$; then we have

and equality holds if $K'$ is linearly disjoint from E over K.

c) We have $[K(E \cup F) : K]_s \leq [E : K]_s \cdot [F : K]_s$, and $[K(E \cup F) : K]_i \leq [E : K]_i \cdot [F : K]_i$, and equality holds if E and F are linearly disjoint over K.

The assertion about the separable degrees in a) follows from (9) (V, p. 32). Since $[F : K] = [F : E] \cdot [E : K]$, the assertion about the inseparable degrees follows from this and (1).

By Cor. 1 of Prop. 13 (V, p. 44) and Prop. 15, we have

(2)

when K' is linearly disjoint from E over K, then E, is linearly disjoint from K' over K and E is linearly disjoint from K'(E_s) over E_s (V, p. 15, Prop. 8). Assertion b) now follows from Prop. 5 (V, p. 14).

By a) we have

by b) we have Hence we obtain the inequality The assertion of c) about the inseparable degrees is proved in a similar fashion.