Proposition 5

Let E and F be two extensions of K contained in $\Omega$.

a) If F has finite degree over K, then the subring of $\Omega$ generated by $E \cup F$ is a field, coinciding with $E(F)$ and the degree of $E(F)$ over E is finite; we have $[E(F):E] \leq [F:K]$, with equality if and only if E and F are linearly disjoint over K. In that case $E(F)$ is E-isomorphic* to $E \otimes_K F$.*

b) If further E is of finite degree over K, then $E(F) = K(E \cup F)$ is of finite degree over K. We have $[K(E \cup F):K] \leq [E:K][F:K]$ with equality if and only if E and F are linearly disjoint over K.

For let C be the subring of $\Omega$ generated by $E \cup F$; if $(b_j)_{1 \leq j \leq n}$ is a basis of F over K, then C is the vector sub-E-space of $\Omega$ generated by the $b_j$ hence C is an algebra of finite rank $\leq n$ over E; since the ring C is contained in a field, it is an integral domain, and hence a field, by the Cor. to Prop. 1 (V, p. 10), whence $C = E(F)$ and $[E(F):E] \leq [F:K]$. The relation $[E(F):E] = [F:K]$ means that the $b_j$ are linearly independent over E, hence that E and F are linearly disjoint over K; this proves part a) of the proposition. Part b) now follows at once because $[E(F):K] = [E(F):E][E:K]$.

Let E and F be extensions of K contained in $\Omega$; if E and F are of infinite degree over K, the subring $C = K[E \cup F]$ is not necessarily a field $^1$; however the field of fractions of C then coincides with $K(E \cup F)$. More generally let A be a subring of E such that E is the field of fractions of A, and let B be a subring of F such that F is the field of fractions of B; then if C is the subring of $\Omega$ generated by $A \cup B$, $K(E \cup F)$ coincides with the field of fractions of C, because the latter field is the least subfield of $\Omega$ containing C and it contains E and F. Moreover: