Proposition 10

— Let L and M be two extensions of a field K and $(E, u, v)$ a composite extension of L and M (V, p. 12). Suppose that the ring $L \otimes_K M$ is an integral domain and that the subextensions $u(L)$ and $v(M)$ of E are algebraically disjoint over K. Then $u(L)$ and $v(M)$ are linearly disjoint over K.

Put $w = u * v$ (V, p. 12), denote by F the field of fractions of the integral domain $L \otimes_K M$ and identify L (resp. M) with a subfield of F by means of the mapping $x \mapsto x \otimes 1$ (resp. $y \mapsto 1 \otimes y$); then the restriction of w to L (resp. M) is u (resp. v). Let B be a transcendence basis of M over K (V, p. 109, Th. 1).

By hypothesis $u(L)$ and $v(M)$ are algebraically disjoint over K; therefore (V, p. 114, Prop. 14), $u(L)$ and $v(K(B))$ are linearly disjoint over K. Thus there exists a K-homomorphism $u': L(B) \to E$ which agrees with u on L and with v ou K(B). By construction L and M are linearly disjoint over K in F; by Prop. 8 (V, p. 15) the subfields $L(B)$ and $M$ of F are linearly disjoint over $K(B)$. It follows that there exists a K-homomorphism $w' : M[L(B)] \to E$ which agrees with $u'$ on $L(B)$ and with $v$ on $M$. But the field $F$ is generated by $M \cup L(B)$ and $M$ is algebraic over $K(B)$; we thus have $M[L(B)] = F$ (V, p. 18, Cor. 2). Hence we conclude that $w'$ is a K-isomorphism of $F$ onto $E$ whose restriction to $L$ (resp. $M$) is $u$ (resp. $v$). This shows $u(L)$ and $v(M)$ to be linearly disjoint over $K$.