Corollary 2

Let E, F be two subalgebras of G, of finite rank over K and such that every element of E commutes with every element of F. For E and F to be linearly disjoint over K, it is necessary and sufficient that the subalgebra H of G generated by $E \cup F$ be such that

This says that the rank over K of the surjective canonical homomorphism h : E \otimes_K F \to H is equal to the rank of E \otimes_K F over K, which is equivalent to saying that this homomorphism is bijective (II, § 7, No. 4, Proposition 9).