Proposition 7

Let G be an algebra over a commutative field K and E, F two subalgebras of G 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 there exist a basis of E over K which is a free subset of G for the right F-module structure on G. When this is so: (i) the canonical homomorphism $h : E \otimes_K F \to G$ is an isomorphism of $E \otimes_K F$ onto the subalgebra of G generated by $E \cup F$; (ii) $E \cap F = K;$ (iii) every free subset of E (resp. F) over K is a free subset of G with its right or left F-module (resp. E-module) structure.

The condition of the statement is obviously necessary, since every basis of E over K is a basis of $E \otimes_K F$ with its right F-module structure (II, § 3, no. 7, Corollary 1 to Proposition 7). To see that the condition is sufficient, note that the image H of $E \otimes_K F$ under h is the set of sums $\sum_i x_i y_i = \sum_i y_i x_i$ in G, with $x_i \in E$ and $y_i \in F$; if $(a_\lambda)$ is a basis of E over K, H is therefore also the submodule of the (right or left) F-module G, generated by $(a_\lambda)$. The condition of the statement therefore means that there exists a basis $(a_\lambda)$ of E which is also a basis of the F-module H; it follows that h is injective. Assertion (iii) follows from the fact that every free subset of E is contained in a basis of E (II, § 7, no. 1, Theorem 2).