In mathematics, algebras A, B over a field k inside some field extension Ω {\displaystyle \Omega } of k are said to be linearly disjoint over k if the following equivalent conditions are met:
(i) The map A ⊗ k B → A B {\displaystyle A\otimes _{k}B\to AB} induced by ( x , y ) ↦ x y {\displaystyle (x,y)\mapsto xy} is injective. (ii) Any k-basis of A remains linearly independent over B. (iii) There exists a k-basis of A which remains linearly independent over B. (iv) If u i , v j {\displaystyle u_{i},v_{j}} are k-bases for A, B, then the products u i v j {\displaystyle u_{i}v_{j}} are linearly independent over k. Note that, since every subalgebra of Ω {\displaystyle \Omega } is a domain, (i) implies A ⊗ k B {\displaystyle A\otimes _{k}B} is a domain (in particular reduced). Conversely if A and B are fields and either A or B is an algebraic extension of k and A ⊗ k B {\displaystyle A\otimes _{k}B} is a domain then it is a field and A and B are linearly disjoint. However, there are examples where A ⊗ k B {\displaystyle A\otimes _{k}B} is a domain but A and B are not linearly disjoint: for example, A = B = k(t), the field of rational functions over k. One also has: A, B are linearly disjoint over k if and only if the subfields of Ω {\displaystyle \Omega } generated by A , B {\displaystyle A,B} , resp. are linearly disjoint over k. (cf. Tensor product of fields) Suppose A, B are linearly disjoint over k. If A ′ ⊂ A {\displaystyle A'\subset A} , B ′ ⊂ B {\displaystyle B'\subset B} are subalgebras, then A ′ {\displaystyle A'} and B ′ {\displaystyle B'} are linearly disjoint over k. Conversely, if any finitely generated subalgebras of algebras A, B are linearly disjoint, then A, B are linearly disjoint (since the condition involves only finite sets of elements.)
See also Tensor product of fields
References
Cohn, Paul (1989). Basic algebra, Volume 2, Chapters 5. Wiley. p. 185. There is a simple criterion for an algebra to be a tensor product which is often useful. Let C {\displaystyle C} be an algebra over a field k {\displaystyle k} , and let U , V {\displaystyle U,V} be subspaces of C {\displaystyle C} ; then U {\displaystyle U} and V {\displaystyle V} are said to be linearly disjoint over K {\displaystyle K} if for any linearly independent elements u i {\displaystyle u_{i}} in U {\displaystyle U} and v j {\displaystyle v_{j}} in V {\displaystyle V} , the elements u i v j {\displaystyle u_{i}v_{j}} in C {\displaystyle C} are linearly independent over k {\displaystyle k} . Clearly this just means that the natural mapping U ⊗ V → C {\displaystyle U\otimes V\to C} induced by the mapping ( u , v ) ↦ u v {\displaystyle (u,v)\mapsto uv} is injective. Now the criterion can be stated as follows: PROPOSITION 5.2 Let C {\displaystyle C} be an algebra over a field k {\displaystyle k} . Given subalgebras A , B {\displaystyle A,B} of C {\displaystyle C} , if (i) A {\displaystyle A} and B {\displaystyle B} are linearly disjoint, (ii) A B = C {\displaystyle AB=C} and (iii) A {\displaystyle A} and B {\displaystyle B} commute elementwise, then C ≅ A ⊗ B {\displaystyle C\cong A\otimes B} .