Linear disjointness

In mathematics, algebras A, B over a field k inside some field extension ${\displaystyle \Omega }${\displaystyle \Omega } of k are said to be linearly disjoint over k if the following equivalent conditions are met:

• (i) The map ${\displaystyle A\otimes _{k}B\to AB}${\displaystyle A\otimes _{k}B\to AB} induced by ${\displaystyle (x,y)\mapsto xy}${\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.^[1]
• (iv) If ${\displaystyle u_{i},v_{j}}${\displaystyle u_{i},v_{j}} are k-bases for A, B, then the products ${\displaystyle u_{i}v_{j}}${\displaystyle u_{i}v_{j}} are linearly independent over k.

Note that, since every subalgebra of ${\displaystyle \Omega }${\displaystyle \Omega } is a domain, (i) implies ${\displaystyle A\otimes _{k}B}${\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 ${\displaystyle A\otimes _{k}B}${\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 ${\displaystyle A\otimes _{k}B}${\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 ${\displaystyle \Omega }${\displaystyle \Omega } generated by ${\displaystyle A,B}${\displaystyle A,B}, resp. are linearly disjoint over k. (cf. Tensor product of fields)

Suppose A, B are linearly disjoint over k. If ${\displaystyle A'\subset A}${\displaystyle A'\subset A}, ${\displaystyle B'\subset B}${\displaystyle B'\subset B} are subalgebras, then ${\displaystyle A'}${\displaystyle A'} and ${\displaystyle B'}${\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.)

• Tensor product of fields

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 ${\displaystyle C}${\displaystyle C} be an algebra over a field ${\displaystyle k}${\displaystyle k}, and let ${\displaystyle U,V}${\displaystyle U,V} be subspaces of ${\displaystyle C}${\displaystyle C}; then ${\displaystyle U}${\displaystyle U} and ${\displaystyle V}${\displaystyle V} are said to be linearly disjoint over ${\displaystyle K}${\displaystyle K} if for any linearly independent elements ${\displaystyle u_{i}}${\displaystyle u_{i}} in ${\displaystyle U}${\displaystyle U} and ${\displaystyle v_{j}}${\displaystyle v_{j}} in ${\displaystyle V}${\displaystyle V}, the elements ${\displaystyle u_{i}v_{j}}${\displaystyle u_{i}v_{j}} in ${\displaystyle C}${\displaystyle C} are linearly independent over ${\displaystyle k}${\displaystyle k}. Clearly this just means that the natural mapping ${\displaystyle U\otimes V\to C}${\displaystyle U\otimes V\to C} induced by the mapping ${\displaystyle (u,v)\mapsto uv}${\displaystyle (u,v)\mapsto uv} is injective. Now the criterion can be stated as follows: PROPOSITION 5.2 Let ${\displaystyle C}${\displaystyle C} be an algebra over a field ${\displaystyle k}${\displaystyle k}. Given subalgebras ${\displaystyle A,B}${\displaystyle A,B} of ${\displaystyle C}${\displaystyle C}, if (i) ${\displaystyle A}${\displaystyle A} and ${\displaystyle B}${\displaystyle B} are linearly disjoint, (ii) ${\displaystyle AB=C}${\displaystyle AB=C} and (iii) ${\displaystyle A}${\displaystyle A} and ${\displaystyle B}${\displaystyle B} commute elementwise, then ${\displaystyle C\cong A\otimes B}${\displaystyle C\cong A\otimes B}.