# Linearly disjoint

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

• (i) The map $A\otimes _{k}B\to AB$ induced by $(x,y)\mapsto xy$ is injective.
• (ii) Any k-basis of A remains linearly independent over B.
• (iii) If $u_{i},v_{j}$ are k-bases for A, B, then the products $u_{i}v_{j}$ are linearly independent over k.

Note that, since every subalgebra of $\Omega$ is a domain, (i) implies $A\otimes _{k}B$ 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 $A\otimes _{k}B$ is a domain then it is a field and A and B are linearly disjoint. However, there are examples where $A\otimes _{k}B$ 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 subfields of $\Omega$ generated by $A,B$ , resp. are linearly disjoint over k. (cf. tensor product of fields)

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