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.)