Linearly disjoint explained

\Omega of k are said to be linearly disjoint over k if the following equivalent conditions are met:

• (i) The map

A ⊗ _{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 ⊗ _{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 is a domain then it is a field and A and B are linearly disjoint. However, there are examples where A ⊗ _{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

\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.)

See also

• Tensor product of fields

References

• P.M. Cohn (2003). Basic algebra