# Linearly disjoint

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

- (i) The map induced by is injective.
- (ii) Any
*k*-basis of*A*remains linearly independent over*B*. - (iii) If are
*k*-bases for*A*,*B*, then the products are linearly independent over*k*.

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

Suppose *A*, *B* are linearly disjoint over *k*. If , are subalgebras, then and 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[edit]

## References[edit]

- P.M. Cohn (2003). Basic algebra