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).

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.)

References

P.M. Cohn (2003). Basic algebra

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Linearly disjoint

See also

References