Linearly disjoint

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In mathematics, algebras A, B over a field k inside some field extension \Omega of k (e.g., universal field) 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.)

See also

References

  • P.M. Cohn (2003). Basic algebra