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2014-01-18T18:34:54Z

Reverted 1 edit by 67.189.121.248 (talk): How does it not? (TW)

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In mathematics, [[algebra over a field|algebra]]s ''A'', ''B'' over a field ''k'' inside some field extension <math>\Omega</math> of ''k'' (e.g., [[universal field]]) are said to be '''linearly disjoint over ''k''''' if the following equivalent conditions are met:
*(i) The map <math>A \otimes_k B \to AB</math> induced by <math>(x, y) \mapsto xy</math> is injective.
*(ii) Any ''k''-basis of ''A'' remains linearly independent over ''B''.
*(iii) If <math>u_i, v_j</math> are ''k''-bases for ''A'', ''B'', then the products <math>u_i v_j</math> are linearly independent over ''k''.

Note that, since every subalgebra of <math>\Omega</math> is a domain, (i) implies <math>A \otimes_k B</math> is a domain (in particular [[reduced ring|reduced]]).

One also has: ''A'', ''B'' are linearly disjoint over ''k'' if and only if subfields of <math>\Omega</math> generated by <math>A, B</math>, resp. are linearly disjoint over ''k''. (cf. [[tensor product of fields]])

Suppose ''A'', ''B'' are linearly disjoint over ''k''. If <math>A' \subset A</math>, <math>B' \subset B</math> are subalgebras, then <math>A'</math> and <math>B'</math> 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

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[[Category:Algebra]]

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en>Plot Spoiler: Reverted 1 edit by 67.189.121.248 (talk): How does it not? (TW)