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Linear disjointness

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) There exists a k-basis of A which remains linearly independent over B.[1]
  • (iv) 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 the 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

References

  1. ^ Bourbaki, N. (1990). Elements of Mathematics, Algebra II, Chapters 4-7. Springer-Verlag. p. A.V. 14.

Cohn, Paul (1989). Basic algebra, Volume 2, Chapters 5. Wiley. p. 185. There is a simple criterion for an algebra to be a tensor product which is often useful. Let be an algebra over a field , and let be subspaces of ; then and are said to be linearly disjoint over if for any linearly independent elements in and in , the elements in are linearly independent over . Clearly this just means that the natural mapping induced by the mapping is injective. Now the criterion can be stated as follows: PROPOSITION 5.2 Let be an algebra over a field . Given subalgebras of , if (i) and are linearly disjoint, (ii) and (iii) and commute elementwise, then .


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