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


  • (i) The map A⊗kB→ABdisplaystyle Aotimes _kBto AB induced by (x,y)↦xydisplaystyle (x,y)mapsto xy is injective.

  • (ii) Any k-basis of A remains linearly independent over B.

  • (iii) If ui,vjdisplaystyle u_i,v_j are k-bases for A, B, then the products uivjdisplaystyle u_iv_j are linearly independent over k.

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


Suppose A, B are linearly disjoint over k. If A′⊂Adisplaystyle A'subset A, B′⊂Bdisplaystyle B'subset B are subalgebras, then A′displaystyle A' and B′displaystyle 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


  • Tensor product of fields


References


  • P.M. Cohn (2003). Basic algebra






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