Linearly disjoint
In mathematics, algebras A, B over a field k inside some field extension [math]\displaystyle{ \Omega }[/math] of k are said to be linearly disjoint over k if the following equivalent conditions are met:
- (i) The map [math]\displaystyle{ A \otimes_k B \to AB }[/math] induced by [math]\displaystyle{ (x, y) \mapsto xy }[/math] is injective.
- (ii) Any k-basis of A remains linearly independent over B.
- (iii) If [math]\displaystyle{ u_i, v_j }[/math] are k-bases for A, B, then the products [math]\displaystyle{ u_i v_j }[/math] are linearly independent over k.
Note that, since every subalgebra of [math]\displaystyle{ \Omega }[/math] is a domain, (i) implies [math]\displaystyle{ A \otimes_k B }[/math] 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 [math]\displaystyle{ A \otimes_k B }[/math] is a domain then it is a field and A and B are linearly disjoint. However, there are examples where [math]\displaystyle{ A \otimes_k B }[/math] 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 [math]\displaystyle{ \Omega }[/math] generated by [math]\displaystyle{ A, B }[/math], resp. are linearly disjoint over k. (cf. Tensor product of fields)
Suppose A, B are linearly disjoint over k. If [math]\displaystyle{ A' \subset A }[/math], [math]\displaystyle{ B' \subset B }[/math] are subalgebras, then [math]\displaystyle{ A' }[/math] and [math]\displaystyle{ 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
References
- P.M. Cohn (2003). Basic algebra
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