Linear disjointness

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

• (i) The map ${\displaystyle A\otimes _{k}B\to AB}$ induced by ${\displaystyle (x,y)\mapsto xy}$ 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.
• (iv) If ${\displaystyle u_{i},v_{j}}$ are k-bases for A, B, then the products ${\displaystyle u_{i}v_{j}}$ are linearly independent over k.

Note that, since every subalgebra of ${\displaystyle \Omega }$ is a domain, (i) implies ${\displaystyle A\otimes _{k}B}$ 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 ${\displaystyle A\otimes _{k}B}$ is a domain then it is a field and A and B are linearly disjoint. However, there are examples where ${\displaystyle A\otimes _{k}B}$ 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 ${\displaystyle \Omega }$ generated by ${\displaystyle A,B}$, resp. are linearly disjoint over k. (cf. Tensor product of fields)

Suppose A, B are linearly disjoint over k. If ${\displaystyle A'\subset A}$, ${\displaystyle B'\subset B}$ are subalgebras, then ${\displaystyle A'}$ and ${\displaystyle B'}$ 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

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 ${\displaystyle C}$ be an algebra over a field ${\displaystyle k}$, and let ${\displaystyle U,V}$ be subspaces of ${\displaystyle C}$; then ${\displaystyle U}$ and ${\displaystyle V}$ are said to be linearly disjoint over ${\displaystyle K}$ if for any linearly independent elements ${\displaystyle u_{i}}$ in ${\displaystyle U}$ and ${\displaystyle v_{j}}$ in ${\displaystyle V}$, the elements ${\displaystyle u_{i}v_{j}}$ in ${\displaystyle C}$ are linearly independent over ${\displaystyle k}$. Clearly this just means that the natural mapping ${\displaystyle U\otimes V\to C}$ induced by the mapping ${\displaystyle (u,v)\mapsto uv}$ is injective. Now the criterion can be stated as follows: PROPOSITION 5.2 Let ${\displaystyle C}$ be an algebra over a field ${\displaystyle k}$. Given subalgebras ${\displaystyle A,B}$ of ${\displaystyle C}$, if (i) ${\displaystyle A}$ and ${\displaystyle B}$ are linearly disjoint, (ii) ${\displaystyle AB=C}$ and (iii) ${\displaystyle A}$ and ${\displaystyle B}$ commute elementwise, then ${\displaystyle C\cong A\otimes B}$.

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