Linear disjointness

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

• (i) The map $A{\otimes }_{k}B\to AB$ induced by $(x,y)\mapsto xy$ 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 $u_i,v_j$ are k-bases for A, B, then the products $u_i{v}_j$ are linearly independent over k.

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

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