# Linearly-disjoint extensions

Two subextensions $A$ and $B$ of an extension $\def\O{\Omega}\O$ of $k$ are called linearly disjoint if the subalgebra generated by $A$ and $B$ in $\O$ is (isomorphic to) the tensor product $A\otimes B$ over $k$ (cf. Extension of a field). Let $A$ and $B$ be arbitrary subrings of an extension $\O$ of $k$, containing $k$, and let $C$ be the subring of $\O$ generated by $A$ and $B$. There is always a ring homomorphism $\phi:A\otimes B \to C$ that associates with an element $x\otimes y\in A\otimes B$, $x\in A$, $y\in B$, the product $xy$ in $C$. The algebras $A$ and $B$ are said to be linearly disjoint over $k$ if $\phi$ is an isomorphism of $A\otimes B$ onto $C$. In this case, $A\cap B = k$. For $A$ and $B$ to be linearly disjoint over $k$ it is sufficient that there is a basis of $B$ over $k$ that is independent over $A$. If $A$ is a finite extension of $k$, then the degree of the extension $[B(A):B]$ does not exceed the degree of extension $A:k$ and equality holds if and only if $A/k$ and $B/k$ are linearly disjoint.